How can I prove this equivalent relation about measurable random variables?

74 Views Asked by At

How can I prove “ random variable $\alpha$ is $\mathcal{F}\vee \sigma(E_k,k=0,1,…,N)$-measurable” if and only if “$\displaystyle{\alpha=\sum_{k=0}^N} \alpha_k\mathbf{1}_{E_k},\ $ where for any k, $\alpha_k$ is $\mathcal{F}$ measurable”?

I would really appreciate it if there are any tips of this proof.

$\textbf{My guess:}$ It's obvious that we can derive the first statement from the latter one. But for the other direction, I don't have any idea. I guess that maybe we can use the Doob-Dynkin theorem? But I don't know how to use it properly.

I read this statement at equation (2.20) in the following paper: https://arxiv.org/pdf/1606.08204.pdf

1

There are 1 best solutions below

2
On

This is indeed not trivial. After having looked at the context of the paper, it seems that $E_k, 0\leqslant k\leqslant N$ forms a partition. Define $$ \mathcal L=\left\{A:\mbox{there exists }\mathcal F-\mbox{measurable functions }\alpha_k,0\leqslant k\leqslant N\mbox{ such that }\mathbf{1}_A=\sum_{k=0}^N\alpha_k\mathbf{1}_{E_k}\right\}. $$ We can check that $\mathcal L$ is a $\lambda$-system which contains the $\pi$ system of sets of the form $B\cap C$, where $B\in\mathcal F$ and $C\in\sigma(E_k,0\leqslant k\leqslant N)$ hence the $\pi$-$\lambda$ theorem ensures that for each $A\in\mathcal F\vee \sigma(E_k,0\leqslant k\leqslant N)$, there exist $\mathcal F$-measurable functions $\alpha_k,0\leqslant k\leqslant N$ for which $\mathbf{1}_A=\sum_{k=0}^N\alpha_k\mathbf{1}_{E_k}$.

Now approximate pointwise a non-negative $\mathcal F\vee\sigma(E_k,0\leqslant k\leqslant N)$-measurable random variable $\alpha$ by an increasing sequence $(\alpha_n)$ of functions of the form $\alpha_n=\sum_{i=1}^{n}c_{n,i}\mathbf{1}_{A_{n,i}}$, where $A_{n,i}\in \mathcal F\vee\sigma(E_k,0\leqslant k\leqslant N)$. From what we have seen, we can find $\mathcal F$-measurable functions $\alpha_{n,i,k}$ such that $\alpha_n=\sum_{k=0}^N\sum_{i=1}^n\alpha_{n,i,k}\mathbf{1}_{E_k}$. Since $\sum_{i=1}^n\alpha_{n,i,k}$ is increasing in $n$, we can define $\alpha_k:=\lim_{n\to\infty}\sum_{i=1}^n\alpha_{n,i,k}$, which is $\mathcal F$-measurable.