Question about conditional independence Assumptions

*

*Let $X,Y,Z$ be random variables such that $Z$ is independent of
$(X,Y)$.


*Let $h$ be a bivariate function such that $h(X,Z)$ is a random
variable.
Question
Is $h(X,Z)$ conditionally independent of $Y$ given $X$?
Attempt
Let $A$ and $B$ be Borel sets. Let $x$ be a real number. Then $$P(h(X,Z)\in A,Y\in B\mid X=x)=P(h(x,Z)\in A,Y\in B\mid X=x).$$ It remains to show that $$P(h(x,Z)\in A,Y\in B\mid X=x)=P(h(x,Z)\in A\mid X=x)P(Y\in B\mid X=x).$$ I guess this equality follows from the  assumption that $Z$ is independent of $(X,Y)$. My reasoning: The assumption implies that $Z$ is independent of $Y$. This implies that $h(x,Z)$ and $Y$ are independent as $h(x,Z)$ is a function of $Z$ only. However, I am wondering if $h(x,Z)$ and $Y$ are also independent when we condition on the event $\{X=x\}$; and if so, why that is the case.
Maybe this argument works. $$\begin{align}P(h(x,Z)\in A,Y\in B\mid X=x)&=\frac{P(h(x,Z)\in A,Y\in B,X=x)}{P(X=x)}\\
&=\frac{P(h(x,Z)\in A)P(Y\in B,X=x)}{P(X=x)}\\
&=P(h(x,Z)\in A)\frac{P(Y\in B,X=x)}{P(X=x)}\\
&=P(h(x,Z)\in A\mid X)\frac{P(Y\in B,X=x)}{P(X=x)}\\
&=P(h(x,Z)\in A\mid X)P(Y\in B\mid X=x).\end{align}$$ This assumes that $P(X=x)>0$. What if $P(X=x)=0$? I guess I can make a similar derivation by looking at densities. Then my derivation would be, with integration over $\mathbb{R}^2$, $$\begin{align}P(h(x,Z)\in A,Y\in B\mid X=x)&=\int1_{A}(h(x,z))1_B(y)f(z,y|x)dzdy\\
&=\cdots\\
&=\int1_{A}(h(x,z))1_B(y)f(z|x)f(y|x)dzdy\\
&=\int1_{A}(h(x,z))f(z|x)dz\int1_B(y)f(y|x)dy\\
&=P(h(x,Z)\in A\mid X)P(Y\in B\mid X=x).\end{align}$$ How can I provide a more general proof? Conditional distributions was what I found hard to understand when I read courses in measure-theoretic probability.

A general proof can be obtained by the tower property of expectations. That is,
\begin{align*}
P(h(x,Z)\in A,Y\in B\mid X)&=\mathrm{E}[\mathrm{E}[\mathbf{1}_{h(x,Z)\in A}\mathbf{1}_{Y\in B}\mid Y,X]\mid X]\\
&=\mathrm{E}[\mathbf{1}_{Y\in B}\mathrm{E}[\mathbf{1}_{h(x,Z)\in A}]\mid X]\\
&=P(h(x,Z)\in A)P(Y\in B\mid X).
\end{align*}
Now, the connection between $\mathrm{E}[\,\cdot\mid X]$ and $\mathrm{E}[\,\cdot\mid X=x]$ allows us to conclude the above also holds when conditioning on $X=x$.
