Probability distribution of functions of random variables? I have a doubt: consider the real valued random variables $X$ and $Z$ both defined on the probability space $(\Omega, \mathcal{F},\mathbb{P})$. 
Let $Y:= g(X,Z)$, where $g(\cdot)$ is a real-valued function. Since $Y$ is a function of random variables it is a random variable. 
Let $x:=X(\omega)$ i.e. a realisation of $X$. 
Is $\mathbb{P}(Y|X=x)=\mathbb{P}(g(X,Z)|X=x)$ equal to $\mathbb{P}(g(x,Z))$?
 A: If $g$ is measurable, then
$$
P(g(X,Z)\in A\mid X=x)=P(g(x,Z)\in A\mid X=x),\quad A\in\mathcal{B}(\mathbb{R})
$$
holds for $P_X$-a.a. $x$. In particular, if $Z$ is independent of $X$, then
$$
P(g(X,Z)\in A\mid X=x)=P(g(x,Z)\in A),\quad A\in\mathcal{B}(\mathbb{R})
$$
holds for $P_X$-a.a. $x$.
This relies on the following general result:

If $U,T$ and $S$ are random variables and $P_S(\cdot \mid T=t)$ denotes a regular conditional probability of $S$ given $T=t$, i.e. $P_S(A \mid T=t)=P(S\in A\mid T=t)$, then
  $$
{\rm E}[U\mid T=t]=\int_\mathbb{R} {\rm E}[U\mid T=t,S=s]\,P_S(\mathrm ds\mid T=t).\tag{*}
$$

Proof: The definition of a regular conditional probability ensures that
$$
{\rm E}[\psi(S,T)]=\int_\mathbb{R}\int_\mathbb{R} \psi(s,t)\,P_S(\mathrm ds\mid T=t)P_T(\mathrm dt)
$$
for measurable and integrable $\psi$. Now let $\psi(s,t)=\mathbf{1}_B(t){\rm E}[U\mid S=s,T=t]$ for some set Borel set $B$. Then
$$
\begin{align}
\int_{T^{-1}(B)} U\,\mathrm dP&={\rm E}[\mathbf{1}_B(T)U]={\rm E}[\mathbf{1}_B(T){\rm E}[U\mid S,T]]={\rm E}[\psi(S,T)]\\
&=\int_{\mathbb{R}}\int_{\mathbb{R}}\psi(s,t)\, P_S(\mathrm ds\mid T=t)P_T(\mathrm dt)\\
&=\int_B\varphi(t)P_T(\mathrm dt)
\end{align}
$$
with 
$$
\varphi(t)=\int_\mathbb{R}{\rm E}[U\mid T=t,S=s]\,P_S(\mathrm ds\mid T=t).
$$
Since $B$ was arbitrary we conclude that $\varphi(t)={\rm E}[U\mid T=t]$.
Now, let $A\in\mathcal{B}(\mathbb{R})$ and use $(*)$ with $U=\psi(X,Z)$, where $\psi(x,z)=\mathbf{1}_{g^{-1}(A)}(x,z)$ and $S=Z$, $T=X$. Then we note that 
$$
{\rm E}[U\mid X=x,Z=z]={\rm E}[\psi(X,Y)\mid X=x,Z=z]=\psi(x,z)
$$
by definition of conditional expectation and hence by $(*)$ we have
$$
\begin{align}
P(g(X,Z)\in A\mid X=x)&={\rm E}[U\mid X=x]=\int_\mathbb{R} \psi(x,z)\,P_Z(\mathrm dz\mid X=x)\\
&=P(g(x,Z)\in A\mid X=x).
\end{align}
$$
