Explore Measurement Concepts with Our Examples and Questions

In the context of the Lévy-Khintchine formula, I have certain integral
$\begin{array}{}\text{(1)}& {\int <!--\; \int \; -->}_{{\mathbb{R}}^p}(e^{i{t}^{\prime }x}-<!--\; -\; -->1-<!--\; -\; -->i\frac{t^{\prime }x}{1-<!--\; -\; -->x^{\prime }x})d\mathrm{\nu <!--\; \nu \; -->}(x),t\in <!--\; \in \; -->{\mathbb{R}}^p\end{array}$
where $\mathrm{\nu <!--\; \nu \; -->}$ is a measure defined on ${\mathbb{R}}^p$.
Suppose that
${\int <!--\; \int \; -->}_{{\mathbb{R}}^p}|x{|}^{2}d\mathrm{\nu <!--\; \nu \; -->}(x)<\mathrm{\infty <!--\; \infty \; -->}$
Define:
$\mathrm{\kappa <!--\; \kappa \; -->}(E)={\int <!--\; \int \; -->}_E|x{|}^{2}d\mathrm{\nu <!--\; \nu \; -->}(x),E\subset <!--\; \subset \; -->{\mathbb{R}}^p.$
I want to get $d\mathrm{\kappa <!--\; \kappa \; -->}(x)$ as a function of $d\mathrm{\nu <!--\; \nu \; -->}(x)$ in order to do a substitution in (1), but I don't know how to differentiate $\mathrm{\kappa <!--\; \kappa \; -->}$:
$d\mathrm{\kappa <!--\; \kappa \; -->}(x)=?d\mathrm{\nu <!--\; \nu \; -->}(x)$

Suppose I have two positive Borel measures on $[0,\mathrm{\infty <!--\; \infty \; -->}]$, say $\mathrm{\mu <!--\; \mu \; -->}$ and $\mathrm{\nu <!--\; \nu \; -->}$ and I know that $\mathrm{\nu <!--\; \nu \; -->}([0,x])=0$ implies $\mathrm{\mu <!--\; \mu \; -->}([0,x])=0$ for any $x$ and that $\mathrm{\nu <!--\; \nu \; -->}((a,b])=0$ implies $\mathrm{\mu <!--\; \mu \; -->}((a,b])=0$ for any pair $a,b$.
Does this imply that $\mathrm{\mu <!--\; \mu \; -->}<<\mathrm{\nu <!--\; \nu \; -->}$? I would think so but I have trouble proving it, I thought that the monotone class theorem was the answer but I don't see how the sets with the property $\mathrm{\nu <!--\; \nu \; -->}(E)=0⟹<!--\; ⟹\; -->\mathrm{\mu <!--\; \mu \; -->}(E)=0$ are closed under monotone intersections.
We may assume that the measures are finite, if that helps anything.

I've encountered something called the "essential supremum" while working with $L^p$ spaces (in particular, for $p=\mathrm{\infty <!--\; \infty \; -->}$).
I tried looking it up on the internet but all the definitions use concepts from Measure Theory, which I'm not familiar with. Is there a way to wrap my head around it without having to deal with measures? I really only need to understand this to show that a piecewise continuous function $f\in <!--\; \in \; -->C(\mathrm{\Omega <!--\; \Omega \; -->})$ is in $L^{\mathrm{\infty <!--\; \infty \; -->}}(\mathrm{\Omega <!--\; \Omega \; -->})$. I can't seem to understand how the $L^{\mathrm{\infty <!--\; \infty \; -->}}$ space works or how it is defined.

Let $A\subset <!--\; \subset \; -->[a,b)$ for $-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}<a<b<\mathrm{\infty <!--\; \infty \; -->}$ be Lebesgue measurable. Show, that $F(x)=\mathrm{\mu <!--\; \mu \; -->}([a,x)\cap <!--\; \cap \; -->A)$ for $a\le <!--\; \le \; -->x\le <!--\; \le \; -->b$ is continous on $[a,b]$
I honestly don't know where to start here. My idea would be to somehow argue that because $\mathrm{\mu <!--\; \mu \; -->}$ is continuous, then $F$ must be continuous too. But this seems too easy.
Another idea I'd have is to do the following: Because we know, that $A\subset <!--\; \subset \; -->[a,b)$, this means, that
$[a,x)\cap <!--\; \cap \; -->A=[a,y]$
for some $a\le <!--\; \le \; -->y\le <!--\; \le \; -->x$. This means, that we can instead of $F(x)$ use
$G(y)=\mathrm{\mu <!--\; \mu \; -->}([a,y])$
$G^{-<!--\; -\; -->1}(\mathrm{\mu <!--\; \mu \; -->}([a,y]))=[a,x]$
So a closed image gives you a closed pre-image, which is one of the conditions for $G$ to be continous. And if $G$ is continuous, then $F$ is, too.
I kind of know that my attempt is not good, because I assume that we should do it in a completely different way.

