Let $h(Y)$ be a random variable such that $E|h(Y)| < \infty$ and $K(X)$ a random variable. Show that $E[h(Y)|X,K(X)] = E[h(Y)|X]$

Question

I have an idea but I don't know how to write it formally. Well, if we are conditioning over $X$ and $K(X)$, this second random variable could be a transformation of $X$ but since we are also conditioning over $X$, then we can forget about $K(X)$

Answer 1

Conditioning on a random variables or a set of them is actually conditioning on the $\sigma$-algebra generated by that random variable or set of them. Presumably $K$ is a Borel measurable function. Then for each $\alpha$, $\{K(X) \le \alpha\} = \{X \in K^{-1}((-\infty, \alpha])\}$ is already in the $\sigma$-algebra generated by $X$, so the $\sigma$-algebra generated by $X$ and $K(X)$ is the same as the $\sigma$-algebra generated by $X$.