Show that $\sum_{n=0}^{\infty}E[X \mathbf{1}_{A_n}] = E[X\mathbf{1}_A]$

Question

Suppose $E[|X|]< \infty$ and that $A_n$ are disjoint sets with $\cup_{n=1}^\infty A_n = A$. Show that $\sum_{n=0}^{\infty}E[X \mathbf{1}_{A_n}] = E[X\mathbf{1}_A]$

Here is what I try with Monotone Convergence Theorem:

\begin{align*} \sum_{n=0}^{\infty}E[X \mathbf{1}_{A_n}] &= \lim_{m \rightarrow \infty}\sum_{n=0}^{m}E[X \mathbf{1}_{A_n}]\\ &= \lim_{m \rightarrow \infty}E[X \sum_{n=0}^{m}\mathbf{1}_{A_n}]\\ &= \lim_{m \rightarrow \infty}E[X \mathbf{1}_{F_n}]\\ &= E[X \lim_{m \rightarrow \infty}\mathbf{1}_{F_n}]\\ &= E[X \mathbf{1}_{A}]\\ \end{align*} where $F_n = \cup_{i=1}^nA_i$, so $\cup_{n=1}^m A_n = \cup_{n=1}^m F_n$, and $\mathbf{1}_{F_n}$ is increasing.

I am not quite sure if I get it right. My main concern is the first step: $\sum_{n=0}^{\infty}E[X \mathbf{1}_{A_n}] = \lim_{m \rightarrow \infty}\sum_{n=0}^{m}E[X \mathbf{1}_{A_n}]$. Do we always have $\sum_{n=0}^\infty = \lim_{m \rightarrow \infty} \sum_{n=0}^m$?

I may also have made other mistakes. Please feel free to point out. Any other hints are also welcome.

Thanks all in advance.

Answer 1

We use DCT, not MCT: $$\begin{aligned} \sum_{n \in \mathbb{N}}E[\mathbf{1}_{A_n}X]&=\lim_{n\to \infty}E\bigg[\bigg(\sum_{k\leq n}\mathbf{1}_{A_k}\bigg)X\bigg]=\\ &=\lim_{n\to \infty}E[\mathbf{1}_{\cup_{k\leq n}A_k}X]=\\ &\stackrel{\textrm{DCT}}{=}E[\mathbf{1}_{A}X] \end{aligned}$$ This is because $\mathbf{1}_{\cup_{k\leq n}A_k}|X|\leq |X|\in \mathcal{L}^1,\,\forall n$ and $\mathbf{1}_{\cup_{k\leq n}A_k}\to \mathbf{1}_A$ pointwise.

Answer 2

As explained in comments and Snoop's answer, the easiest way is to use DCT.

Back to your question, the first step (your main concern) is OK, but where it fails is here:

$$\lim_{m \to \infty}E[X \mathbf{1}_{F_n}] = E[X \lim_{m \rightarrow \infty}\mathbf{1}_{F_n}]$$

The problem is that $X$ might be both positive and negative, thus $\{X\mathbf{1}_{F_n}\}$ is not monotone.

To use MCT, you need to work around it, and decompose $X$ in its positive and negative parts ($X=X_{+}-X_{-}$, with $X_{+},X_{-}\ge 0$), and apply MCT on $X_{+}$ and $X_{-}$ separately.