I`m having some trouble with how to prove that if $\mathrm{\mu <!--\; \mu \; -->}$ is finite then it is sum-finite.
I know that if $\mathrm{\mu <!--\; \mu \; -->}$ is finite on a measurable space (E,A) if $\mathrm{\mu <!--\; \mu \; -->}<\mathrm{\infty <!--\; \infty \; -->}$ and that $\mathrm{\mu <!--\; \mu \; -->}$ is sum-finite on a measurable space (E,A) if $\mathrm{\mu <!--\; \mu \; -->}=\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\mathrm{\mu <!--\; \mu \; -->}}_n$, where ${\mathrm{\mu <!--\; \mu \; -->}}_n$, for $n\in <!--\; \in \; -->\mathbb{N}$, is finite measures. But I dont know how to tie these definitions together.

Given $\Vert <!--\; \Vert \; -->f_n{\Vert <!--\; \Vert \; -->}_{L^2}\le <!--\; \le \; -->2^{-<!--\; -\; -->k},$ where $L^2$ represents the banach space of squared integrable functions, and $g:=\underset{k=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}|f_n|,$ I need to show that $g\in <!--\; \in \; -->L^2,$, i.e. $\int <!--\; \int \; -->|g{|}^{2}d\mathrm{\mu <!--\; \mu \; -->}<\mathrm{\infty <!--\; \infty \; -->}.$
I first thought to use the Hölder inequality for $f_1=g,f_2=g.$
$\int <!--\; \int \; -->ggd\mathrm{\mu <!--\; \mu \; -->}\le <!--\; \le \; -->(\int <!--\; \int \; -->(|f_1|+|f_2|+...{)}^2{)}^{1/2}(\int <!--\; \int \; -->(|f_1|+|f_2|+...{)}^2{)}^{1/2}.$
But this is not leading to anything. It is clear that the geometric series $\sum <!--\; \sum \; -->2^{-<!--\; -\; -->k}$ converges.
Can somebody provide any suggestion or solution proposal ? Thanks.

$u_n=Uniform[0,n]$ where $u_n([a,b])=(b-<!--\; -\; -->a)/n$.
I'm trying to show there isn't such probability measure $u$ such that $u_n\Rightarrow <!--\; \Rightarrow \; -->u$.
Using proof by contradiction,
Suppose there exists such a distribution u, then by the definition of weak convergence, $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{u}_n((-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},x])=u((-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},x])$
I want to show such $\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{u}_n((-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->},x])$ doesn't exist, but confused how to proceed.
Any help would be much appreciated!

