If $E|Y|\lt \infty$, $E[Y|X] = m(X)$ and $(X_1,Y_1)$ come from same distribution as $(X,Y)$. Is it true that $E[Y_1|X_1] = m(X_1)$?

Question

If $X,Y$ are random variables, such that $\mathbb E|Y|\lt \infty$, $\mathbb E[Y|X] = m(X)$ where $m$ is some Borel measurable function and $(X_1,Y_1)$ come from same distribution as $(X,Y)$. Is it true that $\mathbb E[Y_1|X_1] = m(X_1)$?

Attempt I managed to prove $$\mathbb E[Y_1] = \mathbb E[Y] = \mathbb E[\mathbb E[Y|X]] = \mathbb E[m(X)] = \mathbb E[m(X_1)]$$ which leads to $$\mathbb E[\mathbb E[Y_1|X_1]] = \mathbb E[m(X_1)],$$ but I am not sure how to proceed.

Answer 1

It's true. Let $B$ be a Borel subset of the real line. Then $$\int Y \mathbb 1(X\in B)\,dP = \int y \mathbb 1(x \in B)\,\mu(dx,dy) = \int m(x)\mathbb 1(x\in B)\,\mu(dx,dy)$$ , where $\mu$ is the law of $(X,Y)$ and coincidentally $(X_1,Y_1)$. It follows $$\int m(x)\mathbb 1(x\in B)\,\mu(dx,dy) = \int Y_1 \mathbb 1(X_1\in B)\,d\tilde P$$ , since $$\int Y_1 \mathbb 1(X_1\in B)\,d\tilde P = \int y\mathbb 1(x\in B)\,\mu(dx,dy)$$