Why is $E[X1_A]=0$ if $P(A)=0$?

Question

I know this is trivial and intuitive, but I'm not able to convince myself rigorously. If $P(A)=0$, why is it true that $E(X1_A)=0$?

Every book discards it out as an obvious fact. I tried to prove it using the simple functions definition of Lebesgue integration.

Answer 1

Proposition: Let $(\Omega,\mathcal{F},P)$ be a measure space, $f:\Omega\to [0,\infty)$ a measurable function and $A$ a measurable set with $P(A)=0$. Then, $$\int_AfdP=0.$$

Proof: First, suppose $s=\sum_{i=1}^n a_i\chi_{A_i}$ is a simple measurable function (i.e. the range of $s$ is a finite set $\{a_1,\ldots,a_n\}$, and $A_i=s^{-1}(a_i)\in\mathcal{F}$). Then, by definition $$\int_AsdP=\sum_{i=1}^na_iP(A_i\cap A)\leq\sum_{i=1}^na_iP(A)=0.$$ Now, it is a theorem that any measurable function $f:\Omega\to[0,\infty)$ is the pointwise limit of a non-decreasing sequence $s_n$ of simple measurable function. Hence, by Lebesgue monotone convergence theorem, $$\int_A fdP=\lim_{n\to\infty}\int_As_ndP=\lim_{n\to\infty}0=0.$$

Answer 2

Let's prove it for a nonnegative $X$.

A stepfunction $0\leq f\leq X1_A$ can be written as $c_1\times 1_{B_1}+\cdots+c_n\times 1_{B_n}$ with $c_i>0$. This leads to $B_i\subseteq A$ so that $0\leq P(B_i)\leq P(A)=0$.

So $\int fdP=c_1\times P(B_1)+\cdots+c_n\times P(B_n)=0$.

Then $\mathbb EX1_A=0$ as supremum of these values.