Proof using properties of Conditional Expectation with random vectors. Prove that $E(g(Y)|X)=E(g(Z)|X)$

Question

I have this problem with my probability homework.

I have to prove:

Let $(X,Y)$ and $(X,Z)$ be random vectors that have the same joint distribution and $g$ a measurable Borel function that verifies $E|g(X)| \leq \infty.$ Then:

$E(g(Y)|X)=E(g(Z)|X)$

I have to prove for indicator functions, simple functions and positive functions.

For indicator functions:

I took: $1_A=g(Y)$, $1_B=g(Z)$, with $A$ and $B$ $\in F$ with $F$ the $\sigma$ algebra.

I have that:

$E(g(Y)|X)=E(1_A|X)=P(A|X)$

$E(g(Z)|X)=E(1_B|X)=P(B|X)$

I do not know how to continue the proof.

Answer 1

For any bounded measurable function $h$

$$\mathbb{E}[g(Y)h(X)]=\mathbb{E}[g(Z)h(X)]$$

since $(Y,X)$ and $(Z,X)$ have the same (joint) distribution.

On the other hand

$$\begin{align} \mathbb{E}[g(Y)h(X)]&=\mathbb{E}\big[\mathbb{E}[g(Y)|X]h(X)\big]\\ \mathbb{E}[g(Z)h(X)]&=\mathbb{E}\big[\mathbb{E}[g(Z)|X]h(X)\big] \end{align} $$

As these identities hold for any bounded measurable $h$, we obtain by definition of conditional expectation that

$$ \mathbb{E}[g(Y)|X]=\mathbb{E}[g(Z)|X]$$