text:
\begin{document}

\begin{abstract}
Let $X$ be a Banach space and $\Gamma \sub X^*$
a total linear subspace. We study the concept of $\Gamma$-integrability for $X$-valued functions~$f$ defined on a complete probability space,
i.e. an analogue of Pettis integrability by dealing only with the compositions $\langle x^*,f \rangle$ for~$x^*\in \Gamma$.
We show that $\Gamma$-integrability and Pettis integrability are equivalent whenever $X$ has Plichko's property~($\mathcal{D}'$) (meaning that
every $w^*$-sequentially closed subspace of~$X^*$ is $w^*$-closed).
This property is enjoyed by many Banach spaces including all spaces with $w^*$-angelic dual as well as all spaces which
are $w^*$-sequentially dense in their bidual.
A particular case of special interest arises when considering $\Gamma=T^*(Y^*)$
for some injective operator $T:X \to Y$. Within this framework, we show that if $T:X \to Y$ is a semi-embedding, $X$ has property~($\mathcal{D}'$)
and $Y$ has the Radon-Nikod\'{y}m property, then $X$ has the weak Radon-Nikod\'{y}m property. This extends
earlier results by Delbaen (for separable $X$) and Diestel and Uhl (for weakly $\mathcal{K}$-analytic~$X$).
\end{abstract}

\maketitle

\section{Introduction}

A result attributed to Delbaen, which first appeared in a paper by Bourgain and Rosenthal (see \cite[Theorem~1]{bou-ros-2}),
states that if $T:X\to Y$ is a semi-embedding between Banach spaces (i.e. an injective operator such that $T(B_X)$ is closed),
$X$~is separable and $Y$ has the Radon-Nikod\'{y}m property (RNP),
then $X$ has the RNP as well (cf. \cite[Theorem~4.1.13]{bou-J}). That result was also known
to be true if $X$ is weakly $\mathcal{K}$-analytic, see \cite[footnote on p.~160]{die-uhl-2}.
No proof nor authorship info of that generalization was given in \cite{die-uhl-2}.
In our last email exchange I asked Prof. Joe Diestel about that and he told me:
\begin{quote}
{\em ``The result was realized as so, around the time that semi-embeddings were being properly appreciated,
as almost immediate consequences of the more general notions of $\mathcal{K}$-analyticity.
Jerry and I understood what we understood thru Talagrand's papers. So if attribution is the issue, it's Talagrand's fault!''}.
\end{quote}
Loosely speaking, a key point to get such kind of results is to deduce the integrability of an $X$-valued function~$f$
from the integrability of the $Y$-valued composition $T\circ f$, which quite often reflects on the family
of real-valued functions
$$
	\{\langle y^*, T \circ f \rangle: \, y^*\in Y^*\}=
	\{\langle x^*, f \rangle: \, x^*\in T^*(Y^*)\},
$$
i.e. the compositions of~$f$ with elements of the total linear subspace $T^*(Y^*) \sub X^*$.

In this paper we study Pettis-type integration of Banach space valued functions
with respect to a total linear subspace of the dual.
Throughout $(\Omega,\Sigma,\nu)$ is a complete probability space and $X$ is a Banach space. Let $\Gamma \sub X^*$
be a total linear subspace and let $f:\Omega\to X$ be a function. Following~\cite{mus6}, $f$ is said to be: (i)~{\em $\Gamma$-scalarly integrable} if $\langle x^*, f \rangle$ is
integrable for all $x^*\in \Gamma$; (ii) {\em $\Gamma$-integrable} if it is $\Gamma$-scalarly integrable and for every $A\in \Sigma$ there is
an element $\int_A f \, d\nu \in X$ such that $x^*(\int_A f \, d\nu)=\int_A \langle x^*, f \rangle \, d\nu$ for all $x^*\in \Gamma$.
This generalizes the classical Pettis and Gelfand integrals, which are obtained respectively when $\Gamma=X^*$ or $X$ is a
dual space and $\Gamma$ is the predual of~$X$. When does $\Gamma$-integrability imply Pettis integrability?
Several authors addressed this question in the particular case of Gelfand integrability.
For instance, Diestel and Faires (see \cite[Corollary~1.3]{die-fai})
proved that Gelfand and Pettis integrability coincide for any strongly measurable function
$f:\Omega \to X^*$ whenever $X^*$ contains no subspace isomorphic to~$\ell_\infty$,
while Musia{\l} (see \cite[Theorem~4]{mus6}) showed that one can give up strong measurability if
the assumption on~$X$ is strengthened to being $w^*$-sequentially dense in~$X^{**}$.
More recently, $\Gamma$-integrability has been studied in \cite{and-zie,kun}
in connection with semigroups of operators.

This paper is organized as follows.

Section~\ref{section:measurability} contains some preliminaries on scalar measurability of Banach space valued functions and total sets.
Special attention is paid to Banach spaces having Gulisashvili's {\em property~($\cD$)} \cite{gul-J}, i.e. those
for which scalar measurability can be tested by using any total subset of the dual.

In Section~\ref{section:integration} we analyze $\Gamma$-integrability and
discuss its coincidence with Pettis integrability. We show (see Theorem~\ref{theo:Dprime}) that this is the case whenever $X$
has Plichko's {\em property ($\mathcal{D}'$)}~\cite{pli3}, which means that every $w^*$-sequentially closed
subspace of~$X^*$ is $w^*$-closed. By the Banach-Dieudonn\'{e} theorem,
property~($\mathcal{D}'$) is formally weaker than {\em property~($\mathcal{E}'$)} of~\cite{gon3},
which means that every $w^*$-sequentially closed convex bounded subset of~$X^*$ is $w^*$-closed.
The class of Banach spaces having property ($\mathcal{E}'$) includes those with $w^*$-angelic dual
as well as all spaces which are $w^*$-sequentially dense in their bidual
(see \cite[Theorem~5.3]{avi-mar-rod}). On the other hand, we also study conditions ensuring the $\Gamma$-integrability
of a $\Gamma$-scalarly integrable function. This is connected in a natural way to the Mazur property of $(\Gamma,w^*)$ and the completeness of
the Mackey topology $\mu(X,\Gamma)$ associated to the dual pair $\langle X,\Gamma \rangle$, which have been topics of recent
research in \cite{bon-cas,gui-mon,gui-mon-ziz}. If $X$ has property~($\mathcal{E}'$) and $\Gamma$ is norming,
we prove that a function $f:\Omega \to X$ is Pettis integrable whenever the family
$\{\langle x^*, f \rangle: \, x^*\in \Gamma \cap B_{X^*}\}$ is uniformly integrable (Theorem~\ref{theo:ScalarGamma}).

Finally, in Section~\ref{section:operators} we focus on the interplay between the integrability
of a function $f:\Omega \to X$ and that of the composition $T\circ f:\Omega \to Y$, where $T$
is an injective operator from~$X$ to the Banach space~$Y$. We extend
the aforementioned results of Delbaen, Diestel and Uhl by proving that if $T$ is a semi-embedding, $Y$
has the RNP and $X$ has property~($\mathcal{D}'$), then $X$ has the weak Radon-Nikod\'{y}m property (WRNP),
see Theorem~\ref{theo:sequential}. In this statement, the RNP of~$X$ is guaranteed if the assumption on~$X$
is strengthened to being weakly Lindelöf determined (Corollary~\ref{cor:WLD}).

\subsection*{Notation and terminology}

We refer to \cite{bou-J,die-uhl-J} (resp. \cite{Musial,van}) for detailed information on vector measures and the RNP (resp.
Pettis integrability and the WRNP).
All our linear spaces are real. An {\em operator} between Banach spaces is a continuous linear map.
By a {\em subspace} of a Banach space we mean a closed linear subspace.
No closedness is assumed when we just talk about ``linear subspaces''.
We write $B_X=\{x\in X:\|x\|\leq 1\}$ (the closed unit ball of~$X$). The topological dual of~$X$
is denoted by~$X^*$. The evaluation of $x^*\in X^*$ at $x\in X$ is denoted by either $x^*(x)$
or $\langle x^*,x\rangle$. A set $\Gamma \sub X^*$ is {\em total} (over~$X$) if it separates the points of~$X$, i.e. for every $x\in X\setminus \{0\}$
there is $x^*\in \Gamma$ such that $x^*(x)\neq 0$. A linear subspace $\Gamma \sub X^*$ is said to be {\em norming}
if the formula
$$
	|||x|||=\sup\{x^*(x):\, x^*\in \Gamma \cap B_{X^*}\}, \quad x\in X,
$$
defines an equivalent norm on~$X$. The weak topology on~$X$ and the weak$^*$ topology on~$X^*$ are denoted by~$w$ and~$w^*$, respectively.
Given another Banach space~$Z$, we write $X\not\supseteq Z$ if $X$ contains no subspace isomorphic to~$Z$.
The convex hull and the linear span of a set $D \sub X$ are denoted by ${\rm co}(D)$
and ${\rm span}(D)$, respectively. The Banach space~$X$ is said to be {\em weakly Lindel\"{o}f determined} (WLD)
if $(B_{X^*},w^*)$ is a {\em Corson} compact, i.e. it is homeomorphic to a set~$K \subset [-1,1]^I$ (for some non-empty set~$I$),
equipped with the product topology, in such a way that $\{i\in I:k(i)\neq 0\}$ is countable for every $k\in K$.
Every weakly $\mathcal{K}$-analytic (e.g. weakly compactly generated)
Banach space is WLD, but the converse does not hold in general, see e.g. \cite{fab-alt-JJ} for more
information on WLD spaces. Finally, we recall that a locally convex Hausdorff space~$E$ is said to have the {\em Mazur property} if every
sequentially continuous linear functional $\varphi:E \to \mathbb{R}$ is continuous.

