Prove $\mathbb{E}(X|Y,Z) = E(X|Y)$ a.s. given that $Z$ is independent of $Y$ and of $X$.

I am stuck trying to prove the following statement involving conditional expectation of random variables:

Suppose $$Z$$ is independent of $$Y$$ and of $$X$$, and $$E|X|<\infty$$, then $$\mathbb{E}(X|Y,Z) = E(X|Y)$$ a.s..

Here's my attempt: by definition of conditional expectation, it suffices to show the following two conditions:

1. $$\int_A \mathbb{E}(X|Y,Z) dP=\int_A XdP$$ for all $$A\in \sigma(Y)$$.
2. $$\mathbb{E}(X|Y,Z)$$ is $$\sigma(Y)$$ measurable.

The first condition is easy, as for every $$A\in \sigma(Y)$$, $$A\in \sigma(Y,Z)$$. Thus by definition it automatically follows that $$\int_A \mathbb{E}(X|Y,Z) dP=\int_A XdP$$.

However, for the second condition, I have little clue as to what to do. In particular, I haven't figure out a way to utilize the independence assumption to prove measurability. I would appreciate thoughts on whether this is the right approach.

This is false. Let $$A,B,C$$ be pair-wise independent but not jointly independent. Take $$X=I_A,Y=I_B$$ and$$Z=I_C$$. Then the the integral of LHS over $$Y^{-1}(\{1\}) \cap Z^{-1}(\{1\})$$ is $$P(A\cap B \cap C)$$ whereas RHS is $$P(A)$$ so the integral of RHS over $$Y^{-1}(\{1\}) \cap Z^{-1}(\{1\})$$ is $$P(A)P(B \cap C)=P(A)P(B)P(C)$$.