Why can I say $\int_{H} \mathbb E[X |\mathcal{G}]dP=\int_{H} \mathbb E[X| \mathcal{H}]dP$?

Question

Let $(\Omega, \mathcal{F}, P)$ be a probability space and $\mathcal{H}, \mathcal{G}\subseteq\mathcal{F}$ be sub$-\sigma-$algebras where $\mathcal{H}\subseteq \mathcal{G}$.

How do I prove that for $H \in \mathcal{H}$:

$\int_{H}\mathbb E[X|\mathcal{G}]dP=\int_{H}\mathbb E[X|\mathcal{H}]dP$

Since $\mathcal{H}\subseteq \mathcal{G}$ I know that $\mathbb E[X|\mathcal{H}]=\mathbb E[X|A]1_{A}+ E[X|A^{c}]1_{A^{c}}$ if I have sets $A$ and $A^{c}$ in $\mathcal{H}$ but surely I can have other disjoint sets $B,C,D$ where $B\cup C \cup D=\Omega$, and this would lead to a different $E[X|\mathcal{H}]$

Maybe this can also help me understand why

$\int_{H}\mathbb E[X|\mathcal{G}]dP=\int_{H}\mathbb E[X|\mathcal{H}]dP$

Answer 1

By definition of conditional expectation (https://en.wikipedia.org/wiki/Conditional_expectation):

$\int_{H}\mathbb E[X|\mathcal{H}]dP=\int_{H}\mathbb X dP$

$\int_{H}\mathbb E[X|\mathcal{G}]dP=\int_{H}\mathbb X dP$

(note that H is both $\mathcal{G}$ and $\mathcal{H}$ measurable) so the two expressions are both equal to the integral of the original function.

Answer 2

Since $H\in \mathcal H\subset \mathcal G$, $$\boldsymbol 1_H\mathbb E[X\mid \mathcal G]=\mathbb E[\boldsymbol 1_HX\mid \mathcal G],\quad \quad \text{and}\quad \boldsymbol 1_H\mathbb E[X\mid \mathcal H]=\mathbb E[\boldsymbol 1_HX\mid \mathcal H].$$ Therefore $$\mathbb E\big[\mathbb E[\boldsymbol 1_HX\mid \mathcal H]\big]=\mathbb E[\boldsymbol 1_HX]=\mathbb E\big[\mathbb E[\boldsymbol 1_HX\mid \mathcal G]\big].$$