Explore Measurement Concepts with Our Examples and Questions

Consider a sequence of identically distributed random variables $(X_n)$, with $E|X_n|$ finite. Define $Y_n=\sup <!--\; \; -->(|X_1|,\dots <!--\; \dots \; -->,|X_n|)/n$. I must prove that $Y_n$ converges to 0 in $L_1$.
If $\mathrm{\Omega <!--\; \Omega \; -->}$ is the domain of the random variables, then $n{Y}_n(\mathrm{\omega <!--\; \omega \; -->})\le <!--\; \le \; -->x$ if, and only if, $|X_i|(\mathrm{\omega <!--\; \omega \; -->})\le <!--\; \le \; -->x/n$ for $i=1,\dots <!--\; \dots \; -->,n$. Thus, $Y_n^{-<!--\; -\; -->1}((-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},x])=\underset{i=1}{\overset{n}{\bigcap <!--\; \bigcap \; -->}}{X}_i^{-<!--\; -\; -->1}((-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},x/n])$. Then $P(Y_n\le <!--\; \le \; -->x)=P(\underset{i=1}{\overset{n}{\bigcap <!--\; \bigcap \; -->}}{X}_i^{-<!--\; -\; -->1}((-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},x/n]))$. Does it follow from this that $Y_n$ is identically distributed to some of the $|X_i|$? Are there any hypothesis that may be missing?
Edit: there was a typo in the original question

Assume $X$ is a continuous random variable which have a density function. Assume $\underset{\_<!--\; \_\; -->}{Y}$ is a random vector which also have a density function. And finally assume we have a joint density function of the random variable and the random vector.
How can I find a function $\mathrm{\varphi <!--\; \varphi \; -->}$, such that $\mathbb{E}\left[|X-<!--\; -\; -->\mathrm{\varphi <!--\; \varphi \; -->}\left(\underset{\_<!--\; \_\; -->}{Y}\right)||\underset{\_<!--\; \_\; -->}{Y}\right]$ would be minimal?
I'm not sure where to start. I do know that the asnwer should be that $\mathrm{\varphi <!--\; \varphi \; -->}(\underset{\_<!--\; \_\; -->}{Y})$ should be a median of $F_{X|\underset{\_<!--\; \_\; -->}{Y}}$, but Im not sure how to prove it. Any help would be appreciated.
Thanks in advance.

Who knows
If I define algebra $\mathcal{F}(A)$ generated by $A$, collection of subsets of $S$ (the universal set) as the intersection of $\mathcal{F}$, algebra superset of $A$:
$\mathcal{F}(A)=\underset{algebra\mathcal{F}\supseteq <!--\; \supseteq \; -->A}{\bigcap <!--\; \bigcap \; -->}\mathcal{F}$
What if $A$ is an infinite (either countable or uncountable) set? Algebra, unlike $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra, guarantees being closed under finite Boolean operations. Here, finite(in the definition of algebra) and infinite(in the setting) confuses me. e.g. $A$ is the collection of intervals $(-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},x]$($x\in <!--\; \in \; -->\mathbb{R}$) and $S=\mathbb{R}$, what $\mathcal{F}(A)$ be like? Any help will be appreciated!

Let $(\mathrm{\Sigma <!--\; \Sigma \; -->},F,P)$ be a probability space, let $X$ be an $F$-random variable, let $G$ be a sub-sigma algebra of $F$, let $\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}$, let $A$ be the intersection of all the sets in $G$ that contain $\mathrm{\omega <!--\; \omega \; -->}$, and suppose that $A\in <!--\; \in \; -->F$. Then is it true that $E(X|G)(\mathrm{\omega <!--\; \omega \; -->})=E(X|A)$?
I just wrote this statement down based on my intuition of what conditional expectation means, and I want to verify if it’s actually true. Note that sigma algebras need not be closed under uncountable intersections, so $A$ need not be an element of $G$. But I assumed $A$ is at least an element of $F$ so that $E(X|A)$ is meaningful.