Let $Q$ be an oriented smooth manifold. I am trying to construct a finite measure on $Q$, i.e. a top rank positive differential form with finite volume.
My idea is given a collection of local charts $(U_{\mathrm{\alpha <!--\; \alpha \; -->}},{\mathrm{\psi <!--\; \psi \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}})$, we can construct a positive ${\mathrm{\mu <!--\; \mu \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}}$ as ${\mathrm{\mu <!--\; \mu \; -->}}_{\mathrm{\alpha <!--\; \alpha \; -->}}=f_{\mathrm{\alpha <!--\; \alpha \; -->}}d{q}^1\wedge <!--\; \wedge \; -->\cdots <!--\; \cdots \; -->\wedge <!--\; \wedge \; -->d{q}^n$ for any positive smooth $f_{\mathrm{\alpha <!--\; \alpha \; -->}}$ on $U_{\mathrm{\alpha <!--\; \alpha \; -->}}$. Now using the paracompactness of any manifold, we can find a countable collection of regular coordinate balls that is an open locally finite refinement of $U_{\mathrm{\alpha <!--\; \alpha \; -->}}$. Say this collection is ($V_n$). Then we can construct a global top rank form $\mathrm{\mu <!--\; \mu \; -->}$ by a partition of unity subordinate to $V_n$,
$\mathrm{\mu <!--\; \mu \; -->}=\underset{i}{\sum <!--\; \sum \; -->}{\mathrm{\rho <!--\; \rho \; -->}}_i{\mathrm{\mu <!--\; \mu \; -->}}_i.$
I am trying to show that this has finite measure but I am struggling to finish the proof here. So we have
${\int <!--\; \int \; -->}_Q\mathrm{\mu <!--\; \mu \; -->}=\underset{i}{\sum <!--\; \sum \; -->}{\int <!--\; \int \; -->}_{V_i}{\mathrm{\rho <!--\; \rho \; -->}}_i{f}_{i}d{q}^1\wedge <!--\; \wedge \; -->\cdots <!--\; \cdots \; -->\wedge <!--\; \wedge \; -->d{q}^n=\underset{i}{\sum <!--\; \sum \; -->}{\int <!--\; \int \; -->}_{{\mathrm{\psi <!--\; \psi \; -->}}_i(V_i)}({\mathrm{\rho <!--\; \rho \; -->}}_i{f}_i\circ <!--\; \circ \; -->{\mathrm{\psi <!--\; \psi \; -->}}_i^{-<!--\; -\; -->1})d^{n}q$
where $d^{n}q$ is the Lebesgue measure.I think I need to use the compactness of the domains but I can't figure out how to show that this integral can be finite, or come up with a different construction. I would greatly appreciate any help.

Is $f(x)=\{\begin{array}{ll}1& \text{ if }x\in <!--\; \in \; -->\mathbb{Q}\\ 0& \text{ if }x\notin <!--\; \notin \; -->\mathbb{Q}\end{array}$ Lebesgue measurable ?
I haven't study rigorous the chapter of measurable function, so I don't have many tools to work on this.
My idea was to say that since $f^{-<!--\; -\; -->1}(0)=\{\mathbb{Q}\}$ Which is measurable and $f^{-<!--\; -\; -->1}(1)=\{\mathbb{R}\setminus <!--\; \setminus \; -->\mathbb{Q}\}$ which is also measurable then $f$ is measurable, I don't think it's correct. How should I prove this (if $f$ is even measurable)?

in my probability theory course, we defined a sequence of random variables $(X_i{)}_{i=1}^{\mathrm{\infty <!--\; \infty \; -->}}$ to be tight if for all $\mathrm{ϵ<!--\; ϵ\; -->}>0$, there is a constant M s.th. $P(|X_n|>M)<\mathrm{ϵ<!--\; ϵ\; -->}$ for all $n\in <!--\; \in \; -->\mathbb{N}$.
I have seen the following criteria for tightness/non-tightness and I was wondering whether they are true or not:
1. $(X_i{)}_{i=1}^{\mathrm{\infty <!--\; \infty \; -->}}$ tight if there exists M s.th. lim${}_{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}P(|X_n|>M)=0$
2. $(X_i{)}_{i=1}^{\mathrm{\infty <!--\; \infty \; -->}}$ not tight if for all M lim${}_{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}P(|X_n|>M)>0$
I am quite sure that the first one is right (we get the condition that the probability is smaller than ϵ for all but finitely many n and can take the maximum of all the remaining M necessary for the finitely many n). For the second one I am not so sure, I was thinking that maybe we need some uniform bound, i.e. limn${}_{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}P(|X_n|>M)\ge <!--\; \ge \; -->\mathrm{ϵ<!--\; ϵ\; -->}>0$. I know that the criterion is right if the limit is equal to 1.

Let $X,Y,Z$ be integrable random variables on the same probability space.
Is it true that $\mathrm{\sigma <!--\; \sigma \; -->}(E[X|Y],E[X|Z])\subseteq <!--\; \subseteq \; -->\mathrm{\sigma <!--\; \sigma \; -->}(E[X|Y,Z])$ ?
My intuition says yes: the approximation of $X$ on $\mathrm{\sigma <!--\; \sigma \; -->}(Y,Z)$ is richer in information than the one on $\mathrm{\sigma <!--\; \sigma \; -->}(Y)$ and the one on $\mathrm{\sigma <!--\; \sigma \; -->}(Z)$. How to prove this?
If it is not true, what if we add $\mathrm{\sigma <!--\; \sigma \; -->}(Y)\subseteq <!--\; \subseteq \; -->\mathrm{\sigma <!--\; \sigma \; -->}(Z)$ ? Does this make the above true?

