Almost everywhere equality of r.v.'s , based on information on mean values.

Question

Let $X, Y$ be two random variables on a probability space $(\Omega, \mathcal{F},P)$, where $\mathcal{F}=\sigma(\mathcal{E})$. We assume that:

1. $\mathcal{E}$ is closed on intersections, i.e. $A\cap B\in \mathcal{E}$ for $A,B \in \mathcal{E}$,
2. $E(|X|)<+\infty$ and $E(X)=E(Y)$,
3. $E(X1_A)=E(Y1_A)$ for all $A\in \mathcal{E},$ where $1_A$ denotes the characteristic function of the set $A$.

Prove that $X=Y$ almost everywhere (with respect to the probability measure $P$).

Thanks a lot in advance for the help.

Answer 1

1. Use $\pi-\lambda$ theorem, where $\mathcal{E}$ is a $\pi$-system to show that $\mathbb{E}(X1_A)=\mathbb{E}(Y1_A)$ for all $A\in \mathcal{F}$.

2. Use the fact that if $X,Y\in L^1$, then $X=Y$ a.e. iff (1) holds.