Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, let $\mathcal{G}\subset\mathcal{F}$ and $\mathcal{H}\subset\mathcal{F}$ sub-$\sigma$-algebras. Is it true that $$\mathbb{E}[\mathbb{E}[X~|~\mathcal{G}]~|~\mathcal{H}]=\mathbb{E}[X~|~\mathcal{G}\cap\mathcal{H}]$$ holds for any random variable $X$? My attempt at a proof goes as follows:
\begin{align}\mathbb{E}[X~|~\mathcal{G}\cap\mathcal{H}]&=\mathbb{E}[\mathbb{E}[X~|~\mathcal{G}]~|~\mathcal{G}\cap\mathcal{H}]\\ &=\mathbb{E}[\mathbb{E}[\mathbb{E}[X~|~\mathcal{G}]~|~\mathcal{H}]~|~\mathcal{G}\cap\mathcal{H}]\\ &=\mathbb{E}[\mathbb{E}[X~|~\mathcal{G}]~|~\mathcal{H}]. \end{align} All I used is the tower property repeatedly for both $\mathcal{G}\cap\mathcal{H}\subset\mathcal{G}$ and $\mathcal{G}\cap\mathcal{H}\subset\mathcal{H}$. In the last line I used that $\mathbb{E}[\mathbb{E}[X~|~\mathcal{G}]~|~\mathcal{H}]$ must be $\mathcal{G}\cap\mathcal{H}$-measurable. Is my argument correct? I haven't found anything about this on the internet.