Let $(\Omega,\mathcal F,P)$ be a probability space, and let $\mathcal L^0$ denote the space of all measurable real random variables $X$ on $(\Omega,\mathcal F,P)$. For $X\in \mathcal L^0$, let $[X]:=\{Y\in \mathcal L^0 : X=Y, \, P\text{-almost surely}\}$, and let $L^0:=\{[X]:X\in \mathcal L^0\}$. Define addition and multiplication on $ L^0$ as follows:
$$[X]+[Y]:=[X+Y]$$
$$[X][Y]:=[XY]$$
Now suppose we are given an increasing sequence of sets $(A_n)\subset \mathcal F$ with $A_n\uparrow A$, and a sequence $([X_n])\subset L^0$ satisfying
$[1_{A_n^c}X_n]=[0]$ for all $n$.
$[1_{A_n}X_n]=[1_{A_n}X_m]$ whenever $m\geq n$.
We say that $([X_n])$ covers $A$ through $(A_n)$.
Am trying to show the following:
Proposition. Suppose $([X_n])$ covers $A$ through $(A_n)$. Then there exits a unique $[Y]\in L^0$ satisfying $[1_{A^c}Y]=[0]$ and $[1_{A_n}Y]=[X_n]$ for all $n$. We write $[Y]:=\lim_{n\to \infty}[X_n]$.
My attempt:
First we show existence. Choose $X_n\in [X_n]$ for each $n$. From 1. and 2. we have
$1_{A_n^c}X_n=0$ $P$-almost surely for all $n$.
$1_{A_n}X_n=1_{A_n}X_m$ $P$-almost surely whenever $m\geq n$.
Since only countably many conditions are involved, there exists a big null set $N\in \mathcal F$ outside of which the following holds
$1_{A_n^c}X_n=0$ for all $n$.
$1_{A_n}X_n=1_{A_n}X_m$ whenever $m\geq n$.
Define
$$Y(\omega):=\begin{cases} \lim_{n\to \infty} X_n(\omega)& \text{if } \omega\in \cup_{n=1}^\infty A_n\setminus N \\ 0 & \text{otherwise} \end{cases}$$
This is well-defined, for if $ \omega\in A_n\setminus N$ for some $n$, then by 2. we have $X_n(\omega)=X_m(\omega)$ whenever $m\geq n$, so that $(X_n(\omega))$ is eventually constant. Moreover $Y\in \mathcal L^0$, because pointwise limits of measurable maps are measurable. On $N^c$ we have that $1_{A^c}Y=0$ and $1_{A_n}Y=X_n$ for each $n$, so $[Y]\in L^0$ is the desired element.
Second we show uniqueness. Suppose $[Y']\in L^0$ satisfies $[1_{A^c}Y']=[0]$ and $[1_{A_n}Y']=[X_n]$ for all $n$. Then
- $1_{A^c}Y'=0$ $P$-almost surely.
- $1_{A_n}Y'=X_n$ $P$-almost surely for each $n$.
As before there exists a null set $N'\in\mathcal F$ outside of which the following holds
- $1_{A^c}Y'=0$.
- $1_{A_n}Y'=X_n$ for each $n$.
But then we see that $Y'=Y$ outside of $N'\cup N$. Therefore $[Y']=[Y]$.
Proposition. Suppose $([X_n])$ covers $A$ through $(A_n)$, and $([Y_n])$ covers $B$ through $(B_n)$. Suppose also that
$P[A\triangle B]=0$
$[1_{A_n\cap B_n}X_n]=[1_{A_n \cap B_n}Y_n]$ for all $n$.
Then $\lim_{n\to \infty}[X_n]=\lim_{n\to \infty}[Y_n]$.
My attempt:
From the previous proposition we have $\lim_{n\to \infty}[X_n]=[X]$ and $\lim_{n\to \infty}[Y_n]=[Y]$ where
$$X(\omega):=\begin{cases} \lim_{n\to \infty} X_n(\omega)& \text{if } \omega\in \cup_{n=1}^\infty A_n\setminus N_1 \\ 0 & \text{otherwise} \end{cases}$$
$$Y(\omega):=\begin{cases} \lim_{n\to \infty} Y_n(\omega)& \text{if } \omega\in \cup_{n=1}^\infty B_n\setminus N_2 \\ 0 & \text{otherwise} \end{cases}$$
and $N_1,N_2$ are two null sets. Moreover from 1. and 2. there exists a null set $N_3$ outside of which the following holds:
$1_A=1_B$
$1_{A_n\cap B_n}X_n=1_{A_n \cap B_n}Y_n$ for all $n$.
Therefore we see that $X=Y$ outside of $N_1\cup N_2\cup N_3$. This means that $[X]=[Y]$.
Is this correct? Thanks for your help.