Does $X=^d Y$ imply $\mathbb{E}[X1_A]=\mathbb{E}[Y1_A]$ for all measurable $A$?

Question

A brief question just to be sure: If I assume that two real-valued random variables $X,Y$ have the same distribution, is it true that for any measurable set $A$, $$ \mathbb{E}[X1_A]=\mathbb{E}[Y1_A] $$ ?

Answer 1

As pointed out by Kavi Rama Murthy's counterexample, what you state is not true. In fact, the only way for two random variables to satisfy this condition is if they are equal almost everywhere.
In other words, you have the following

Lemma : For any two real-valued random variables $X$ and $Y$ on $(\Omega,\mathcal F,\mathbb P)$, $$\mathbb E[X\mathbf1_A] =\mathbb E[Y\mathbf1_A]\ \ \forall A\in\mathcal F\iff X=Y \ \ \ \mathbb P-\text{a.e.} $$

The proof goes something like this :
$\bullet \ \Longleftarrow\, $ This is the easy direction : if $X-Y = 0$ almost everywhere, then by the definition of Lebesgue integral, on any set $A\in\mathcal F$,$$\ \mathbb E[(X-Y)\mathbf 1_A] = \int_A(X-Y)d\mathbb P = \sup_{s\ \text{simple}}\left\{\int_Asd\mathbb P\ ,\ 0\le s\le(X-Y)\right\} = \sup\{0\}=0$$ And we then conclude by linearity of expectation.

$\bullet \ \Longrightarrow\, $ For the other direction, we can do it by contradiction : assume that the set $A^+:=\{X-Y>0\}$ is such that $\mathbb P(A^+)>0$. Then it follows that $$\mathbb E[(X-Y)\mathbf 1_{A^+}]> 0 \implies \mathbb E[X\mathbf 1_{A^+}] - \mathbb E[Y\mathbf 1_{A^+}] > 0 \implies \mathbb E[X\mathbf 1_{A^+}]> \mathbb E[Y\mathbf 1_{A+}] $$ This contradicts the assumption that $\mathbb E[X\mathbf1_A] =\mathbb E[Y\mathbf1_A]\ \ \forall A\in\mathcal F$, we thus conclude that $\mathbb P(A^+) = 0$. By the same argument we have that if we let $A^- :=\{X-Y<0\}$, then $\mathbb P(A^-) = 0$.

Therefore, if we let $A :=\{X-Y\ne0\} = A^+\cup A^-$, we have that $\mathbb P(A) = \mathbb P(A^+) + \mathbb P(A^-) = 0.\ \ \square$