Let $T$ be a stopping time. We define the stopped $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra ${\mathcal{F}}_T$ as
${\mathcal{F}}_T:=\{A\in <!--\; \in \; -->\mathcal{F}\mid <!--\; \mid \; -->\mathrm{\forall <!--\; \forall \; -->}n\in <!--\; \in \; -->\mathbb{N}:A\cap <!--\; \cap \; -->\{T=n\}\in <!--\; \in \; -->{\mathcal{F}}_n\}$
How can this $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra be interpreted in "the real world"? For example, what would ${\mathcal{F}}_T$ mean for a gambler?
Thank you already!

how to show that the set
$B_{\mathrm{\delta <!--\; \delta \; -->}}=\{x\notin <!--\; \notin \; -->D_g:\mathrm{\exists <!--\; \exists \; -->}y\in <!--\; \in \; -->S\text{ such that }{d}_S(x,y)<\mathrm{\delta <!--\; \delta \; -->}\text{ but }{d}_{S^{\mathrm{\prime <!--\; \prime \; -->}}}(g(x),g(y))>\mathrm{ϵ<!--\; ϵ\; -->}\}$
is measurable with respect to the Borel $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra of $S$?
I changed the notation a bit since the result is stated generally for an a.e. continuous mapping $g:S\to <!--\; \to \; -->S^{\mathrm{\prime <!--\; \prime \; -->}}$ between two (separable, I presume) metric spaces, $S$ and $S^{\mathrm{\prime <!--\; \prime \; -->}}$.

Suppose $E[|X|]<\mathrm{\infty <!--\; \infty \; -->}$ and that $A_n$ are disjoint sets with ${\cup <!--\; \cup \; -->}_{n=1}^{\mathrm{\infty <!--\; \infty \; -->}}{A}_n=A$. Show that $\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}E[X{\mathbf{1}}_{A_n}]=E[X{\mathbf{1}}_A]$
Here is what I try with Monotone Convergence Theorem:
$\begin{array}{rl}\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}E[X{\mathbf{1}}_{A_n}]& =\underset{m\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\underset{n=0}{\overset{m}{\sum <!--\; \sum \; -->}}E[X{\mathbf{1}}_{A_n}]\\ & =\underset{m\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}E[X\underset{n=0}{\overset{m}{\sum <!--\; \sum \; -->}}{\mathbf{1}}_{A_n}]\\ & =\underset{m\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}E[X{\mathbf{1}}_{F_n}]\\ & =E[X\underset{m\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\mathbf{1}}_{F_n}]\\ & =E[X{\mathbf{1}}_A]\end{array}$
where $F_n={\cup <!--\; \cup \; -->}_{i=1}^n{A}_i$, so ${\cup <!--\; \cup \; -->}_{n=1}^m{A}_n={\cup <!--\; \cup \; -->}_{n=1}^m{F}_n$, and ${\mathbf{1}}_{F_n}$ is increasing.
I am not quite sure if I get it right. My main concern is the first step: $\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}E[X{\mathbf{1}}_{A_n}]=\underset{m\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\underset{n=0}{\overset{m}{\sum <!--\; \sum \; -->}}E[X{\mathbf{1}}_{A_n}]$. Do we always have $\underset{n=0}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}=\underset{m\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\underset{n=0}{\overset{m}{\sum <!--\; \sum \; -->}}$?
I may also have made other mistakes. Please feel free to point out. Any other hints are also welcome.

Let $E$ be a metric space and $\mathcal{M}(E)$ denote the set of finite signed measures on $(E,\mathcal{B}(E))$. Remember that $\mathcal{C}\subseteq <!--\; \subseteq \; -->C_b(E)$ is called separating for $\mathcal{F}\subseteq <!--\; \subseteq \; -->\mathcal{M}(E)$ if
$\begin{array}{}\text{(1)}& \mathrm{\forall <!--\; \forall \; -->}\mathrm{\mu <!--\; \mu \; -->},\mathrm{\nu <!--\; \nu \; -->}\in <!--\; \in \; -->\mathcal{F}:(\mathrm{\forall <!--\; \forall \; -->}f\in <!--\; \in \; -->\mathcal{C}:\int <!--\; \int \; -->f\mathrm{d}(\mathrm{\mu <!--\; \mu \; -->}-<!--\; -\; -->\mathrm{\nu <!--\; \nu \; -->})=0)\Rightarrow <!--\; \Rightarrow \; -->\mathrm{\mu <!--\; \mu \; -->}=\mathrm{\nu <!--\; \nu \; -->}.\end{array}$
Let ${\mathcal{M}}_{+}(E)$ and ${\mathcal{M}}_1(E)$ denote the set of finite and probability measures on $(E,\mathcal{E})$, respectively.