\section{Preliminaries on scalar measurability and total sets}\label{section:measurability}

Given any set $\Gamma \sub X^*$, we denote by $\sigma(\Gamma)$ the smallest $\sigma$-algebra on~$X$ for which each $x^*\in \Gamma$ is measurable
(as a real-valued function on~$X$).
According to a result of Edgar (see~\cite[Theorem~2.3]{edgar1}),
$\sigma(X^*)$ coincides with the Baire $\sigma$-algebra of~$(X,w)$, that is, $\sigma(X^*)=\Ba(X,w)$
(cf. \cite[Theorem~2.1]{Musial}).
The following Banach space property was introduced by Gulisashvili~\cite{gul-J}:

\begin{defi}\label{defi:D}
The Banach space $X$ is said to have {\em property ($\cD$)} if the equality $\Ba(X,w)=\sigma(\Gamma)$
holds for every total set $\Gamma\sub X^*$.
\end{defi}

Equivalently, $X$ has property ($\cD$) if and only if for every total set $\Gamma \sub X^*$ and every measurable space $(\tilde{\Omega},\tilde{\Sigma})$ we have:
\begin{quote}
a function $f:\tilde{\Omega}\to X$ is {\em $\Gamma$-scalarly measurable} (i.e. $\langle x^*,f\rangle$ is measurable for every $x^*\in \Gamma$)
if and only if it is {\em scalarly measurable} (i.e. $\langle x^*,f\rangle$ is measurable for every $x^*\in X^*$).
\end{quote}

Any Banach space with $w^*$-angelic dual has property~($\cD$), see \cite[Theorem~1]{gul-J}.
The converse fails in general: this is witnessed by the Johnson-Lindenstrauss space $JL_2(\mathcal{F})$ associated
to any maximal almost disjoint family $\mathcal{F}$ of infinite subsets of~$\mathbb{N}$ (see the introduction of \cite{avi-mar-rod}
and the references therein). More generally, {\em property~($\cD'$) implies property~($\cD$)}, see \cite[Proposition~12]{pli3}.

\begin{lem}\label{lem:subspaces}
If $X$ has property ($\cD$), then any subspace of~$X$ has property ($\cD$).
\end{lem}
\begin{proof}
Let $Y \sub X$ be a subspace. Given any set $\Gamma \sub Y^*$, for each $\gamma \in \Gamma$
we take $\tilde{\gamma}\in X^*$ such that $\tilde{\gamma}|_Y=\gamma$. Define $\tilde{\Gamma}:=\{\tilde{\gamma}:\gamma\in \Gamma\}\sub X^*$.
The following three statements hold true without the additional assumption on~$X$:
\begin{enumerate}
\item[(i)] $\Ba(Y,w)=\{Y\cap A: A\in \Ba(X,w)\}$, as an application of the aforementioned result of Edgar
(see e.g. \cite[Corollary~2.2]{Musial}).
\item[(ii)] $\sigma(\Gamma)=\{Y\cap A: A\in \sigma(\tilde{\Gamma}\cup Y^\perp)\}$. Indeed, since every
element of $\tilde{\Gamma}\cup Y^\perp$ is $\sigma(\Gamma)$-measurable when restricted to~$Y$, the inclusion
operator from~$Y$ into~$X$ is ($\sigma(\Gamma)$-($\sigma(\tilde{\Gamma}\cup Y^\perp)$)-measurable, that is,
$$
	\mathcal{A}:=\{Y\cap A: \, A\in \sigma(\tilde{\Gamma}\cup Y^\perp)\} \sub \sigma(\Gamma).
$$
On the other hand, $\mathcal{A}$ is a $\sigma$-algebra on~$Y$ for which every element of~$\Gamma$ is $\mathcal{A}$-measurable,
hence $\sigma(\Gamma)=\mathcal{A}$.
\item[(iii)] If $\Gamma$ is total (over~$Y$), then $\tilde{\Gamma}\cup Y^\perp$ is total (over~$X$). Indeed, just bear in mind that
$Y^{\perp\perp}\cap X=Y$ (by the Hahn-Banach separation theorem).
\end{enumerate}
So, if $X$ has property~($\cD$) and $\Gamma$ is total, then  (i), (ii) and~(iii)
ensure that $\Ba(Y,w)=\sigma(\Gamma)$.
This shows that $Y$ has property~($\cD$).
\end{proof}

It is easy to check that any WLD Banach space has $w^*$-angelic dual (so it has property~($\cD$)).
On the other hand, a non-trivial fact is that any WLD Banach space has a Markushevich basis (see e.g. \cite[Theorem~5.37]{fab-alt-JJ}).
The following result might be compared with that of Vanderwerff, Whitfield and Zizler
that a Banach space having a Markushevich basis and Corson's property~($\mathcal{C}$) is WLD
(see \cite[Theorem~3.3]{van-alt}, cf. \cite[Theorem~5.37]{fab-alt-JJ}).

\begin{pro}\label{pro:Mbasis}
If $X$ has a Markushevich basis and property~($\cD$), then $X$ is WLD.
\end{pro}
\begin{proof}
Let $\{(x_i,x_i^*):i\in I\}\sub X\times X^*$ be a Markushevich basis of~$X$. Then $\{x_i^*:i\in I\}$ is total and property ($\cD$) ensures that
$\Ba(X,w)=\sigma(\{x_i^*:i\in I\})$.
Take any $x^*\in X^*$. By \cite[Lemma~3.5]{rod-ver}, there is a countable
set $I_{x^*}\sub I$ such that $x^* \in \overline{{\rm span}(\{x_i^*: \, i\in I_{x^*}\})}^{w^*}$,
hence $x^*(x_i)=0$ for every $i\in I\setminus I_{x^*}$. This implies that $X$ is WLD
(see e.g. \cite[Theorem~5.37]{fab-alt-JJ}).
\end{proof}

Bearing in mind that $\ell_1(\omega_1)$ is not WLD, Lemma~\ref{lem:subspaces} and Proposition~\ref{pro:Mbasis}
yield the following known result (cf. \cite[Lemma~11]{pli3}).

\begin{cor}\label{cor:Plichko-Gulisashvili-l1}
A Banach space having property~($\cD$) contains no subspace isomorphic to~$\ell_1(\omega_1)$.
\end{cor}

To the best of our knowledge, the next question remains open:

\begin{question}[Gulisashvili, \cite{gul-J}]\label{question:G}
Is $(X^{**},w^*)$ angelic whenever $X^*$ has property~($\cD$)?
\end{question}

This question has affirmative answer if $X$ is separable (see \cite[Theorem~2]{gul-J}).
Indeed, the Odell-Rosenthal and Bourgain-Fremlin-Talagrand theorems ensure
that $(X^{**},w^*)$ is angelic whenever $X$ is a separable Banach space such that $X\not \supseteq \ell^1$
(see e.g. \cite[Theorem~4.1]{van}). On the other hand, property~($\cD$) of~$X^*$ implies that $X\not\supseteq \ell^1$, even
for a non-separable~$X$. Indeed, this follows from the following result (see \cite[Proposition~3.9]{rod-ver}):

\begin{fact}\label{fact:properties-dual-D}
Let us consider the following statements:
\begin{enumerate}
\item[(i)] $X^*$ has property~($\mathcal{D}$).
\item[(ii)] $\Ba(X^*,w)=\sigma(X)$.
\item[(iii)] $X$ is $w^*$-sequentially dense in~$X^{**}$.
\item[(iv)] $X\not\supseteq \ell_1$.
\end{enumerate}
Then (i)$\impli$(ii)$\Leftrightarrow$(iii)$\impli$(iv).
\end{fact}

It is known that~$X^*$ is WLD if and only if $X$ is Asplund
and $X$ is $w^*$-sequentially dense in~$X^{**}$, see \cite[Theorem~III-4, Remarks~III-6]{dev-god} and \cite[Corollary~8]{ori1}.
Recall that $X$ is said to be {\em Asplund} if every subspace of~$X$ has separable dual or, equivalently, $X^*$ has the RNP
(see e.g. \cite[p.~198]{die-uhl-J}). Therefore, Question~\ref{question:G} has affirmative answer for Asplund spaces:

\begin{cor}\label{cor:Asplund}
If $X$ is Asplund and $X^*$ has property~($\cD$), then $X^*$ is WLD.
\end{cor}

In particular, since a Banach lattice is Asplund if (and only if) it contains no subspace isomorphic to~$\ell_1$
(see \cite[p.~95]{die-uhl-J} and \cite[Theorem~7]{gho-saa-J}), we have:

\begin{cor}\label{cor:lattice}
If $X$ is a Banach lattice and $X^*$ has property~($\cD$), then $X^*$ is WLD.
\end{cor}

\section{Integration and total sets}\label{section:integration}

Throughout this section $\Gamma\sub X^*$ is a total linear subspace.
Given a function $f:\Omega \to X$, we write
$$
	Z_{f,D}:=\{\langle x^*, f \rangle: \, x^*\in D\}
$$
for any set $D \sub X^*$. Note that if $f$ is $\Gamma$-integrable, then the map
$$
	I_f: \Sigma \to X,
	\quad I_f(A):=\int_A f\, d\nu,
$$
is a finitely additive vector measure vanishing on $\nu$-null sets. It is countably additive (and it is called the {\em indefinite Pettis integral}
of~$f$) in the particular case $\Gamma=X^*$.

Statement~(i) of our next lemma is an application of a classical result of Diestel and Faires~\cite{die-fai},
while part~(iii) is similar to \cite[Lemma~3.1]{cas-rod-2}.
Recall first that a set $H\sub L_1(\nu)$ is called {\em uniformly integrable} if it is bounded
and for every $\epsilon>0$ there is $\delta>0$ such that $\sup_{h\in H}\int_A |h| \, d\nu\leq \epsilon$
for every $A\in \Sigma$ with $\nu(A)\leq\delta$.

\begin{lem}\label{lem:UniformlyIntegrableZf}
Let $f:\Omega \to X$ be a $\Gamma$-integrable function.
\begin{enumerate}
\item[(i)] If $X \not \supseteq \ell_\infty$, then $I_f$ is countably additive.
\item[(ii)]  If $I_f$ is countably additive, then $Z_{f,\Gamma \cap B_{X^*}}$ is uniformly integrable.
\item[(iii)] If $I_f$ is countably additive and there is a partition $\Omega=\bigcup_{n\in \Nat} A_n$ into countably many measurable sets
such that each restriction $f|_{A_n}$ is Pettis integrable, then $f$ is Pettis integrable.
\end{enumerate}
\end{lem}
\begin{proof} (i) follows from \cite[Theorem~1.1]{die-fai} (cf. \cite[p.~23, Corollary~7]{die-uhl-J}).

