Let $(\Omega, \mathcal{F}, P)$ denote a probability space. Let $X: (\Omega, \mathcal{F}) \to (\mathbb{R}^n, \mathcal{B}^n)$ and $Y: (\Omega, \mathcal{F}) \to (\Psi, \mathcal{G})$ denote two random variables, where $\mathcal{B}^n$ denotes the Borel sets over $\mathbb{R}^n$ and $(\Psi, \mathcal{G})$ is a measurable space. Assume that $\mathbb{E}\left(|X|\right)< \infty$. Suppose there exists a measurable function $g:(\Psi, \mathcal{G}) \to (\mathbb{R}^m, \mathcal{B}^m)$ such that
$$ \mathbb{E}\left(X|Y\right) = \mathbb{E}\left(X|g(Y)\right) $$ almost surely.
Show that for any measurable function $h: (\Psi, \mathcal{G}) \to (\mathbb{R}^k, \mathcal{B}^k)$ we have $$ \mathbb{E}\left(X|g(Y)\right) = \mathbb{E}\left(X|g(Y),h(Y)\right) $$ almost surely.
$\textbf{My attempt}$:
Using the Tower Property, we have
\begin{align} \mathbb{E}(X|g(Y),h(Y)) &= \mathbb{E}(\mathbb{E}(X|g(Y),h(Y),Y)|g(Y),h(Y)) \\ &= \mathbb{E}(\mathbb{E}(X|Y)|g(Y),h(Y)) \\ &= \mathbb{E}(\mathbb{E}(X|g(Y))|g(Y),h(Y)) \\ &= \mathbb{E}(X|g(Y)) \end{align}