If $\mathcal{C}$ is separating for ${\mathcal{M}}_1(E)$, does it already follow that $\mathcal{C}$ is separating for $\mathcal{M}(E)$?

I'm not sure whether I'm missing something or not, but I think this should be true, since every finite signed measures can be written as a difference of two finite measures and every nontrivial finite measure can be normalized to a probability measure.

So I am reading the construction of non-measurable set $\mathcal{N}$. This is done so by considering an enumeration $\{r_k\}$ of $\mathbb{Q}\cap <!--\; \cap \; -->[-<!--\; -\; -->1,1]$ I am stuck on seeing the obviousness in why (by monotonicity) The following is a contradiction. So we claim and show the following Inclusion
$[0,1]\subset <!--\; \subset \; -->\underset{k}{\bigcup <!--\; \bigcup \; -->}{\mathcal{N}}_{\mathcal{k}}\subset <!--\; \subset \; -->[-<!--\; -\; -->1,2]$
Where ${\mathcal{N}}_k=\mathcal{N}+r_k$
I get that by monotonicity, we have
$1\le <!--\; \le \; -->\underset{k\in <!--\; \in \; -->\mathbb{N}}{\sum <!--\; \sum \; -->}m({\mathcal{N}}_k)\le <!--\; \le \; -->3$
But I do not see the contradiction as to why since neither $m(\mathcal{N})=0$ nor $m(\mathcal{N})>0$ we are done by contradiction, can't the measure be strictly between 1 and 3? so 2?

Let X be a locally compact Hausdorff space and $C_b$(X) the set of all continuous functions with support compact. The dual of $C_b$ is isometric isomorphism of M(X), with M(X) the set of all regular Borel measures on X? Do the measures need to be finite as well?

Let $T:X\to <!--\; \to \; -->X$ be a measure preserving invertible transformation and let $F$ be as sub sigma algebra. My book said it is clear that $E({\mathrm{\chi <!--\; \chi \; -->}}_{T^{-<!--\; -\; -->1}A}|T^{-<!--\; -\; -->1}F)(x)=E({\mathrm{\chi <!--\; \chi \; -->}}_A|F)(Tx)$ but I am not sure why, especially considering how the right hand side doesn't seem to be $T^{-<!--\; -\; -->1}F$ measurable.
Edit: The book doesn't mention it, but I think it is implicitly assuming that $T^{-<!--\; -\; -->1}F\subset <!--\; \subset \; -->F$. This is from an ergodic theory textbook, and this result is used to prove that the conditional entropy satisfies $H(T^{-<!--\; -\; -->1}A|T^{-<!--\; -\; -->1}F)=H(A|F)$

For a discrete set $X$ with $n$ elements, there can be finite many $\mathrm{\Sigma <!--\; \Sigma \; -->}=\{{\mathrm{\sigma <!--\; \sigma \; -->}}_1,{\mathrm{\sigma <!--\; \sigma \; -->}}_2,\cdots <!--\; \cdots \; -->,{\mathrm{\sigma <!--\; \sigma \; -->}}_m\}$ algebra defined on it.
Given a $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra $\mathrm{\sigma <!--\; \sigma \; -->}$, let $S(\mathrm{\sigma <!--\; \sigma \; -->})=\{A\in <!--\; \in \; -->\mathrm{\sigma <!--\; \sigma \; -->}\mid <!--\; \mid \; -->A\cap <!--\; \cap \; -->B=\mathrm{\varnothing <!--\; \varnothing \; -->},\mathrm{\forall <!--\; \forall \; -->}B\in <!--\; \in \; -->\mathrm{\sigma <!--\; \sigma \; -->}\setminus <!--\; \setminus \; -->\{A,X\}\}$. And how can I prove that $𝑆(\mathrm{\sigma <!--\; \sigma \; -->})=𝑆({\mathrm{\sigma <!--\; \sigma \; -->}}^{\prime })$ implies $\mathrm{\sigma <!--\; \sigma \; -->}=\mathrm{\sigma <!--\; \sigma \; -->}\prime$?

