Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $\mathcal F$ be a $\sigma$-algebra on $\Omega$ with $\mathcal
F\subseteq\mathcal A$ - $X_n,X$ be non-negative random variables on $(\Omega,\mathcal A,\operatorname P)$
The monotone convergence theorem for the conditional expectation states, that if $X_n\uparrow X$ almost surely, then $$\operatorname E\left[X_n\mid\mathcal F\right]\stackrel{n\to\infty}\to\operatorname E\left[X\mid\mathcal F\right]\;.$$
Clearly, by the monotonicity of the conditional expectation, $$Z:=\lim_{n\to\infty}\operatorname E\left[X_n\mid\mathcal F\right]$$ exists. Since each $\operatorname E\left[X_n\mid\mathcal F\right]$ is $\mathcal F$-measurable by definition, $Z$ is $\mathcal F$-measurable, too.
Why is it not that clear, that $$\operatorname E\left[X_n\mid\mathcal F\right]\uparrow Z\;?\tag{1}$$ All proofs I've read so far only state, that there is a modification (version) of $Z$ with $(1)$.
Clearly, the conditional expectation is only almost everywhere uniquely determined. So, the monotonicity only yields $$\operatorname E\left[X_n\mid\mathcal F\right]\le \operatorname E\left[X_{n+1}\mid\mathcal F\right]$$ on $\Omega\setminus N_n$ for some $\operatorname P$-null set $N_n\subseteq\Omega$. However, since $$N:=\bigcup_nN_n$$ is a $\operatorname P$-null set, too, we should be able to immediately conclude, that $(1)$ holds on $\Omega\setminus N$, i.e. almost everywhere.
So, what am I missing?
Best Answer
As pointed out by John Dawkins, we have to prove that $Z=\mathbb E\left[X\mid\mathcal F\right]$, which is not clear a priori. This relies on $$\mathbb E\left[Z\mathbf 1_{F}\right]= \lim_{n\to +\infty}\mathbb E\left[\mathbb E\left[X_n\mid\mathcal F\right]\mathbf 1_{F}\right]=\lim_{n\to +\infty}\mathbb E\left[X_n\mathbf 1_{F}\right]=\mathbb E\left[X\mathbf 1_{F}\right]$$ for $F\in\mathcal F$. The first and third equal sign are due to the monotone convergence, the second by definition of conditional expectation.