What does it mean that a function in Sobolev spaces (for example) to be defined only upto a set of measure zero?

Pratt's Lemma is : $\mathrm{\xi <!--\; \xi \; -->},\mathrm{\eta <!--\; \eta \; -->},\mathrm{\zeta <!--\; \zeta \; -->}$ and ${\mathrm{\xi <!--\; \xi \; -->}}_n,{\mathrm{\eta <!--\; \eta \; -->}}_n,{\mathrm{\zeta <!--\; \zeta \; -->}}_n$ such that:
${\mathrm{\xi <!--\; \xi \; -->}}_n\to <!--\; \to \; -->\mathrm{\xi <!--\; \xi \; -->},{\mathrm{\eta <!--\; \eta \; -->}}_n\to <!--\; \to \; -->\mathrm{\eta <!--\; \eta \; -->},{\mathrm{\zeta <!--\; \zeta \; -->}}_n\to <!--\; \to \; -->\mathrm{\zeta <!--\; \zeta \; -->},\text{convergence in probability}$
and ${\mathrm{\eta <!--\; \eta \; -->}}_n\le <!--\; \le \; -->{\mathrm{\xi <!--\; \xi \; -->}}_n\le <!--\; \le \; -->{\mathrm{\zeta <!--\; \zeta \; -->}}_n$, $E{\mathrm{\zeta <!--\; \zeta \; -->}}_n\to <!--\; \to \; -->E\mathrm{\zeta <!--\; \zeta \; -->},E{\mathrm{\eta <!--\; \eta \; -->}}_n\to <!--\; \to \; -->E\mathrm{\eta <!--\; \eta \; -->}$, and $E\mathrm{\zeta <!--\; \zeta \; -->},E\mathrm{\eta <!--\; \eta \; -->},E\mathrm{\xi <!--\; \xi \; -->}$ are finite, prove :
If ${\mathrm{\eta <!--\; \eta \; -->}}_n\le <!--\; \le \; -->0\le <!--\; \le \; -->{\mathrm{\zeta <!--\; \zeta \; -->}}_n$, then $E|{\mathrm{\xi <!--\; \xi \; -->}}_n-<!--\; -\; -->\mathrm{\xi <!--\; \xi \; -->}|\to <!--\; \to \; -->0$.

I know how to prove $E{\mathrm{\xi <!--\; \xi \; -->}}_n\to <!--\; \to \; -->E\mathrm{\xi <!--\; \xi \; -->}$, but $E|{\mathrm{\xi <!--\; \xi \; -->}}_n-<!--\; -\; -->\mathrm{\xi <!--\; \xi \; -->}|\to <!--\; \to \; -->0$ seems not easy proved from it.
My first question is how to prove $E|{\mathrm{\xi <!--\; \xi \; -->}}_n-<!--\; -\; -->\mathrm{\xi <!--\; \xi \; -->}|\to <!--\; \to \; -->0$.
And I don't know why should emphasize the condition "${\mathrm{\eta <!--\; \eta \; -->}}_n\le <!--\; \le \; -->0\le <!--\; \le \; -->{\mathrm{\zeta <!--\; \zeta \; -->}}_n$", so my second question is: Is there any example that $E|{\mathrm{\xi <!--\; \xi \; -->}}_n-<!--\; -\; -->\mathrm{\xi <!--\; \xi \; -->}|\to <!--\; \to \; -->0$ is wrong if the condition is not fullfilled.