(ii) Since $I_f$ is countably additive, it is bounded
and $\lim_{\nu(A)\to 0} \|I_f\|(A)=0$, where $\|I_f\|(\cdot)$
denotes the semivariation of~$I_f$ (see e.g. \cite[p.~10, Theorem~1]{die-uhl-J}).
Now, the uniform integrability of $Z_{f,\Gamma \cap B_{X^*}}$ follows from the inequality
$$
	\sup_{x^*\in \Gamma \cap B_{X^*}} \int_A |\langle x^*, f \rangle| \, d\nu
	\leq \|I_f\|(A)
	\quad \mbox{for all }A\in \Sigma.
$$

(iii) We write $\Sigma_{A}:=\{B\cap A:B\in \Sigma\}$ for every $A\in \Sigma$.
Clearly, $f$ is scalarly measurable. We first prove that $f$ is {\em scalarly integrable}. Fix $x^*\in X^*$. Take any $N\in \Nat$ and
set $B_N:=\bigcup_{n=1}^N A_n$. Note that $f|_{B_N}$ is Pettis integrable and its indefinite Pettis integral
coincides with $I_f$ on $\Sigma_{B_N}$. Hence
$$
	\int_{B_N} |\langle x^*, f \rangle| \, d\nu\leq \|I_f\|(B_N)\leq \|I_f\|(\Omega) < \infty.
$$
As $N\in \Nat$ is arbitrary, $\langle x^*, f \rangle$ is integrable. This shows that $f$ is scalarly integrable.

Fix $E\in \Sigma$. Since $I_f$ is countably additive, the series $\sum_{n\in \Nat} I_f(E\cap A_n)$ is unconditionally convergent in~$X$ and for each $x^*\in X^*$ we have
$$
	x^*\Big(\sum_{n\in \Nat} I_f(E\cap A_n)\Big)=
	\sum_{n\in \Nat} x^*\big(I_f(E\cap A_n)\big)\stackrel{(*)}{=}
	\sum_{n\in \Nat} \int_{E\cap A_n} \langle x^*, f \rangle \, d\nu=
	\int_E \langle x^*, f \rangle \, d\nu,
$$
where equality~($*$) holds because $I_f|_{\Sigma_{A_n}}$ is the indefinite Pettis integral of $f|_{A_n}$ for every $n\in \Nat$.
This proves that $f$ is Pettis integrable.
\end{proof}

The Banach space $X$ is said to have the {\em $\nu$-Pettis Integral Property} ($\nu$-PIP)
if every scalarly measurable and scalarly bounded function $f:\Omega \to X$ is Pettis integrable.
Recall that a function $f:\Omega \to X$ is called {\em scalarly bounded} if there is a constant $c>0$
such that for each $x^*\in X^*$ we have $|\langle x^*,f\rangle|\leq c\|x^*\|$ $\nu$-a.e. (the exceptional set depending on~$x^*$).
A Banach space is said to have the PIP if it has the $\nu$-PIP with respect to any complete probability measure~$\nu$. In general:
\begin{center}
$X$ has property~($\cD'$) \ $\impli$ \ $(X^*,w^*)$ has the Mazur property \ $\impli$ \ $X$ has the PIP.
\end{center}
Indeed, the first implication is easy to check, while the second one goes back to~\cite{edgar2} (cf. Theorem~\ref{theo:ScalarGamma}(i) below).

\begin{theo}\label{theo:Dprime}
Suppose $X$ has property~($\cD'$). Then every $\Gamma$-integrable
function $f:\Omega \to X$ is Pettis integrable.
\end{theo}
\begin{proof}
Since $X$ has property~($\cD$), $f$ is scalarly measurable. Then there is
a partition $\Omega=\bigcup_{n\in \Nat} A_n$ into countably many measurable sets
such that $f|_{A_n}$ is scalarly bounded for every $n\in \Nat$ (see e.g. \cite[Proposition~3.1]{Musial}).
On the other hand, $X$~has the PIP acording to the comments preceeding the theorem. Therefore, each $f|_{A_n}$ is Pettis integrable.
Bearing in mind that a Banach space having property~($\cD$) cannot contain subspaces isomorphic to~$\ell_\infty$
(by Corollary~\ref{cor:Plichko-Gulisashvili-l1} and the fact that $\ell_1(\mathfrak{c})$ embeds
isomorphically into~$\ell_\infty$), an appeal to Lemma~\ref{lem:UniformlyIntegrableZf}
allows us to conclude that $f$ is Pettis integrable.
\end{proof}

\begin{rem}
From the proof of Theorem~\ref{theo:Dprime} it follows that the result still holds true if $X$ has property~($\cD$) and the $\nu$-PIP.
There exist Banach spaces having the PIP but failing property~($\cD$), like $\ell_1(\omega_1)$ (see \cite[Theorem~5.10]{edgar2},
cf. \cite[Proposition~7.2]{Musial}).
However, we do not know an example of a Banach space having property~($\cD$) but failing the PIP.
\end{rem}

\begin{question}\label{question:PIP}
Does property~($\cD$) imply the PIP?
\end{question}

The previous question has affirmative answer for dual spaces. Indeed, if $X^*$
has property~($\cD$), then $X$ is $w^*$-sequentially dense in~$X^{**}$
(Fact~\ref{fact:properties-dual-D}) and so $(X^{**},w^*)$ has the Mazur property, as
it can be easily checked. As a consequence:

\begin{cor}\label{cor:dual}
Suppose $X^*$ has property~($\cD$) and let $\tilde{\Gamma} \sub X^{**}$ be a total linear subspace.
Then every $\tilde{\Gamma}$-integrable function $f:\Omega \to X^*$ is Pettis integrable.
\end{cor}

Given any set $D \sub X^*$, we denote by $S_1(D) \sub X^*$ the set of all limits of $w^*$-convergent
sequences contained in~$D$. More generally, for any ordinal $\alpha\leq \omega_1$, the $\alpha$-th $w^*$-sequential closure $S_\alpha(D)$
is defined by transfinite induction as follows: $S_0(D):=D$, $S_{\alpha}(D):=S_1(S_\beta(D))$ if $\alpha=\beta+1$ and $S_{\alpha}(D):=\bigcup_{\beta<\alpha}S_\beta(D)$
if $\alpha$ is a limit ordinal. Then $S_{\omega_1}(D)$ is the smallest $w^*$-sequentially closed subset of~$X^*$ containing~$D$.
Clearly, the Banach space $X$ has property ($\cD'$) (resp. ($\mathcal{E}'$))
if and only if $S_{\omega_1}(D)=\overline{D}^{w^*}$ for every subspace (resp. convex bounded set) $D \sub X^*$.

\begin{lem}\label{lem:ScalarGamma}
Let $f:\Omega \to X$ be a $\Gamma$-scalarly integrable function. For each $A\in \Sigma$, let $\varphi_{f,A}: \Gamma \to \mathbb{R}$ be the linear functional
defined by
$$
	\varphi_{f,A}(x^*):=\int_A \langle x^*, f \rangle \, d\nu \quad\mbox{for all }x^*\in \Gamma.
$$
The following statements hold:
\begin{enumerate}
\item[(i)] $f$ is $\Gamma$-integrable if and only if $\varphi_{f,A}$ is $w^*$-continuous for every $A\in \Sigma$.
\item[(ii)] Let $D \sub \Gamma$. If $Z_{f,D}$ is uniformly integrable, then:
\begin{itemize}
\item[(ii.1)] $\varphi_{f,A}$ is $w^*$-sequentially continuous on~$D$ for every $A\in \Sigma$;
\item[(ii.2)] $Z_{f,S_{\omega_1}(D)}$ is a uniformly integrable subset of~$L_1(\nu)$.
\end{itemize}
\end{enumerate}
\end{lem}
\begin{proof} (i) The ``only if'' part is obvious, while the ``if'' part follows from the fact that any
$w^*$-continuous linear functional $\varphi:\Gamma \to \mathbb{R}$ is induced by some $x\in X$
via the formula $\varphi(x^*)=x^*(x)$ for all $x^*\in \Gamma$ (see e.g. \cite[\S10.4]{kot}).

(ii.1) Let $(x_n^*)_{n\in \Nat}$ be a sequence in~$D$
which $w^*$-converges to some~$x^*\in D$. Then $(\langle x_n^*,f\rangle)_{n\in \Nat}$ is uniformly integrable
and $\lim_{n\to \infty}\langle x_n^*,f\rangle=\langle x^*,f\rangle$ pointwise on~$\Omega$. By
Vitali's convergence theorem, we have $\lim_{n\to \infty}\|\langle x_n^*,f \rangle-\langle x^*,f\rangle\|_{L_1(\nu)}=0$.
Hence, $\lim_{n\to \infty}\varphi_{f,A}(x_n^*)=\varphi_{f,A}(x^*)$ for every $A\in \Sigma$.

(ii.2) Fix $c>0$ such that $Z_{f,D}\sub cB_{L_1(\nu)}$ and
a function $\delta: (0,\infty)\to (0,\infty)$ such that
$$
	\sup_{\substack{A\in \Sigma\\\nu(A)\leq \delta(\epsilon)}} \sup_{x^*\in D}
	\int_A |\langle x^*,f \rangle| \, d\nu\leq \epsilon
	\quad \mbox{for every }\epsilon>0.
$$
We will prove, by transfinite induction, that for each $\alpha\leq \omega_1$ we have
\begin{eqnarray*}
	(p_\alpha) &\quad \qquad& Z_{f,S_\alpha(D)} \sub c B_{L_1(\nu)} \qquad \mbox{ and}\\
	(q_\alpha) &\quad \qquad& \sup_{\substack{A\in \Sigma\\\nu(A)\leq \delta(\epsilon)}} \sup_{x^*\in S_\alpha(D)}
	\int_A |\langle x^*,f \rangle| \, d\nu\leq \epsilon
	\quad \mbox{for every }\epsilon>0.
\end{eqnarray*}
The case $\alpha=0$ is obvious. Suppose now that $0<\alpha\leq \omega_1$ and that $(p_\beta)$
and~$(q_\beta)$ hold true for every $\beta<\alpha$. If $\alpha$ is a limit,
then $S_\alpha(D)=\bigcup_{\beta<\alpha}S_\beta(D)$ and so $(p_\alpha)$
and~$(q_\alpha)$ also hold. Suppose on the contrary that $\alpha=\beta+1$.
Fix an arbitrary $x^*\in S_\alpha(D)=S_1(S_\beta(D))$. Then
there is a sequence $(x_n^*)_{n\in \Nat}$ in~$S_\beta(D)$ which $w^*$-converges to~$x^*$,
hence $\lim_{n\to \infty}\langle x_n^*,f \rangle=\langle x^*,f\rangle$ pointwise on~$\Omega$.
Since $Z_{f,S_\beta(D)}$ is uniformly integrable
(by $(p_\beta)$ and~$(q_\beta)$), we can apply Vitali's convergence theorem
to conclude that $\langle x^*,f\rangle\in L_1(\nu)$ and $\lim_{n\to \infty}\|\langle x_n^*,f \rangle-\langle x^*,f\rangle\|_{L_1(\nu)}=0$.
Clearly, this shows that $(p_\alpha)$ and~$(q_\alpha)$ hold.
\end{proof}

\begin{theo}\label{theo:ScalarGamma}
Let $f:\Omega \to X$ be a $\Gamma$-scalarly integrable function such that $Z_{f,\Gamma \cap B_{X^*}}$
is uniformly integrable.
\begin{enumerate}
\item[(i)] If $(\Gamma,w^*)$ has the Mazur property, then $f$ is $\Gamma$-integrable.
\item[(ii)] If $X$ has property~($\mathcal{E}'$) and $\Gamma$ is norming, then $f$ is Pettis integrable.
\end{enumerate}
\end{theo}
\begin{proof}
(i) follows at once from Lemma~\ref{lem:ScalarGamma}.

