This is probably a very simple question, but I’m getting confused on how to show this. This comes from the proof of the Tower Property in conditional Expectation. So, one knows that for sub-$\sigma$-algebras $\mathcal H\subset \mathcal G \subset F$ $$ E(X\mid \mathcal H) =E(E(X\mid\mathcal G)\mid \mathcal H) $$
Now, a corollary for this is that for $\mathcal H=\{\varnothing, \Omega\}$, and $\mathcal G = \sigma(Y)$, where $Y$ is a r.v. Then
$$E(X) = E(E(X\mid Y))$$
My question is how to show that $E(X)=E(X\mid \mathcal H)$? I’m mostly confused about the measurability of $E(X\mid \mathcal H)$.
Lemma: If $X: \Omega \to \Bbb{R}$ is $\{\emptyset, \Omega\}$-measurable, then $X$ is constant.
Proof: Suppose to the contrary that $X$ is not constant. Then there are $\omega \neq \omega'$ with $X(\omega) \neq X(\omega')$. However, then $$\omega \in X^ {-1}(\{X(\omega)\}); \quad \omega' \notin X^ {-1}(\{X(\omega)\})$$ which is impossible since $X^{-1}(\{X(\omega)\}) \in \{\emptyset, \Omega\}$. $\quad \square$
Alright, let us apply the lemma now to our special situation. Recall that $E(X\mid \mathcal{H})$ is $\mathcal{H}$-measurable, so by our lemma $E(X\mid \mathcal{H})$ is constant. If $Y$ is a constant random variable, then $Y = EY$ and thus $$E(X\mid \mathcal{H})= E(E(X \mid \mathcal{H})) = E(X)$$ and we can conclude.