Let $E_n=\{x\in <!--\; \in \; -->\mathbb{X}\mid <!--\; \mid \; -->(n-<!--\; -\; -->1)\le <!--\; \le \; -->|f(x)|<n\}$, where $(\mathbb{X},\mathcal{A},\mathrm{\mu <!--\; \mu \; -->})$ is a finite measure space (i.e., $\mathrm{\mu <!--\; \mu \; -->}(\mathbb{X})<\mathrm{\infty <!--\; \infty \; -->}$) and $f$ is a measurable function. Prove that $f\in <!--\; \in \; -->L^p⟺<!--\; ⟺\; -->\sum <!--\; \sum \; -->n^p\mathrm{\mu <!--\; \mu \; -->}(E_n)<\mathrm{\infty <!--\; \infty \; -->}$.
One of the implications is very straightforward to me. It is clear that $\mathbb{X}=\bigcup <!--\; \bigcup \; -->E_n$, where the union is disjoint. Therefore, we have:
${\int <!--\; \int \; -->}_{\mathbb{X}}|f{|}^{p}d\mathrm{\mu <!--\; \mu \; -->}=\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\int <!--\; \int \; -->}_{E_n}|f{|}^{p}d\mathrm{\mu <!--\; \mu \; -->}$
By the definition of the sets, we get the inequalities:
$\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\int <!--\; \int \; -->(n-<!--\; -\; -->1{)}^p{\mathbb{1}}_{E_n}d\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\int <!--\; \int \; -->}_{E_n}|f{|}^{p}d\mathrm{\mu <!--\; \mu \; -->}<\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}\int <!--\; \int \; -->n^p{\mathbb{1}}_{E_n}d\mathrm{\mu <!--\; \mu \; -->}$
These, in turn, simplify to
$\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}(n-<!--\; -\; -->1{)}^p\mathrm{\mu <!--\; \mu \; -->}(E_n)\le <!--\; \le \; -->{\int <!--\; \int \; -->}_{\mathbb{X}}|f{|}^{p}d\mathrm{\mu <!--\; \mu \; -->}<\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{n}^p\mathrm{\mu <!--\; \mu \; -->}(E_n)$
Thus, it is immediately apparent that, if the sum converges, then so does the integral, which implies $f\in <!--\; \in \; -->L^p$.
My problem is in the other implication: $f\in <!--\; \in \; -->L^p$ implies convergence
I haven't used finite measure yet, so maybe it comes to play. I first thought about expanding $(n-<!--\; -\; -->1{)}^p$, but this would be a mess even if $p$ was an integer (unless in the case where $p=1$, in which finite measure would imply the result easily).
Does anyone have a hint as per how to proceed?
Thanks in advance!

How does the equation come up:
${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\mathrm{\theta <!--\; \theta \; -->}{e}^{\mathrm{\theta <!--\; \theta \; -->}x}Q(x,\mathrm{\infty <!--\; \infty \; -->})\mathrm{\lambda <!--\; \lambda \; -->}(dx)={\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}(e^{\mathrm{\theta <!--\; \theta \; -->}x}-<!--\; -\; -->1)Q(dx)$
with $Q$ is a measure and $\mathrm{\lambda <!--\; \lambda \; -->}$ is the Lebesgue measure.
I calculated:
${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\mathrm{\theta <!--\; \theta \; -->}{e}^{\mathrm{\theta <!--\; \theta \; -->}x}{\int <!--\; \int \; -->}_{\mathbb{R}}{\mathbf{1}}_{(x,\mathrm{\infty <!--\; \infty \; -->})}(y)Q(dy)\mathrm{\lambda <!--\; \lambda \; -->}(dx)$
$={\int <!--\; \int \; -->}_{\mathbb{R}}{\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}\mathrm{\theta <!--\; \theta \; -->}{e}^{\mathrm{\theta <!--\; \theta \; -->}x}{\mathbf{1}}_{(x,\mathrm{\infty <!--\; \infty \; -->})}(y)\mathrm{\lambda <!--\; \lambda \; -->}(dy)Q(dx)$
$={\int <!--\; \int \; -->}_{\mathbb{R}}{\int <!--\; \int \; -->}_x^{\mathrm{\infty <!--\; \infty \; -->}}\mathrm{\theta <!--\; \theta \; -->}{e}^{\mathrm{\theta <!--\; \theta \; -->}x}\mathrm{\lambda <!--\; \lambda \; -->}(dy)Q(dx)$
but don’t know how to come to ${\int <!--\; \int \; -->}_0^{\mathrm{\infty <!--\; \infty \; -->}}(e^{\mathrm{\theta <!--\; \theta \; -->}x}-<!--\; -\; -->1)Q(dx)$.
KR, toni