(ii) The fact that $\Gamma$ is norming is equivalent to saying that $\overline{\Gamma \cap B_{X^*}}^{w^*}
\supseteq c B_{X^*}$ for some $c>0$ (by the Hahn-Banach separation theorem). Since $X$ has property~($\mathcal{E}'$), we have
$S_{\omega_1}(\Gamma \cap B_{X^*}) =\overline{\Gamma \cap B_{X^*}}^{w^*}\supseteq cB_{X^*}$.
An appeal to Lemma~\ref{lem:ScalarGamma}(ii.2) ensures that $f$ is scalarly integrable and $Z_{f,B_{X^*}}$ is
uniformly integrable. Since $(X^*,w^*)$ has the Mazur property (by property~($\cD'$) of~$X$),
statement~(i) (applied to~$\Gamma=X^*$) implies that $f$ is Pettis integrable.
\end{proof}

\begin{rem}\label{rem:CascalesRodriguez}
Theorem~\ref{theo:ScalarGamma}(ii) generalizes an earlier analogous result for spaces with $w^*$-angelic dual,
see \cite[pp.~551--552]{cas-rod-2}.
\end{rem}

The Mackey topology $\mu(X,\Gamma)$ is defined as the (locally convex Hausdorff) topology on~$X$ of uniform convergence on absolutely convex $w^*$-compact subsets of~$\Gamma$.
When $\Gamma$ is norm-closed, $(X,\mu(X,\Gamma))$ is complete if (and only if) it is quasi-complete (see e.g. \cite{avi-gui-mar-rod}).
This completeness assumption was used by Kunze~\cite{kun} to find conditions ensuring the $\Gamma$-integrability
of a $\Gamma$-scalarly integrable function provided that $\Gamma$ is norming and norm-closed. After Kunze's work, the completeness of $(X,\mu(X,\Gamma))$
has been discussed in \cite{avi-gui-mar-rod,bon-cas,gui-mon,gui-mon-ziz}. For instance, $(X,\mu(X,\Gamma))$ is complete whenever $(X^*,w^*)$ is angelic
and $\Gamma$ is norming and norm-closed (see \cite[Theorem~4]{gui-mon-ziz}). There is also a connection between the completeness
of $(X,\mu(X,\Gamma))$ and the Mazur property of~$(\Gamma,w^*)$, see \cite{gui-mon-ziz}.

The following result is a refinement of \cite[Theorem~4.4]{kun}. We denote by
$\sigma(X,\Gamma)$ the (locally convex Hausdorff) topology on~$X$ of pointwise convergence on~$\Gamma$.

\begin{theo}\label{theo:Kunze}
Suppose $(X,\mu(X,\Gamma))$ is complete. Let $f:\Omega \to X$ be a $\Gamma$-scalarly integrable function such that
$Z_{f,\Gamma\cap B_{X^*}}$ is uniformly integrable and there is a
$\sigma(X,\Gamma)$-separable linear subspace $X_0 \sub X$ such that $f(\omega)\in X_0$ for $\mu$-a.e. $\omega \in \Omega$.
Then $f$ is $\Gamma$-integrable.
\end{theo}

To deal with Theorem~\ref{theo:Kunze} we need the following folk lemma:

\begin{lem}\label{lem:Separable}
If $X$ is $\sigma(X,\Gamma)$-separable, then every $w^*$-compact subset of~$\Gamma$ is $w^*$-metrizable.
\end{lem}
\begin{proof} Let $(x_n)_{n\in \Nat}$ be a $\sigma(X,\Gamma)$-dense sequence in~$X$. The map
$$
	\Gamma \ni x^* \mapsto (x^*(x_n))_{n\in \Nat} \in \mathbb{R}^\mathbb{N}
$$
is ($w^*$-pointwise)-continuous and injective. Hence any $w^*$-compact subset of~$\Gamma$ is homeomorphic to a (compact)
subset of the metrizable topological space $\mathbb{R}^\mathbb{N}$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{theo:Kunze}]
We will prove that $f$ is $\Gamma$-integrable by checking that $\varphi_{f,A}$ is $w^*$-continuous
for every $A\in \Sigma$ (see Lemma~\ref{lem:ScalarGamma}(i)).

Note that $X_1:=\overline{X_0}^{\sigma(X,\Gamma)}$ is a subspace of~$X$.
We denote by $r:X^*\to X_1^*$ the restriction operator, i.e. $r(x^*):=x^*|_{X_1}$ for all $x^*\in X^*$. Then
$$
	\Gamma_1:=r(\Gamma)=\{x^*|_{X_1}: \, x^*\in \Gamma\}\sub X_1^*
$$
is a total linear subspace (over~$X_1$). Of course, we can assume without loss of generality
that $f(\Omega)\sub X_1$, so $f$ can be seen as an $X_1$-valued $\Gamma_1$-scalarly integrable function.

Fix $A\in \Sigma$. Since $(X,\mu(X,\Gamma))$ is complete, in order to check that $\varphi:=\varphi_{f,A}$ is $w^*$-continuous it suffices
to show that the restriction $\varphi|_K$ is $w^*$-continuous for each absolutely convex $w^*$-compact set~$K \sub \Gamma$
(see e.g. \cite[\S21.9]{kot}). Since $r$ is ($w^*$-$w^*$)-continuous, $r(K)$ is a $w^*$-compact
subset of~$\Gamma_1$. Note that $\varphi|_{K}$ factors as
$$
	\xymatrix{K\ar_{r|_K}[d]\ar[rr]^{\varphi|_K}&&\mathbb{R}\\
	r(K)\ar_{\psi}[urr]&& }
$$
where $\psi(y^*):=\int_A \langle y^*,f \rangle \, d\nu$ for all $y^*\in r(K)$.
Observe that $Z_{f,K}$ is uniformly integrable
(because $K$ is a bounded subset of~$\Gamma$ and $Z_{f,\Gamma \cap B_{X^*}}$
is uniformly integrable). From Lemma~\ref{lem:ScalarGamma}(ii.1) (applied to~$f$ as an $X_1$-valued $\Gamma_1$-scalarly
integrable function and $D=r(K)$) it follows that $\psi$ is $w^*$-sequentially continuous on~$r(K)$.
Since $r(K)$ is $w^*$-metrizable (bear in mind the $\sigma(X_1,\Gamma_1)$-separability of~$X_1$ and Lemma~\ref{lem:Separable}),
we conclude that $\psi$ is $w^*$-continuous, and so is~$\varphi|_K$.
The proof is finished.
\end{proof}

\section{Integration and operators}\label{section:operators}

Throughout this section $Y$ is a Banach space.
Any operator $T:X\to Y$ is ($\Ba(X,w)$-$\Ba(Y,w)$)-measurable, in the sense that
$T^{-1}(B) \in \Ba(X,w)$ for every $B\in \Ba(Y,w)$.
In fact, we have the following lemma, whose proof
is included for the sake of completeness.

\begin{lem}\label{lem:sigma-adjoint-total}
Let $T:X \to Y$ be an operator. Then
$$
	\sigma(T^*(Y^*))=\big\{T^{-1}(B): \, B\in \Ba(Y,w)\big\}.
$$
\end{lem}
\begin{proof}
Note that $\mathcal{A}:=\{T^{-1}(B): \, B\in \Ba(Y,w)\}$ is a $\sigma$-algebra on~$X$.
Since $T^*(y^*)=y^*\circ T: X \to \mathbb{R}$ is $\mathcal{A}$-measurable for every $y^*\in Y^*$,
the inclusion $\sigma(T^*(Y^*)) \sub \mathcal{A}$ holds. On the other hand, define
$$
	\mathcal{B}:=\big\{B\in \Ba(Y,w): \, T^{-1}(B)\in \sigma(T^*(Y^*))\big\},
$$
which is clearly a $\sigma$-algebra on~$Y$. Since each $y^*\in Y^*$ is $\mathcal{B}$-measurable,
we have $\Ba(Y,w)=\mathcal{B}$ and so $\sigma(T^*(Y^*))=\mathcal{A}$.
\end{proof}

\begin{cor}\label{cor:characterizationWBE}
Let $T:X \to Y$ be an operator. The following statements are equivalent:
\begin{enumerate}
\item[(i)] $\Ba(X,w)=\{T^{-1}(B): \, B\in \Ba(Y,w)\}$.
\item[(ii)] For every measurable space $(\tilde{\Omega},\tilde{\Sigma})$ and every function $f:\tilde{\Omega} \to X$
we have: $f$ is scalarly measurable if (and only if) $T\circ f$ is scalarly measurable.
\end{enumerate}
\end{cor}

An injective operator satisfying the statements of Corollary~\ref{cor:characterizationWBE}
is said to be  a {\em weak Baire embedding}. This concept was considered by Andrews~\cite[p.~152]{and} in the particular case of adjoint operators.
Plainly, {\em if $T:X \to Y$ is an injective operator, then $T^*(Y^*)$ is a total linear subspace of~$X^*$.}
Therefore:

\begin{cor}\label{cor:Dtilde}
If $X$ has property~($\cD$), then every injective operator $T:X \to Y$ is a weak Baire embedding.
\end{cor}

As to strong measurability, in \cite[Proposition 4.3(ii)]{dev-rod} it was shown that if $X$ is WLD and $T:X\to Y$ is an injective operator, then a function
$f:\Omega \to X$ is strongly measurable if (and only if) $T\circ f:\Omega \to Y$ is strongly measurable.
The following proposition extends that result:

\begin{pro}\label{pro:Dfunctions}
Suppose $X$ is weakly measure-compact and has property~($\cD$).
Let $T:X \to Y$ be an injective operator and let $f:\Omega \to X$ be a function. If $T\circ f$ is strongly measurable, then so is~$f$.
\end{pro}

