Probability measure induced by a measurable function of a random variable

Question

Suppose there are two random variables X and Z: $(\Omega, \Sigma, \mathbb{P}) \rightarrow (\mathbb{R}, \mathbb{B}) $, a measurable function $g: \mathbb{R} \times \mathbb{R}\rightarrow \mathbb{R}$ and assume $\mathbb{P}^{(X,Z)} =\mathbb{P}^{(Z,X)} $.

I read in some proof, that it follows that $\mathbb{P}^{g(X,Z)} =\mathbb{P}^{g(Z,X)}$ but I don't understand why.

Thanks!

Answer 1

To say that $\mathbb{P}^{(X,Z)} =\mathbb{P}^{(Z,X)}$ means that for every Borel set $A\subseteq\mathbb{R}^2$, $$\mathbb{P}((X,Z)\in A)=\mathbb{P}((Z,X)\in A).$$ To say that $\mathbb{P}^{g(X,Z)} =\mathbb{P}^{g(Z,X)}$ means that for every Borel set $B\subseteq\mathbb{R}$, $$\mathbb{P}(g(X,Z)\in B)=\mathbb{P}(g(Z,X)\in B).$$ To prove the second statement from the first, just observe that $g(X,Z)\in B$ iff $(X,Z)\in g^{-1}(B)$ and similarly for $(Z,X)$, so you can just set $A=g^{-1}(B)$.