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Prof. Leif Döring, Felix Benning

^†^†course: Wahrscheinlichkeitstheorie 1^†^†semester: FSS 2022^†^†tutorialDate: 21.02.2022^†^†dueDate: 10:15 in the exercise on Monday 21.02.2022

For any set $E$ , we denote by $\mathcal{P}(E)$ the powerset of $E$ , i.e. the set of all subsets of $E$ .

Exercise 1.

Consider two measurable spaces $(\Omega,\mathcal{F})$ and $(E,\mathcal{E})$ . Suppose that $\mathcal{E}=\sigma(\mathcal{A})$ with some $\mathcal{A}\subset\mathcal{P}(E)$ . Show that a mapping $f\colon(\Omega,\mathcal{F})\to(E,\mathcal{E})$ is measurable, if

f^{-1}(A)\in\mathcal{F}\quad\text{for every}~{}A\in\mathcal{A}.

Exercise 2.

Let $E\subset\Omega$ . Show that $\mathcal{F}_{E}:=\{A\cap E\colon A\in\mathcal{F}\}$ is a sigma-algebra. If $\mathcal{F}=\sigma(\mathcal{E})$ , that is $\mathcal{F}$ is generated by for $\mathcal{E}$ , where $\mathcal{E}$ is a a collection of subsets of $\Omega$ . Then prove the identity $\mathcal{F}_{E}=\sigma(\{A\cap E\colon A\in\mathcal{E}\})$ .

Exercise 3 (Factorization lemma).

Let $Y\colon(\Omega,\mathcal{F})\to(E,\mathcal{E})$ be measurable. Show that, for every random variable $X\colon(\Omega,\sigma(Y))\to(\bar{\mathbb{R}}:=[-\infty,+\infty],\mathcal{B}(% \bar{\mathbb{R}}))$ , there exists a measurable function $g\colon(E,\mathcal{E})\to(\bar{\mathbb{R}},\mathcal{B}(\bar{\mathbb{R}}))$ , such that $X=g(Y)$ .

Solution.

Reference of this exercise: Corollary 1.97 in Klenke.

We start with the case that $X$ is a $\sigma(Y)$ simple function: that is $X=\sum_{i=1}^{k}\lambda_{k}\mathbf{1}_{A_{k}}$ , where $\lambda_{k}\geq 0$ and $A_{k}\in\sigma(Y)$ . By the definition of the sigma-algebra $\sigma(Y)$ , the fact that $A_{k}\in\sigma(Y)$ implies that there exists $B_{k}\in\mathcal{E}$ such that $Y^{-1}(B_{k})=A_{k}$ . Define $g:=\sum_{i=1}^{k}\lambda_{k}\mathbbm{1}_{B_{k}}$ , which is $(E,\mathcal{E})\to(\bar{\mathbb{R}},\mathcal{B}(\bar{\mathbb{R}}))$ -measurable. Then we have the identity, for every $\omega\in\Omega$ :

X(\omega)=\sum_{i=1}^{k}\lambda_{k}\mathbbm{1}_{A_{k}}(\omega)=\sum_{i=1}^{k}% \lambda_{k}\mathbbm{1}_{Y^{-1}(B_{k})}(\omega)=\sum_{i=1}^{k}\lambda_{k}% \mathbbm{1}_{B_{k}}(Y(\omega))=g(Y(\omega)).

This is to say $X=g(Y)$ .

Now consider a non-negative $\sigma(Y)$ -measurable function $X$ . Then there exists a sequence of simple function $(X_{n},n\in\mathbb{N})$ , such that $X_{n}\leq X_{n+1}$ for each $n\in\mathbb{N}$ and $\lim_{n\to\infty}X_{n}=X$ . Applying the statement in the previous step, we have, for each $n\in\mathbb{N}$ , a $(E,\mathcal{E})\to(\bar{\mathbb{R}},\mathcal{B}(\bar{\mathbb{R}}))$ -measurable function $g_{n}$ , such that $X_{n}=g_{n}(Y)$ . We define a function $g:(E,\mathcal{E})\to(\bar{\mathbb{R}},\mathcal{B}(\bar{\mathbb{R}}))$ by

g(y)=\begin{cases}\lim_{n\to\infty}g_{n}(y),&\text{if exists or is }+\infty\\ 0,&\text{otherwises}.\end{cases}

Then $g$ is a $(E,\mathcal{E})\to(\bar{\mathbb{R}},\mathcal{B}(\bar{\mathbb{R}}))$ -measurable function. Moreover, we have

X(\omega)=\lim_{n\to\infty}X_{n}(\omega)=\lim_{n\to\infty}g_{n}(Y(\omega))=g(Y% (\omega)),\quad\omega\in\Omega.

So we identify $X=g(Y)$ .

Finally, we conclude for every $\sigma(Y)$ -measurable function $X$ by using the decomposition $X=X^{+}-X^{-}$ . ∎