Proving that some truncations have the same expected value (using Lebesgue integral)

Question

Let $X_1,X_2\colon(\Omega,\mathcal F,\mathbb P)\to (\mathbb R,\mathcal B(\mathbb R))$ be two random real variables, where $(\Omega,\mathcal F,\mathbb P)$ denotes a probability space. And $(\mathbb R,\mathcal B(\mathbb R))$ denotes the real number and the borel sets.

If $X_1, X_2$ have the same distribution, i.e for any $B\in\mathcal B(\mathbb R)$ we have:

$\mathbb P(X_1\in B)=\mathbb P(X_2 \in B)$.

Prove that $\mathbb EX_11_{|X_1|\le n}=\mathbb EX_21_{|X_2|\le n}$.

Here $\mathbb EX=\int_{\Omega} X \mathrm d\mathbb P$ denotes the Lebesgue integral with respect to the measure P and $1_A$ denotes the characteristic function of the set $A$.

Please I need to prove this to prove a result about a weak law of large numbers.

Answer 1

Some useful facts:

1. If $U$ and $V$ are integrable random variables with the same distribution, they have the same expectation.

2. If $U$ and $V$ have the same distribution and $\phi\colon \mathbb R\to\mathbb R$ is Borel-measurable, then $\phi(U)$ and $\phi(V)$ have the same distribution.

3. The map $\phi(x):=x\chi_{\{|x|\leqslant n\}}$ is Borel-measurable.