Let $X\ge 0$ be a random variable with $\Bbb E[X] = \infty$. I have to show that there exists a unique $\mathcal F$-measurable random variable $0\le Y \le \infty$, so that $$\int_A X \mathrm dP = \int_A Y \mathrm dP,\:\: \forall A \in \mathcal F$$ $\mathcal F$ isn't further specified.
I don't really know where to start for showing this. Does anyone have any hints or tips on how to show this? Thanks in advance!
I don't have time, so I can only give a sketch of proof.
Existence: For each $n\in\mathbb{N}$, define $X_{n}=\min(X,n)$. Let $Y_{n}=E[X_{n}\mid\mathcal{F}]$. Let $Y=\limsup_{n}Y_{n}$.
Uniqueness: Suppose that $Y_{1}$ and $Y_{2}$ have that property. Let $A=[Y_{1}<Y_{2}<\infty]$. We assert that $P(A)=0$. Define $A_{n}=[\frac{1}{n}\leq Y_{2}-Y_{1}\leq n]$. Note that $A=\cup_{n}A_{n}$. Therefore, it suffices to show that $P(A_{n})=0$. Define $B=[Y_1<Y_2=\infty]$. By considering $B_n=[Y_1\leq n]\cap[Y_=\infty]$ and note that $B=\cup_n B_n$, we can show that $P(B)=0$. Therefore $P([Y_1<Y_2])=0.$