I`m trying to find an answer, but i have some problems, help.
Let ${\mathbb{P}}_{\mathrm{\theta <!--\; \theta \; -->}}=U[0,\mathrm{\theta <!--\; \theta \; -->}]$.
For $h,{\mathrm{\theta <!--\; \theta \; -->}}_0>0$ and $Z\sim <!--\; \sim \; -->\mathrm{e}\mathrm{x}\mathrm{p}\left(\frac{1}{{\mathrm{\theta <!--\; \theta \; -->}}_0}\right)$ I have to show that:
$\frac{\mathrm{d}{\mathbb{P}}_{{\mathrm{\theta <!--\; \theta \; -->}}_0-<!--\; -\; -->h/n}^n}{\mathrm{d}{\mathbb{P}}_{{\mathrm{\theta <!--\; \theta \; -->}}_0}^n}\stackrel{d,{\mathbb{P}}_{{\mathrm{\theta <!--\; \theta \; -->}}_0}^n}{⟶<!--\; ⟶\; -->}{e}^{\frac{h}{{\mathrm{\theta <!--\; \theta \; -->}}_0}}{\mathbb{1}}_{\{Z\ge <!--\; \ge \; -->h\}}$
I already proved that for $Z_n=n({\mathrm{\theta <!--\; \theta \; -->}}_0-<!--\; -\; -->max\{X_1,\dots <!--\; \dots \; -->,X_n\})$ with $X_1,\dots <!--\; \dots \; -->X_n\sim <!--\; \sim \; -->{\mathbb{P}}_{{\mathrm{\theta <!--\; \theta \; -->}}_0}$ holds $Z_n\stackrel{D}{\to <!--\; \to \; -->}Z$ and this task seems like I have to prove that the pdf is converging too. I'm not sure which technical steps I need to show this and I'm not sure which kind of convergence is meant by $d,{\mathbb{P}}_{{\mathrm{\theta <!--\; \theta \; -->}}_0}^n$.

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Often in physics we integrate by parts

by:

I have a really simple question, how can we assume that $[f(x)\mathrm{\delta <!--\; \delta \; -->}(x-<!--\; -\; -->y){]}_{x_0}^{x_1}=0$?

Intuitively the delta function is zero except for at $x=y$, but what if either $x_0$ or $x_1$ was equal to y?

Is the answer simply 'we must assume separately that $x_0,x_1\ne <!--\; \ne \; -->y$, or is there something obvious that I'm missing, or is there some measure theory reason why we can say it is zero?

Given the step function
$g(x,y)=\{\begin{array}{ll}\frac{1}{x^2}& 0<y<x<1\\ -<!--\; -\; -->\frac{1}{y^2}& 0<x<y<1\\ 0& \text{otherwise}\end{array}$
I want to show that
${\int <!--\; \int \; -->}_{(0,1)}{\int <!--\; \int \; -->}_{(0,1)}g(x,y)d\mathrm{\lambda <!--\; \lambda \; -->}(x)d\mathrm{\lambda <!--\; \lambda \; -->}(y)\ne <!--\; \ne \; -->{\int <!--\; \int \; -->}_{(0,1)}{\int <!--\; \int \; -->}_{(0,1)}g(x,y)d\mathrm{\lambda <!--\; \lambda \; -->}(y)d\mathrm{\lambda <!--\; \lambda \; -->}(x)$
However, when trying to use the relation with Riemann integrability I end up with divergent integrals such as
${\int <!--\; \int \; -->}_0^x{\int <!--\; \int \; -->}_0^1\frac{1}{x^2}dxdy-<!--\; -\; -->{\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_0^y\frac{1}{y^2}dxdy$
I also considered using limits such as following one, but again I end up with divergence.
$\underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_{(\frac{1}{k},1)}\underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_{(\frac{1}{k},1)}g(x,y)dxdy$
I would really appreciate any help!

Let $f:\mathrm{\Omega <!--\; \Omega \; -->}\to <!--\; \to \; -->B$ be Bochner-measurable, i.e. the point-wise limit of a sequence of simple (i.e. countably-valued measurable) functions $(s_n)$. I know that if
${\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}||f(\mathrm{\omega <!--\; \omega \; -->})||d\mathrm{\mu <!--\; \mu \; -->}(\mathrm{\omega <!--\; \omega \; -->})<\mathrm{\infty <!--\; \infty \; -->}(\ast <!--\; \ast \; -->)$
then one can show that f is Bochner-integrable, i.e. there is even a Bochner-integrable sequence of simple functions $({\stackrel{~<!--\; ~\; -->}{s}}_n)$ such that
$\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}||f(\mathrm{\omega <!--\; \omega \; -->})-<!--\; -\; -->{\stackrel{~<!--\; ~\; -->}{s}}_n(\mathrm{\omega <!--\; \omega \; -->})||d\mathrm{\mu <!--\; \mu \; -->}(\mathrm{\omega <!--\; \omega \; -->})=0(\ast <!--\; \ast \; -->\ast <!--\; \ast \; -->)$
I am unsure, however, how to derive this sequence when the measure space $\mathrm{\Omega <!--\; \Omega \; -->}$ is not finite. Apparently one is supposed to make use of the fact that due to $(\ast <!--\; \ast \; -->)$, the set
$A:=\{f\ne <!--\; \ne \; -->0\}=\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\bigcup <!--\; \bigcup \; -->}}\underset{=:A_n}{\underset{⏟<!--\; ⏟\; -->}{\{||f||>\frac{1}{n}\}}}$
is $\mathrm{\sigma <!--\; \sigma \; -->}$-finite, as each $A_n$ has finite measure. But how to proceed? Setting ${\stackrel{~<!--\; ~\; -->}{s}}_n=1_A{s}_n$ would not be enough to make the functions Bochner-integrable, as the entire set $A$ might still have infinite measure. Setting ${\stackrel{~<!--\; ~\; -->}{s}}_n=1_{A_n}{s}_n$ would do it and maintain pointwise convergence, but then I don't know how to show the convergence in $(\ast <!--\; \ast \; -->\ast <!--\; \ast \; -->)$ anymore...any tips are much appreciated.