Recall that the Banach space $X$ is said to be {\em weakly measure-compact} if every probability measure $P$ on~$\Ba(X,w)$
is {\em $\tau$-smooth}, i.e. for any net $h_\alpha:X \to \mathbb{R}$ of bounded weakly continuous functions
which pointwise decreases to~$0$, we have $\lim_\alpha \int_X h_\alpha \, dP=0$. Every weakly Lindelöf (e.g. WLD) Banach space
is weakly measure-compact, see \cite[Section~4]{edgar1}.

\begin{proof}[Proof of Proposition~\ref{pro:Dfunctions}]
By Corollary~\ref{cor:Dtilde}, $f$ is scalarly measurable.
Since $X$ is weakly measure-compact, a result of Edgar (see~\cite[Proposition~5.4]{edgar1}, cf. \cite[Theorem~3.3]{Musial})
ensures that $f$ is {\em scalarly equivalent} to a strongly measurable function $f_0:\Omega \to X$.
Scalar equivalence means that for each $x^*\in X^*$ we have $\langle x^*,f\rangle=\langle x^*,f_0\rangle$ $\nu$-a.e.
(the exceptional set depending on~$x^*$).
Note that $T\circ f$ and $T\circ f_0$ are scalarly equivalent and strongly measurable, hence $T\circ f = T\circ f_0$ $\nu$-a.e.
The injectivity of~$T$ implies that $f=f_0$ $\nu$-a.e., so that $f$ is strongly measurable.
\end{proof}

\begin{rem}\label{rem:improvesDeville}
The class of weakly measure-compact Banach spaces having property~$(\mathcal{D})$
is strictly larger than that of WLD spaces.
Indeed, there are compact Hausdorff topological spaces $K$ with the following properties:
\begin{enumerate}
\item[(i)] $K$ is scattered of height~$3$,
\item[(ii)] $C(K)$ is weakly Lindelöf,
\item[(iii)] $K$ is not Corson,
\end{enumerate}
see \cite{pol2} (cf. \cite{fer-kos-kub}). Condition~(i) implies that
$(B_{C(K)^*},w^*)$ is sequential (see \cite[Theorem~3.2]{gon3}), thus $C(K)$
has property~($\mathcal{E}'$) and so property~($\cD$). (ii)~implies that $C(K)$ is weakly measure-compact.
On the other hand, $C(K)$ is not WLD by~(iii).
\end{rem}

Obviously, if $T:X \to Y$ is an injective operator and the function $f:\Omega \to X$
is $T^*(Y^*)$-integrable, then $T\circ f$ is Pettis integrable. In the opposite direction, we have the following result:

\begin{pro}\label{pro:TX-integrability}
Let $T:X \to Y$ be a semi-embedding and let $f:\Omega \to X$ be a function for which there is a constant $c>0$ such that,
for each $y^*\in Y^*$, we have
\begin{equation}\label{eqn:scalarboundedness}
	|\langle T^*(y^*),f \rangle| \leq c \|T^*(y^*)\| \quad \nu\mbox{-a.e.}
\end{equation}
If $T\circ f$ is Pettis integrable, then $f$ is $T^*(Y^*)$-integrable.
\end{pro}
\begin{proof} Since $T$ is a semi-embedding, $T(c B_X)$ is closed. We claim that
\begin{equation}\label{eqn:inclusion}
	\frac{1}{\nu(A)}\int_A T\circ f \, d\nu \in T(c B_X)
\end{equation}
for every $A\in \Sigma$ with $\nu(A)>0$. Indeed, suppose this is not the case. Then the Hahn-Banach separation theorem ensures the
existence of some $y^*\in Y^*$ such that
$$
	\frac{1}{\nu(A)}\int_A \langle y^*,T\circ f \rangle \, d\nu=
	y^*\Big(\frac{1}{\nu(A)}\int_A T\circ f \, d\nu\Big) >
	\sup_{x\in B_X} y^*(T(cx)) =c \|T^*(y^*)\|.
$$
But, on the other hand, we have
$$
	\frac{1}{\nu(A)}\int_A \langle y^*,T\circ f \rangle \, d\nu
	=
	\frac{1}{\nu(A)}\int_A \langle T^*(y^*), f \rangle \, d\nu
	\stackrel{\eqref{eqn:scalarboundedness}}{\leq}
	c\|T^*(y^*)\|,
$$
which is a contradiction. This proves~\eqref{eqn:inclusion}. Therefore, $\int_A T\circ f \, d\nu \in T(X)$ for every $A\in \Sigma$, which clearly implies
that $f$ is $T^*(Y^*)$-integrable.
\end{proof}

\begin{rem}\label{rem:tauberian}
Let $T:X\to Y$ be a tauberian operator (i.e. $(T^{**})^{-1}(Y)=X$).
In this case, the argument of \cite[Proposition~8]{edg7} shows that
a scalarly measurable and scalarly bounded function $f:\Omega \to X$
is Pettis integrable whenever $T\circ f$ is Pettis integrable. We stress that an operator
$T:X \to Y$ is injective and tauberian if and only if the restriction $T|_Z$ is a semi-embedding for every subspace $Z \sub X$,
see \cite{nei-ros}.
\end{rem}

We arrive at the main result of this section:

\begin{theo}\label{theo:sequential}
Let $T:X \to Y$ be a semi-embedding. If $X$ has property~($\mathcal{D}'$) and $Y$ has the $\nu$-RNP,
then $X$ has the $\nu$-WRNP.
\end{theo}
\begin{proof}
Let $I:\Sigma\to X$ be a countably additive vector measure of $\sigma$-finite variation
such that $I(A)=0$ for every $A\in \Sigma$ with $\nu(A)=0$.
We will prove the existence of a Pettis
integrable function $f:\Omega\to X$ such that $I(A)=\int_A f \, d\nu$ for all $A\in \Sigma$.
We can assume without loss of generality that there is a constant $c>0$
such that $\|I(A)\|\leq c \nu(A)$ for every $A\in \Sigma$ (see e.g. \cite[proof of Lemma~5.9]{van}).

Define
$$
	\tilde{I}:\Sigma \to Y,
	\quad
	\tilde{I}(A):=T(I(A)),
$$
so that $\tilde{I}$ is a countably additive vector measure satisfying
$\|\tilde{I}(A) \|\leq \|T\|c \nu(A)$ for every $A\in \Sigma$.
Since $Y$ has the $\nu$-RNP, there is a Bochner integrable function $g:\Omega \to Y$ such that
\begin{equation}\label{eqn:derivative}
	T(I(A))=\tilde{I}(A)=\int_A g\, d\nu
	\quad\mbox{for all }A\in \Sigma.
\end{equation}
The Bochner integrability of~$g$ implies that
$$
	g(\omega) \in H:=\overline{\Big\{\frac{1}{\nu(A)}\int_A g \, d\nu: \, A\in \Sigma, \, \nu(A)>0\Big\}}^{\|\cdot\|}
	\quad\mbox{for }\nu\mbox{-a.e. }\omega \in \Omega,
$$
see e.g. \cite[Lemma~4.3]{kup} (cf. \cite[Lemma~3.7]{cas-kad-rod-5}),
so we can assume without loss of generality that $g(\Omega) \sub H$.
On the other hand, \eqref{eqn:derivative} yields
$$
	H\sub \overline{T(c B_X)}^{\|\cdot\|}=T(c B_X),
$$
where the equality holds because $T$ is a semi-embedding. Hence $g(\Omega) \sub T(c B_X)$, so that
there is a function $f:\Omega \to cB_X$ such that $T\circ f=g$.

We claim that $f$ is Pettis integrable. Indeed, $f$ is $T^*(Y^*)$-integrable by Proposition~\ref{pro:TX-integrability}.
Bearing in mind that $X$ has property~($\cD'$), an appeal to Theorem~\ref{theo:Dprime} ensures that $f$ is Pettis integrable.
Since $T$ is injective and
$$
	T\Big(\int_A f\, d\nu\Big)=\int_A T\circ f\, d\nu=\int_A g\, d\nu\stackrel{\eqref{eqn:derivative}}{=}T(I(A))
	\quad
	\mbox{for all }A\in \Sigma,
$$
we have $\int_A f\, d\nu=I(A)$ for every $A\in \Sigma$. This proves that
$X$ has the $\nu$-WRNP.
\end{proof}

\begin{rem}\label{rem:g}
In the previous proof, an alternative way to check the Pettis integrability of~$f$ is as follows. Since
$f$ is $T^*(Y^*)$-scalarly measurable and $X$ has property~($\cD$), $f$ is scalarly measurable.
On the other hand, $f$ is bounded, hence the $\nu$-PIP of~$X$ ensures that $f$ is Pettis integrable.
\end{rem}

At this point it is convenient to recall that the RNP (resp. WRNP) is equivalent to the $\lambda$-RNP (resp. $\lambda$-WRNP), where
$\lambda$ is the Lebesgue measure on $[0,1]$, see e.g. \cite[p.~138, Corollary~8]{die-uhl-J} (resp. \cite[Theorem~11.3]{Musial}).

\begin{rem}\label{rem:WRNP-1}
The conclusion of Theorem~\ref{theo:sequential} might fail if $Y$ is only assumed to have the WRNP.
Indeed, let $Y$ be any Banach space having the WRNP but failing the RNP
(e.g. $Y=JT^*$, where $JT$ is the James tree space). Since the RNP is a separably determined property
(see e.g. \cite[p.~81, Theorem~2]{die-uhl-J}), there is a separable subspace $X\sub Y$
without the RNP. Then $X$ has property~($\cD'$) and the inclusion operator from~$X$ into~$Y$
is a semi-embedding, but $X$ fails the WRNP (which is equivalent to the RNP for separable Banach spaces).
\end{rem}

\begin{rem}\label{rem:WRNP-2}
Theorem~\ref{theo:sequential} cannot be improved by replacing the WRNP of~$X$ by the RNP,
even if $X$ has $w^*$-angelic dual. Indeed, let $Z$ be a separable Banach space not containing subspaces isomorphic to~$\ell_1$ such that
$X:=Z^*$ is not separable (e.g. $Z=JT$). Then $(X^*,w^*)$ is angelic (as we already pointed out in Section~\ref{section:measurability}),
$X$~fails the RNP and there is a semi-embedding $T:X \to \ell_2$
(because $X$ is the dual of a separable Banach space, see e.g. \cite[Lemma 4.1.12]{bou-J}).
\end{rem}

The $\nu$-WRNP and the $\nu$-RNP are equivalent for weakly measure-compact Banach spaces
(thanks to \cite[Proposition~5.4]{edgar1}). Since every WLD Banach space is weakly measure-compact
and has property~$(\cD'$), from Theorem~\ref{theo:sequential} we get:

\begin{cor}\label{cor:WLD}
Let $T:X \to Y$ be a semi-embedding. If $X$ is WLD and $Y$ has the $\nu$-RNP,
then $X$ has the $\nu$-RNP.
\end{cor}

\subsection*{Acknowledgements}
The author wishes to thank A.~Avil\'{e}s, A.J.~Guirao and G.~Mart\'{i}nez-Cervantes
for valuable discussions related to the topic of this paper.

config: arxiv


TITLE: Poynting vector and Rindler flux under time inversion
QUESTION [0 upvotes]: This question is about some reply by John Baez on sci.physics.research
the post is this: https://groups.google.com/d/msg/sci.physics.research/F6x5GkFt0ic/fxsfuNl9d8gJ
the article he is talking about is this one: http://xxx.lanl.gov/abs/gr-qc/0106075
the question itself is pretty simple. In that reply, which i'll quote for completeness, reads:
Unruh says that Haisch and Rueda's calculations are wrong, and
that a correct calculation shows a uniformly accelerating observer
zipping through the vacuum state of a quantized electromagnetic
field on Minkowski spacetime sees a *perfectly thermalized* bath
of photons.