The following is a conclusion in a Chinese textbook "Lecture of Measure Theory" and the proof of it is left as an exercise.
$(\mathrm{\Omega <!--\; \Omega \; -->},\mathcal{F})$ is a measurable space, and $\mathcal{C}$ is an algebra consisting of subsets of $\mathrm{\Omega <!--\; \Omega \; -->}$ such that $\mathrm{\sigma <!--\; \sigma \; -->}(\mathcal{C})=\mathcal{F}$. Given $\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}$, define
${\mathcal{F}}_{\mathrm{\omega <!--\; \omega \; -->}}=\{B\in <!--\; \in \; -->\mathcal{F}\mid <!--\; \mid \; -->\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->B\}\text{and}{\mathcal{C}}_{\mathrm{\omega <!--\; \omega \; -->}}=\{B\in <!--\; \in \; -->\mathcal{C}\mid <!--\; \mid \; -->\mathrm{\omega <!--\; \omega \; -->}\in <!--\; \in \; -->B\}.$
It is obvious that $A(\mathrm{\omega <!--\; \omega \; -->}):=\underset{B\in <!--\; \in \; -->{\mathcal{F}}_{\mathrm{\omega <!--\; \omega \; -->}}}{\bigcap <!--\; \bigcap \; -->}B\subset <!--\; \subset \; -->\underset{B\in <!--\; \in \; -->{\mathcal{C}}_{\mathrm{\omega <!--\; \omega \; -->}}}{\bigcap <!--\; \bigcap \; -->}B$, but how to prove the opposite inclusion relation, i.e.
$\underset{B\in <!--\; \in \; -->{\mathcal{C}}_{\mathrm{\omega <!--\; \omega \; -->}}}{\bigcap <!--\; \bigcap \; -->}B\subset <!--\; \subset \; -->\underset{B\in <!--\; \in \; -->{\mathcal{F}}_{\mathrm{\omega <!--\; \omega \; -->}}}{\bigcap <!--\; \bigcap \; -->}B.$

I can`r do this, need help
Given a locally integrable function $\mathrm{\gamma <!--\; \gamma \; -->}:{\mathbb{R}}_{\ge <!--\; \ge \; -->0}\to <!--\; \to \; -->\mathbb{R}$, we define the absolutely continuous function $\mathrm{\Gamma <!--\; \Gamma \; -->}(t):=\underset{u\le <!--\; \le \; -->t}{max}{\int <!--\; \int \; -->}_u^t\mathrm{\gamma <!--\; \gamma \; -->}\mathrm{d}\mathrm{\lambda <!--\; \lambda \; -->}$.
I want to show, that $\mathrm{\gamma <!--\; \gamma \; -->}(t)\le <!--\; \le \; -->0$ holds for almost all $t$ with $\mathrm{\Gamma <!--\; \Gamma \; -->}(t)=0$. In other words, I want to show that the set
$A:=\{t\in <!--\; \in \; -->{\mathbb{R}}_{\ge <!--\; \ge \; -->0}|\mathrm{\Gamma <!--\; \Gamma \; -->}(t)=0\text{and}\mathrm{\gamma <!--\; \gamma \; -->}(t)>0\}$
is a Lebesgue-null set, i.e. $\mathrm{\lambda <!--\; \lambda \; -->}(A)=0$.
All my attempts have failed. Nevertheless, I was able to show, that if $\mathrm{\Gamma <!--\; \Gamma \; -->}$ vanishes on a (proper) interval $[a,b]$ with $a<b$, then $\mathrm{\lambda <!--\; \lambda \; -->}(A\cap <!--\; \cap \; -->[a,b])=0$. This however does not lead to a proof of the more general claim $\mathrm{\lambda <!--\; \lambda \; -->}(A)=0$.

I have the following exercise, but i can't find the answer.
Let $(\mathit{X},\mathcal{A},\mathrm{\mu <!--\; \mu \; -->})$ be a measure space. Show that the measure $\mathrm{\mu <!--\; \mu \; -->}$ is $\mathrm{\sigma <!--\; \sigma \; -->}$-finite, if there exists a function $f\in <!--\; \in \; -->{\mathcal{M}}^{+}(\mathit{X})$ with $f(x)><!--\; >\; -->0$ for all $x\in <!--\; \in \; -->\mathit{X}$ and ${\int <!--\; \int \; -->}_{\mathit{X}}fd\mathrm{\mu <!--\; \mu \; -->}<<!--\; <\; -->\mathrm{\infty <!--\; \infty \; -->}$.
From what I understand is that the function $f$ has to be from the set ${\mathcal{M}}^{+}$, so the set of strictly positive numeric functions (as defined in the script of the professor).

