This is exercise 4.1.3 from Durrett $3^{rd}$, stating as follows:
Let $(\Omega,\mathcal{F}_{0},\mathbb{P})$ be a probability space with $\mathcal{F}\subset\mathcal{F}_{0}$. Suppose $X\in\mathcal{F}_{0}$ with $X\geq 0$ but $\mathbb{E}X=\infty$. (There is nothing to prove when $\mathbb{E}X<\infty).$ Show that there is a unique $\mathcal{F}-$measurable $Y$ with $0\leq Y\leq\infty$, so that $$\int_{A}Xd\mathbb{P}=\int_{A}Yd\mathbb{P}\ \text{for all}\ A\in\mathcal{F}.$$
I have some attempt but I could not prove it.
Below is my attempt:
As we always did for infinity mean random variable, we write $X_{M}:=\min(X,M)$ and $Y_{M}:=\mathbb{E}(X_{M}|\mathcal{F})$.
Note that we can define such $Y_{M}$ since $\mathbb{E}|X_{M}|<\infty$, and thus everything follows from the existence and uniqueness of conditional expectation. In particular, $Y_{M}$ is unique and $\mathcal{F}-$measurable for each $M$.
Let $A\in\mathcal{F}$, then by definition of conditional expectation, for each $M$, we have $$\int_{A}X_{M}d\mathbb{P}=\int_{A}Y_{M}d\mathbb{P}.$$
Then since $X_{M}\nearrow X$, and each $X_{M}\geq 0$, it follows from the monotone convergence theorem that as $M\longrightarrow\infty$, $$\int_{A}X_{M}d\mathbb{P}\longrightarrow\int_{A}Xd\mathbb{P}.$$
However, I do not know how to deal with RHS. The problem here is that $\mathbb{E}X=\infty$, so we cannot use the continuity of conditional expectation, i.e.
If $X_{n}\geq 0$, $X_{n}\nearrow X$ and $\mathbb{E}X<\infty$, then $\mathbb{E}(X_{n}|\mathcal{F})\nearrow\mathbb{E}(X|\mathcal{F})$;
What should I do? Thank you!
Edit 1:
I edit this post to add a complete proof of this exercise, which follows the suggestion "kccu" provided.
The key point here is that we don't need the continuity of conditional expectation, and in fact this exercise does not really imply that $Y$ is a conditional expectation of $X$ with respect to $\mathcal{F}$, since $\mathbb{E}X=\infty$ which directly put you away from the definition of conditional expectation (which requires $X$ to be integrable).
This exercise is not about showing which one is a conditional expectation of the other with respect to a specific $\sigma-$algebra, but is using conditional expectation as an "intermediate agent" to prove something else.
Proof:
Write $X_{M}:=\min(X,M)$ and $Y_{M}:=\mathbb{E}(X_{M}|\mathcal{F})$.
Note that we can define such $Y_{M}$ since $\mathbb{E}|X_{M}|<\infty$, and thus everything follows from the existence and uniqueness of conditional expectation. In particular, $Y_{M}$ is unique and $\mathcal{F}-$measurable for each $M$.
Let $A\in\mathcal{F}$, then by definition, for each $M$, we have $$\int_{A}X_{M}d\mathbb{P}=\int_{A}Y_{M}d\mathbb{P}.$$
Note that each $X_{M}\geq 0$ and $X_{M}\nearrow X$, and thus it follows from monotone convergence theorem that, as $M\longrightarrow\infty$, we have $$\int_{A}X_{M}d\mathbb{P}\longrightarrow\int_{A}Xd\mathbb{P}.$$
On the other hand, since $X_{M}\nearrow X$, $Y_{M}$ is increasing and $Y_{m}\nearrow Y_{\infty}$, for some $Y_{\infty}$ (it can be infinite). Also, since $X_{M}\geq 0$, then for each $M$, we have $$Y_{M}:=\mathbb{E}(X_{M}|\mathcal{F})\geq \mathbb{E}(0|\mathcal{F})=0.$$
Thus, it follows from the monotone convergence theorem that as $M\longrightarrow\infty$, we have $$\int_{A}Y_{M}d\mathbb{P}\longrightarrow\int_{A}Y_{\infty}d\mathbb{P}.$$
Therefore, for all $A\in\mathcal{F}$, we have $$\int_{A}Xd\mathbb{P}=\int_{A}Y_{\infty}d\mathbb{P},$$ in particular, since $Y_{M}$ is unique and $\mathcal{F}-$measurable for each $M$, $Y_{\infty}$ is unique and $\mathcal{F}-$measurable.
Therefore, $Y_{\infty}$ is the desired $Y$.
Best Answer
Since $X_M$ is increasing, so is $\mathbb{E}(X_M \mid \mathcal{F})$. Define $Y=\lim_{M \to \infty} \mathbb{E}(X_M \mid \mathcal{F})$, which exists everywhere but may be infinity. You need to show $Y=\mathbb{E}(X \mid \mathcal{F})$, i.e., for every $A \in \mathcal{F}$, $$\int_A Y \ d\mathbb{P} = \int_A X \ d\mathbb{P}$$ or equivalently, $$\int_A \lim_{M \to \infty} \mathbb{E}(X_M \mid \mathcal{F}) \ d\mathbb{P} = \int_AX \ d\mathbb{P}.$$ Do you see how to proceed?