In particular, this means such an observer will see no "Rindler flux" -
i.e., the expectation value of the Poynting vector is zero.  Or in
less fancy language: there will be, on average, no net flux of momentum
in the photons seen by the accelerating observer.

He gives a very simple argument showing that the expectation value
of the Poynting vector *must* be zero: the whole situation is
time-symmetric, and time reversal flips the direction of the
Poynting vector!

Now, i don't understand at all that last symmetry argument. Yes, time inversion will reverse direction of real vectors, so? an accelerated particle has a definite acceleration vector, that is reversed under time inversion, and is not zero. Why would any Poynting vector be zero?

REPLY [0 votes]: Your mistake is simple. The vector of acceleration is clearly even (unchanged), not odd (flipping sign), under the time reversal because it is the second derivative of the position, $d^2 \vec x/ dt^2$, and "second" is derived from two, an even number. Two minus signs cancel.
On the other hand, the Poynting vector is odd because it is essentially the density of energy times its velocity and velocity is the first derivative. So they can't be proportional to one another.


TITLE: Does this question require double u-substitution?
QUESTION [2 upvotes]: $$\int_0^1\frac1{(1+\sqrt x)^4}\,dx$$
The first substitution I thought to do was $u=\sqrt x$ and $du=\frac 12x^{-1/2}\,dx$. So what I have my integral change into is
$$\int_0^1\frac1{(1+\sqrt u)^4}\,du$$
I am not sure how to use the $du$ and where to go from there.

REPLY [1 votes]: Or:
$$1+\sqrt{x}=u \Rightarrow x=(u-1)^2 \Rightarrow dx=2(u-1)du.$$
$$\int_1^2 \frac{2(u-1)du}{u^4}=\int_1^2 2u^{-3}du -\int_1^2 2u^{-4}du=-u^{-2}|_1^2+\frac23u^{-3}|_1^2=...$$


TITLE: Show that a function of a markov chain is not a markov chain
QUESTION [2 upvotes]: How do you show that a function, $Y(n)=g(X(n))$ if some Markov chain $X(n)$ cannot be a Markov chain?

REPLY [1 votes]: Some functions of Markov chains are Markov chains. For example, if you're working on a finite state space and $Y$ simply is a permutation of the states, then the result is still a Markov chain.
In general, a function of a Markov chain as  described forms a Hidden Markov Model. To see if the function is not a Markov chain, simply check the Markov property: Does $P(Y(n)=y_n | Y(n-1)=y_{n-1}, Y(n-2)=y_{n-2}, \ldots) = P(Y(n)=y_n | Y(n-1)=y_{n-1})$? If so, it is a Markov chain. If not, it is not a Markov chain.


TITLE: Explanation for difference between groups of finite set under $+, \times$ operation.
QUESTION [1 upvotes]: Explanation for difference between groups of finite set under operations:$+,\times$ for prime value of $n$.
All values in the set $\{1,\dots, n-1\}$ are generators of $\lbrace \mathbb{Z}_n, + \rbrace$ for prime $n$.
The order of elements are:

Order $=7, 1+1+1+1+1+1+1\equiv 0\pmod 7$
Order $=7,2\times 7 = 14\equiv 0\pmod 7$
Order $=7,3\times 7 = 21 \equiv 0\pmod 7$
Order $=7,4\times 7 = 28 \equiv 0\pmod 7$
Order $=7,5\times 7 = 35 \equiv 0\pmod 7$
Order $=7,6\times 7 = 42 \equiv 0\pmod 7$

The set of elements generated by each element are:

$\{1,2,3,4,5,6,0\}$,
$\{2,4,6,1,3,5,0\}$,
$\{3,6,2,5,1,4,0\}$,
$\{4,1,5,2,6,3,0\}$,
$\{5,3,1,6,4,2,0\}$,
$\{6,5,4,3,2,1,0\}$,

So, is there only one subgroup = the group $\lbrace \mathbb{Z}_n, + \rbrace$?
But, the ordering of elements corresponding to each are unique, though as a set it doesn't matter.
But, when change the operation to multiplication, then same sized subgroup is not there, for all elements. Though, the multiplicative group has to exclude element $0$, as closure property then is not satisfied.
For the generators of $\lbrace \mathbb{Z}_7, \times \rbrace$ the set of values for elements in the set: $\{1,2,\dots,n-1\}$ are:
$1> \{1\}$
$2> \{2,4,1\}$
$3> \{3,2,6,4,5,1\}$
$4> \{4,2,1\}$
$5> \{5,4,6, 2,3,1\}$
$6> \{6,1\}$
The subgroups of $\lbrace \mathbb{Z}_7 \rbrace^{\times}$ are : $\{1\},
\{1,6\},\{1,2,4\},\{1,2,3,4,5,6\}$.
Also, why it is not necessary in case of latter that for prime $n$, all elements in the set $\{1,2,\dots,n-1\}$ are generators?
Say, above only elements $3,5$ are generators.
Similarly, for $\lbrace \mathbb{Z}_{23} \rbrace^{\times}$, the elements $2,3$ are not generators as $|2|=|3|=11, |6|=5$, while $|7|=|5|=23$.

REPLY [2 votes]: Here $(\mathbb{Z}_{p}, +) $ is finite cyclic group of order $p$.
Also $0\neq a\in \mathbb{Z}_p$ implies $|a|=p$ (Lagrange's theorem and uniqueness of element of order $1$ in a group).
Hence every non identity elements of $\mathbb{Z}_p$ are generators.
In case of $U_{p}=\mathbb{Z}_{p}^{\star}=\{1,2,3,\ldots ,p-1\}.$
We have that $(U_p, \times)$ is a group of order $\varphi(p)=p-1$
See the difference, any non-identity element of $U_p$ may not have order $p-1$.
To avoid triviality assume $p>3$ and $\varphi(n)$ is even forall $p>3$. Hence $2\mid\text{ord}(U_p) $. Then by Cauchy's theorem, $U_p$ has an element of order $2$ which can't generate the group $U_p (p>3)$.
Again since $U_p$ is a finite abelian group, for all divisor of $\varphi(p) $, there exists an element of that order. Hence for all elements whose order divide $\varphi(p) $ and less than $\varphi(p) $ doesn't generate $U_p$ (such element exists for $p>3$ as $\varphi(p)$ is composite).
In your example $|U_{23}|=\varphi(23)=22$.
Then elements of order $2$, $11$ are not generators.


TITLE: Cartesian product of countable number of finite sets is countable.
QUESTION [3 upvotes]: Cartesian product of countable number of finite sets is countable. True or false?
I think this statement is false. Because consider
$A = \{1,0\}$ . If we compute countable product of $A$ with itself, i.e, $A \times A \times \dots $ then it contains every sequence consists of 0 and 1, which is uncountable. Am I correct?

REPLY [3 votes]: Yes, you are right, except that, when you wrote $A\times A\times\cdots\times A$, you should have written just $A\times A\times\cdots$. Yes, this is the set of all sequence of $0$'s and $1$'s, which is uncountable.


text:
\subsection{Proof of Lemma~\ref{lem:connection-cover-time}}
\label{sec:proof-lemma-connection-cover-time}

For notational convenience, set $t_{l}:=\tcovertime l$, and define
\[
\mathcal{H}_{l}:=\Big\{\exists(s,a)\in\mathcal{S\times\mathcal{A}} ~\text{that is not visited within }\big(t_{l},t_{l+1}\big]\,\Big\}
\]
for any integer $l\geq0$. In view of the definition of $\tcovertime$,
we see that for any given $(s',a')\in\mathcal{S}\times\mathcal{A}$,
\begin{equation}
\mathbb{P}\left\{ \mathcal{H}_{l}\mid(s_{t_{l}},a_{t_{l}})=(s',a')\right\} \leq\frac{1}{2} .\label{eq:Hl-prob-UB}
\end{equation}
Consequently,
for any integer $L>0$, one can invoke the Markovian property to obtain
\begin{align*}
 & \mathbb{P}\left\{ \mathcal{H}_{1}\cap\cdots\cap\mathcal{H}_{L}\right\} =\mathbb{P}\left\{ \mathcal{H}_{1}\cap\cdots\cap\mathcal{H}_{L-1}\right\} \mathbb{P}\left\{ \mathcal{H}_{L}\mid\mathcal{H}_{1}\cap\cdots\cap\mathcal{H}_{L-1}\right\} \\
 & \quad=\mathbb{P}\left\{ \mathcal{H}_{1}\cap\cdots\cap\mathcal{H}_{L-1}\right\} \sum_{s',a'}\mathbb{P}\left\{ \mathcal{H}_{L}\mid(s_{t_{l}},a_{t_{l}})=(s',a')\right\} \mathbb{P}\left\{ (s_{t_{l}},a_{t_{l}})=(s',a')\mid\mathcal{H}_{1}\cap\cdots\cap\mathcal{H}_{L-1}\right\} \\
 & \quad\leq\frac{1}{2}\mathbb{P}\left\{ \mathcal{H}_{1}\cap\cdots\cap\mathcal{H}_{L-1}\right\} \sum_{s',a'}\mathbb{P}\left\{ (s_{t_{l}},a_{t_{l}})=(s',a')\mid\mathcal{H}_{1}\cap\cdots\cap\mathcal{H}_{L-1}\right\} \\
 & \quad=\frac{1}{2}\mathbb{P}\left\{ \mathcal{H}_{1}\cap\cdots\cap\mathcal{H}_{L-1}\right\} ,
\end{align*}
where the inequality follows from (\ref{eq:Hl-prob-UB}). Repeating
this derivation recursively, we deduce that
\[
\mathbb{P}\left\{ \mathcal{H}_{1}\cap\cdots\cap\mathcal{H}_{L}\right\} \leq\frac{1}{2^{L}}.
\]

This tells us that
\begin{align*}
\mathbb{P}\left\{ \exists(s,a)\in\mathcal{S\times\mathcal{A}}\text{ that is not visited between }(0,\tcoverall]\right\}  & \leq\mathbb{P}\left\{ \mathcal{H}_{1}\cap\cdots\cap\mathcal{H}_{\log_{2}\frac{T}{\delta}}\right\} \leq\frac{1}{2^{\log_{2}\frac{T}{\delta}}}=\frac{\delta}{T},
\end{align*}

which in turn establishes the advertised result by applying the union bound.