The following is the definition of a random variable:
Let $(\mathrm{\Omega <!--\; \Omega \; -->},\mathbb{F},P)$ be a probability space. A random variable is a real-valued function X on $\mathrm{\Omega <!--\; \Omega \; -->}$ such that for all $x\in <!--\; \in \; -->\mathbb{R},\{\mathrm{\omega <!--\; \omega \; -->}:X(\mathrm{\omega <!--\; \omega \; -->})\in <!--\; \in \; -->B\}\in <!--\; \in \; -->\mathbb{F},\forall B\in B(\mathbb{R})$
I don't understand how this makes sense if our choice of $F$ can be arbitrary.
Let's say we are rolling a biased dice such that each side has a different probability, and the information we know about the system is $F=\{\mathrm{\varnothing <!--\; \varnothing \; -->},\{1\},\{2,...,6\},\mathrm{\Omega <!--\; \Omega \; -->}\}$ -- then for this $F$, according to the definition, $X$ is not a random variable (because you could construct a (disjoint) Borel set which covers the probabilities of e.g. $\{1,3,5\}$, and $\{1,3,5\}$ is not in $F$). How does this make sense?

I have a question: if $\{f_n\}\subset <!--\; \subset \; -->L^2(X,d\mathrm{\mu <!--\; \mu \; -->})$ with $||f_n|{|}_{L^2}\le <!--\; \le \; -->1$, then $f_n(x)/n\to <!--\; \to \; -->0$ for a.e. x. Here we are considering a measure space $(X,\mathcal{M},\mathrm{\mu <!--\; \mu \; -->})$. Intuitively, I would think we prove this by
$||\frac{f_n}{n}|{|}_{L^2}=({\int <!--\; \int \; -->}_X|\frac{f_n(x)}{n}{|}^{2}d\mathrm{\mu <!--\; \mu \; -->}(x){)}^{\frac{1}{2}}=\frac{1}{|n|}({\int <!--\; \int \; -->}_X|f_n(x){|}^{2}d\mathrm{\mu <!--\; \mu \; -->}(x){)}^{\frac{1}{2}}\le <!--\; \le \; -->\frac{1}{|n|}$
$\Rightarrow <!--\; \Rightarrow \; -->\underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\frac{f_n}{n}=0$
However, this can't be true because the same proof could be used to show that the result holds in the case where we consider the $L^1$ norm, which is false. What am I missing here?

Let $f$ be a (Lebesgue) measurable function defined on ${\mathbb{R}}^n$. Given a vector $x_0$ in ${\mathbb{R}}^n$, I would like to know whether the function $f(x+x_0)$ is measurable or not. I know $\mathrm{\Phi <!--\; \Phi \; -->}\circ <!--\; \circ \; -->g$ is measurable whenever $\mathrm{\Phi <!--\; \Phi \; -->}$ is continuous and $g$ is measurable, and a book warns me of an example of a measurable function $g$ and a continuous function $\mathrm{\Phi <!--\; \Phi \; -->}$ such that $g\circ <!--\; \circ \; -->\mathrm{\Phi <!--\; \Phi \; -->}$ is not measurable. However, I have no idea how to prove or disprove measurability of $f(x+x_0)$. Can someone please give me a hand? Thank you very much.