Let $X=Y$ uncountable, and let $A=B$ be the $\mathrm{\sigma <!--\; \sigma \; -->}$ algebra of countable-cocountable subsets of $X$,$Y$ respectively. Let $C$ be the countable-cocountable $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra on $X\times <!--\; \times \; -->Y$.
Is $C=\mathrm{\sigma <!--\; \sigma \; -->}(A\times <!--\; \times \; -->B)$?
$\subseteq <!--\; \subseteq \; -->$ is obvious, but I am not sure about the other direction. I know that in general the two do not have to be equal, so I am trying to find a counterexample i.e find an element in C which is not in $\mathrm{\sigma <!--\; \sigma \; -->}(A\times <!--\; \times \; -->B)$. Perhaps AC is involved, which makes it tricky? I am not sure.
Also, is it true that $\mathrm{\sigma <!--\; \sigma \; -->}(A\times <!--\; \times \; -->B)=A\times <!--\; \times \; -->B?$ This may help simplify the question.

Consider $f$ a bounded, measurable function defined on $[1,\mathrm{\infty <!--\; \infty \; -->}]$ and define $a_n:={\int <!--\; \int \; -->}_{[n,n+1)}f$. Does the fact that $f$ is integrable imply $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_n$ converges?
I would like to say that the argument
$\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_n={\int <!--\; \int \; -->}_1^{\mathrm{\infty <!--\; \infty \; -->}}f$
made in that post is not correct.
My attempt was the following: Since $f$ is integrable, $|f|$ is also integrable. Therefore,
${\int <!--\; \int \; -->}_{[1,\mathrm{\infty <!--\; \infty \; -->})}|f|=\underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_1^k|f|<\mathrm{\infty <!--\; \infty \; -->}.$
Moreover, we have the following:
$\begin{array}{rl}\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_n& =\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\int <!--\; \int \; -->}_n^{n+1}f\\ & \le <!--\; \le \; -->\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}|{\int <!--\; \int \; -->}_n^{n+1}f|\\ & \le <!--\; \le \; -->\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\int <!--\; \int \; -->}_n^{n+1}|f|\\ & =\underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\underset{n=1}{\overset{k}{\sum <!--\; \sum \; -->}}{\int <!--\; \int \; -->}_n^{n+1}|f|\end{array}$
However, I cannot proceed further from here. Can anyone help?

Let $X$ be an arbitrary space, and let $\mathrm{\mu <!--\; \mu \; -->}$,$\mathrm{\nu <!--\; \nu \; -->}$ be two probability measures on $X$. Suppose there is a common set $A$ on which $\mathrm{\mu <!--\; \mu \; -->}(A)=\mathrm{\nu <!--\; \nu \; -->}(A)=1$.
Now suppose there are $B,C$ such that $\mathrm{\mu <!--\; \mu \; -->}(B)=\mathrm{\nu <!--\; \nu \; -->}(C)=1$. Can anything be said of $\mathrm{\mu <!--\; \mu \; -->}(C)$ and $\mathrm{\nu <!--\; \nu \; -->}(B)$, or $\mathrm{\mu <!--\; \mu \; -->}(B\cap <!--\; \cap \; -->C)$ and $\mathrm{\nu <!--\; \nu \; -->}(B\cap <!--\; \cap \; -->C)$?
My initial intuition was that the answer should be yes since $B\cap <!--\; \cap \; -->A$ and $C\cap <!--\; \cap \; -->A$ both have full $\mathrm{\mu <!--\; \mu \; -->}$ and full $\mathrm{\nu <!--\; \nu \; -->}$ measure respectively, and so there should be "enough" overlap for $B\cap <!--\; \cap \; -->C$ to have at least positive measure. But I can't seem to prove this - so I'm interested in whether anything can be said of these quantities (either alone or possibly together, e.g. is it possible for $\mathrm{\mu <!--\; \mu \; -->}(C)=\mathrm{\nu <!--\; \nu \; -->}(B)=0$) and whether there is any clever counterexample for which $\mathrm{\mu <!--\; \mu \; -->}(C)$ and $\mathrm{\nu <!--\; \nu \; -->}(B)$ can be any arbitrary number in [0,1].