TITLE: Construct a cubic polynomial that is irreducible over $\mathbb{Z}_{5}$
QUESTION [1 upvotes]: It is known that a polynomial with degree less than $4$ is irreducible if and only if it has no roots. Suppose we have the following cubic polynomial, $ax^3+bx^2+cx+d=0$ with $a \neq 0$ and $a,b,c,d \in \mathbb{Z}_{5}$. $a$ therefore can take 4 elements and $b,c,d$ can take $5$ elements and our total number of polynomials is thus $500$. Now, we need to narrow down the irreducible polynomial.
Where do I go from here?

REPLY [0 votes]: You can divide it by $a$, thereby getting a polynomial of the type $x^3+b'x^2+c'x+d'$. And, of course, you cannot take $d'=0$. So, now you only have $100$ cases to consider.
Actually, you can suppose that $b'=0$, using the same trick that works for solving cubic equations over $\mathbb C$ ($x\rightarrow x-\frac{b'}3$). So, now you only have $20$ possibilities. That's not a lot.
By the way, $x^3+x+1$ will do.


TITLE: Number of ways a group element of a finite group can be written as a given word
QUESTION [6 upvotes]: I had previously asked about the number of ways a group element in a finite group could be written as a commutator (the question is still open for a proof, by the way) In how many ways can a group element in a finite group be written as a commutator?
Let $G$ be a finite group. Suppose we have a word on $k$ letters and we want to write a given $g \in G$ using that word (for example, if $k=2$, we'd be looking at the number of ways $g$ can be written as a product, which should be $|G|$).
For the case where the word is a product of $m$ commutators (so that $k=2m$), the following procedure should give the number of ways we can write the identity element as that word:
Let $\delta$ be the regular character of $G$ and $w$ be the word (which is a product of $m$ commutators). So $\delta$ takes value $|G|$ at the identity and $0$ at every other element. So $\sum_{\vec{g} \in G^{2m}} \delta(w(\vec{g}))=|G| \gamma(w)$, where $\gamma(w)$ is the number of ways we can write the identity with $w$. So we are able to calculate $\gamma$(w).
If $w$ is a product of $m$ commutators, one may also prove that $\gamma(w)=|G|^{2m-1}\zeta(2m-2)$, where $\zeta$ is Witten's zeta function ($\zeta(k)=\sum_{\chi} \chi(1)^{-k}$, where the sum is taken over all irreducible characters).
Is it possible to get the number of ways to write a general element $g$, from this result for the identity (or an analogous method)? At least for the case where $w$ is a product of commutators?
Thank you very much!

REPLY [6 votes]: Nevermind, I've sorted it out. In the end you get that the number of ways you can write a group element $g$ with a word $w$ is given by $\sum_{\chi} \frac{\overline{\chi(g)}}{|G|} \sum_{\vec{g} \in G^n} \chi(w(\vec{g}))$. For that, just look at the class function $f$ that takes value $|G|$ at elements in the conjugacy class of $g$ and $0$ everywhere else (then do the analogous procedure). This is nice, since it also proves the statement I wanted to prove in the commutator question (by taking $w=[a,b]$).
Thanks anyways!


TITLE: Exact sequence splits iff it kind of weakly splits through other sequence
QUESTION [0 upvotes]: This is an exercise from Assem's book "Algèbre et module" (more precisely, III.50). I couldn't solve it after thinking for a while. Let
$\require{AMScd}$
\begin{CD}
   0 @>>> L @>i>> P @>p>> V @>>> 0\\
   @.    @|   @VuVV   @VVvV  \\
   0 @>>> L @>>i'> M @>>p'> N @>>> 0
  \end{CD}
be a commutative diagram with exact lines. If $v$ factors through $M$, then the first line splits. Call that morphism say $h: V \longrightarrow M$ and assume that it commutes inside the two respective triangles in the right square (if you can do it only assuming commutativity of the lower triangle, that's even better).
You can assume that everything is a module over a ring. Still, I would prefer a proof inside abelian categories or something more general like pseudo-abelian with possibly more assumptions.
By factoring through the images of $u$ and $v$, I can reduce to the case where $u$ and $v$ are epi. I could find some relations between $\ker (u)$, $\ker (v)$ and $\ker (h)$. Nothing very illuminating though. But maybe I'm just overseeing something trivial or maybe there's a nasty trick around.
Thanks in advance.

REPLY [1 votes]: Take the fiber product $P':=M\times_NV$. Note that the map $P'\to V$, as a fiber product of a surjective map, is also surjective. Letting $L':=\ker(P'\to V)$, we obtain two maps of exact sequences:
\begin{CD}
0@>>>L@>>>P@>>>V@>>>0\\
@.@VVV@VVV@|@.\\
0@>>>L'@>>>P'@>>>V@>>>0\\
@.@VVV@VVV@VVV@.\\
0@>>>L@>>>M@>>>N@>>>0
\end{CD}
Note that the vertical map $L'\to L$ is an isomorphism by definition of $P'$, thus the leftest vertical arrows are all isomorphisms. By the five lemma, the middle vertical map $P\to P'$ is also an isomorphism. Then the lift $V\to M$ of $V\to N$ along with the identity map $V\to V$ induces, by definition of $P'$, a map $V\to P'$ which splits $P'\to V$.


TITLE: What is the standard deviation of this random variable? (I want to check my calculation)
QUESTION [3 upvotes]: Consider the random variable (orientation angle) $0\le\theta\le 2\pi$ with the following PDF where $\theta_0$ is the mean orientation angle: $(n\in\mathbb Z , n\ge 0)$
$$p(\theta,\theta_0,n)=\frac{\cos^{2n}(\theta-\theta_0)}{\int_0^{2\pi}\cos^{2n}(\theta-\theta_0) \, d\theta}$$
What is the variance of the random variable $\theta$ in terms of $n$
I have taken these steps:
First of all we should calculate $A_n=\int_0^{2\pi}\cos^{2n}(\theta-\theta_0)\,d\theta$
From the Integration by substitution method assuming that $x=\theta-\theta_0$ we have:
$$A_n=\int_{-\theta_0}^{-\theta_0+2\pi}\cos^{2n}x\,dx$$
From page $30$ in Elementary Functions we know:
$$\cos^{2n}x=\frac{1}{2^{2n}}\binom{2n}{n}+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}\binom{2n}{k}\cos 2(n-k)x$$
Then
$$A_n=\int_{-\theta_0}^{-\theta_0+2\pi}\frac{1}{2^{2n}}\binom{2n}{n}\,dx+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}\binom{2n}{k}\int_{-\theta_0}^{-\theta_0+2\pi}\cos 2(n-k)x\,dx$$
About the second integral in the above expression, we know that the period of the function $\cos 2(n-k)x$ is $\frac{\pi}{n-k}$
So the length of the interval $[-\theta_0,-\theta_0+2\pi]$ is $2(n-k)$ times the period which is an integer coefficient of the period, that means the answer of this integral will be $0$
So $$A_n=\frac{\pi}{2^{2n-1}}\binom{2n}{n}$$
The PDF mentioned at the begining will be converted to:
$$p(\theta,\theta_0,n)=\frac{\cos^{2n}(\theta-\theta_0)}{A_n}$$
So the variance will be:
$$\sigma^2(\theta)=\frac{1}{A_n}\int_0^{2\pi}(\theta-\theta_0)^2\cos^{2n}(\theta-\theta_0)\,d\theta$$
Again from the Integration by substitution method and the aforementioned relation we have:
$$\begin{align}
\int_0^{2\pi}(\theta-\theta_0)^2\cos^{2n}(\theta-\theta_0)\,d\theta
&=\int_{-\theta_0}^{-\theta_0+2\pi}x^2\cos^{2n}x\,dx\\
&=\int_{-\theta_0}^{-\theta_0+2\pi}x^2(\frac{1}{2^{2n}}\binom{2n}{n}+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}\binom{2n}{k}\cos 2(n-k)x)\,dx\\
&=\frac{\binom{2n}{n}}{2^{2n}}\underbrace{\int_{-\theta_0}^{-\theta_0+2\pi}x^2\,dx}_1\\
&+\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}\binom{2n}{k}\underbrace{\int_{-\theta_0}^{-\theta_0+2\pi}x^2\cos 2(n-k)x\,dx}_2
\end{align}$$
About the first integral in the above equation, we have
$$(1):\int_{-\theta_0}^{-\theta_0+2\pi}x^2\,dx=\left[\frac{x^3}{3}\right]_{-\theta_0}^{-\theta_0+2\pi}=\frac{8\pi^3-12\pi^2\theta_0+6\pi\theta_0^2}{3}$$
And about the second integral:
$$(2):\int_{-\theta_0}^{-\theta_0+2\pi}x^2\cos 2(n-k)x\,dx$$
From integration by parts method:
$$
\begin{array}{c|c}
\text{Derivation}&\text{Integration}\\
\hline
{} \oplus x^2 & \cos 2(n-k)x  \\
{} \ominus 2x & \frac{1}{2(n-k)}\sin 2(n-k)x \\
{} \oplus 2 & \frac{-1}{2^2(n-k)^2}\cos 2(n-k)x \\
{} \ominus 0 & \frac{-1}{2^3(n-k)^3}\sin 2(n-k)x
\end{array}
$$
$$\begin{align}
(2):\int_{-\theta_0}^{-\theta_0+2\pi}x^2\cos 2(n-k)x\,dx &=\left[\frac{x^2}{2(n-k)}\sin 2(n-k)x\right]_{-\theta_0}^{-\theta_0+2\pi} \tag 1\\
&+\left[\frac{x}{2(n-k)^2}\cos 2(n-k)x\right]_{-\theta_0}^{-\theta_0+2\pi} \tag 2 \\
&-\left[\frac{1}{2^2(n-k)^3}\sin 2(n-k)x\right]_{-\theta_0}^{-\theta_0+2\pi} \tag 3
\end{align}$$
For $(1)$ regarding the facts $\sin (2n\pi+\theta)=\sin\theta$ and $\sin(-\theta)=-\sin\theta$ after some simplifications, we have $$(1)=\frac{2\pi(\theta_0-\pi)}{n-k}\sin 2(n-k)\theta_0$$
For $(2)$ regarding the facts $\cos (2n\pi+\theta)=\cos\theta$ and $\cos(-\theta)=\cos\theta$ after some simplifications, we have $$(2)=\frac{\pi}{(n-k)^2}\cos 2(n-k)\theta_0$$
Also after some simplifications, we conclude that $(3)=0$
So:$$\int_{-\theta_0}^{-\theta_0+2\pi}x^2\cos 2(n-k)x\,dx=\frac{2\pi(\theta_0-\pi)}{n-k}\sin 2(n-k)\theta_0+\frac{\pi}{(n-k)^2}\cos 2(n-k)\theta_0$$
Then
$$\begin{align}
&\int_0^{2\pi}(\theta-\theta_0)^2\cos^{2n}(\theta-\theta_0)\,d\theta=\frac{\binom{2n}{n}}{2^{2n}}\frac{8\pi^3-12 \pi^2 \theta_0 + 6\pi\theta_0^2}{3}+{}\\
&\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}\binom{2n}{k}\left(\frac{2\pi(\theta_0-\pi)}{n-k}\sin 2(n-k)\theta_0+\frac{\pi}{(n-k)^2}\cos 2(n-k)\theta_0\right)
\end{align}$$
And the variance will be:
$$\frac{2^{2n-1}}{\binom{2n}{n}\pi}\left(\begin{align}
&\frac{\binom{2n}{n}}{2^{2n}}\frac{8\pi^3-12\pi^2\theta_0+6\pi\theta_0^2}{3}+{}\\
&\frac{1}{2^{2n-1}}\sum_{k=0}^{n-1}\binom{2n}{k}\left(\frac{2\pi(\theta_0-\pi)}{n-k}\sin 2(n-k)\theta_0+\frac{\pi}{(n-k)^2}\cos 2(n-k)\theta_0\right)
\end{align}\right)$$
With the final form:
$$\sigma^2(\theta)=\frac{4\pi^2-6\pi\theta_0+3\theta_0^2}{3}+\sum_{k=0}^{n-1}\frac{n!n!}{k!(2n-k)!}\left(\frac{2(\theta_0-\pi)}{n-k}\sin 2(n-k)\theta_0+\frac{1}{(n-k)^2}\cos 2(n-k)\theta_0\right)$$
So it seems that the variance is dependent to $\theta_0$(mean orientation angle) and even if we assume $\theta_0=0$
$$\sigma(\theta)=\sqrt{\frac{4\pi^2}{3}+\sum_{k=0}^{n-1}\frac{n!n!}{k!(2n-k)!}.\frac{1}{(n-k)^2}}$$
But in the papers that I'm studying:
A General Characterization for Polarimetric Scattering From Vegetation Canopies
Adaptive Model-Based Decomposition of Polarimetric SAR Covariance Matrices
It has been pointed out that:

Also the limit of the expression that I get finally is not $0$ an $n\to +\infty$ which is not consistent with our hypothesis
I'm really confused

REPLY [1 votes]: Too long for a comment: without getting into the details of your calculation, I think there is a more fundamental problem here. On one side, you are treating $\theta$ as a scalar variable that takes values in the real interval $[0,2\pi]$, and then you define a pdf over that interval and compute the mean, variance, etc by integrating over that interval. But that conflicts with your interpretation of $\theta$ as an angle. For example, suppose you had a pdf for $\theta$ that is concentrated (evenly) at $[0,\epsilon]$ and $[2\pi - \epsilon, 2\pi]$, for very small $\epsilon$. Seeing $\theta $ as an angle, we'd say that it's variance is very small (and it's mean is zero). Computing it as pdf, we wouldn't get that result.
Also, you assume that $\theta_0$ is the mean of your density, but that, again, is incompatible with you formulation. Think of $\theta_0 =0$...
Further, take for example $n=1$, $\theta_0=1$. The density function (graph) has period $2\pi$ (as it should), and its integral gives $1$ over the interval $[0,2\pi]$ as well as over (say) $[-\pi,3/2\, \pi ]$  (as it should).

But if we compute the mean and variance by the standard formulas we'd get, integrating over $[0,2\pi]$ : $\mu=2.687$ (already a meaningless result!) and $\sigma^2=2.8750$. If we take $[-\pi,3/2\, \pi ]$ (it should not matter), it gives instead $\mu=2.02545$ and $\sigma^2=2.79192$
All this needs rethinking. That's not the way to deal with circular statistics, I'm afraid. http://www.theanalysisfactor.com/circular-statistics/
https://en.wikipedia.org/wiki/Directional_statistics


TITLE: How to find the standard matrix for H(θ) by finding the images of the standard basis vectors?
QUESTION [0 upvotes]: According to the figure here, how to find the standard matrix for H(θ) without using formulas?
Another question is that for example 6 and 7, the question is exactly the same, but  why the answers are different? What's the difference between the two questions ? And how do I know when two use formula 16 and when to use formula 18?

REPLY [0 votes]: The statement of the problem in Example 7 is mistaken (I believe the authors have copied & pasted the one in Example 6 without making the necessary changes). The correct version should read (the modifications are in boldface):

"Find the reflection of the vector $\Bbb x = (1,5)$ about the line through the origin that makes an angle of $\pi /6 (= 30 ^\circ)$ with the $x$-axis."

When you are required to compute orthogonal projections on a line, use $P_\theta$ from formula 16 (easy to remember: $P$ comes from "projection"). When required to compute reflections about a line, use $H_\theta$ from formula 18 (and no, $H$ doesn't come from anything this time...).
Concerning your first question, the reflection of $\Bbb x$ about the line of slope $\tan \theta$ will obviously depend on $\theta$, dependence encoded in formulae. You cannot not have formulae...
Edit: To answer the question in the comment, note that $(1,5) = 1 \cdot e_1 + 5 \cdot e_2$, therefore $H_\theta (1,5) = 1 \cdot H_\theta e_1 + 5 \cdot H_\theta e_2$. Therefore, once we find out how the vectors $e_1$ and $e_2$ reflect about the line $L$, we are done. Fortunately, we can do this without complicated formulae; take a pencil and a sheet of paper and draw while reading this answer. Call the reflections of $e_1$ and $e_2$ respectively $r_1$ and $r_2$.
The $e_1$ vector is below $L$, forming an angle of $30 ^\circ$ with it, therefore its reflection will be above $L$, at the same angle, so $r_1$ will be at an angle of $30 ^\circ + 30 ^\circ = 60 ^\circ$ above the $x$ axis. Now, look at the right triangle formed by $0$ (the origin), the tip of the reflection if $r_1$, and the projection of $r_1$ on the $x$ axis; the length of the horizontal cathetus (i.e. leg) will give the $x$ coordinate of $r_1$, and the length of the vertical cathetus will give the $y$ coordinate of $r_1$. But the length of the horizontal cathetus is the length of the hypothenuse times $\cos 60 ^\circ$; since the length of the hypothenuse is the length of $r_1$, which is the length of $e_1$, which is $1$, the length of the horizontal cathetus will be $\cos 60 ^\circ = \frac 1 2$. Similarly, the $y$ cordinate of $r_1$ will be the length of the other cathetus, i.e. $\sin 60 ^\circ = \frac {\sqrt 3} 2$. Therefore, $r_1 = (\frac 1 2, \frac {\sqrt 3} 2)$.
The same discussion now for $e_2$: it is above $L$, making $60 ^\circ$ with it, therefore $r_2$ will be below $L$, at the same angle, which means that it will be below the $x$ axis, making $30 ^\circ$ with it. Again, investigating the right triangle formed (under the $x$ axis) by $0$, the tip of $r_2$ and the projection of $r_2$ on the $x$ axis, you will get $r_2 = (\frac {\sqrt 3} 2, - \frac 1 2)$ (the minus comes from being below the $x$ axis, where $y$ is negative).
Putting these together, you get $H_\theta (1,5) = 1 \cdot (\frac 1 2, \frac {\sqrt 3} 2) + 5 \cdot (\frac {\sqrt 3} 2, - \frac 1 2) = (\frac {5 \sqrt 3 + 1} 2, \frac {\sqrt 3 -5} 2)$ in a purely graphical way, without matrix multiplications (if this is what you meant by "witout formulae"). Please note, though, that some formulae are still necessary, you cannot expect to do mathematics without them save for in some extremely simple situations.


TITLE: Determine the area of $\phi (A) $
QUESTION [1 upvotes]: Let $ A=[0, 1] \times[0, 1] $. Let $ h $ be a continous function on $\mathbb{R}$ and let $\phi $ be defined by
$$\phi (x, y)=(x+h (x+y), y-h (x+y))$$
Determine the area of $\phi (A)$.

REPLY [1 votes]: Hint: Note that $\mu(A\cap l)=\mu(\phi(A)\cap l)$ for all lines $l$ of the form $x+y=\text{const}$.


TITLE: Last digit in base $10$ for $3^{22}$?
QUESTION [0 upvotes]: Find the last digit (in base 10) for the number: $$ {3}^{22}$$

This is a question in my course in discrete math. I have checked my notes from the lectures but can't seem to solve this one.

REPLY [1 votes]: If you do not know modular arithmetic you may use the binomial formula
$$3^{22} = (10-1)^{11} = 10\cdot \underbrace{m}_{m \in \mathbb{N}} + (- 1)^{11}= 10(m-1)+ \boxed{9}$$
So, $9$ is the last digit in base $10$.