Equality of two conditional expectations

Question

I would like to show that for any random variable $X$ and $Z$ such that $X$ and $Z$ are independent and for any measurable functions $f$ and $g$,

$$ \mathbb E \left[ f(g(X),Z) | g(X) \right] = \mathbb E \left[ f(g(X),Z) | X \right] $$

Answer 1

It follows, almost surely, from two observations (assuming integrability of $f(g(X),Z)$):

(i) $$\sigma\left(g\left(X\right)\right)\subset\sigma\left(X\right)\,,$$

therefore $\,\,\mathbb E \left[ f(g(X),Z) | g(X) \right]\,\,$ is $\,\,\sigma\left(X\right)$-measurable.

(ii) For $A\in\sigma\left(X\right)$, $$ \begin{eqnarray*} \mathbb{E}\bigg[\,{\bf{1}}_A\,\mathbb{E}\left[ f(g(X),Z) | g(X) \right]\,\bigg]&{}={}&\mathbb{E}\bigg[\,{\bf{1}}_A\,\mathbb{E}\bigg[\,\, \mathbb{E}_Z\left[f(g(X),Z)\right]\,\, \bigg|\, g(X) \bigg]\,\bigg]\newline &{}={}&\mathbb{E}\bigg[\,{\bf{1}}_A\,\,\mathbb{E}_Z\left[f(g(X),Z)\right]\bigg]\newline &{}={}&\mathbb{E}\bigg[\,{\bf{1}}_A\,\mathbb{E}\bigg[\,\, \mathbb{E}_Z\left[f(g(X),Z)\right]\,\, \bigg|\, X \bigg]\,\bigg]\newline &{}={}&\mathbb{E}\bigg[\,{\bf{1}}_A\,\mathbb{E}\bigg[\,\,f(g(X),Z)\,\, \bigg|\, X \bigg]\,\bigg]\,. \end{eqnarray*} $$