It is claimed that for $\mathcal{G}=\{0,\Omega\}$ (smalles $\sigma$-field) the $\int_GE[X|\mathcal{G}]dP=\int_GXdP$ for $G\equiv\Omega$ implies that $E[X|\mathcal{G}]=E[X] a.e.$
I am having trouble understanding why this is so. I understood for the largest $\sigma$-field based on https://stats.stackexchange.com/questions/319941/int-gex-mathcalgdp-int-gxdp-implies-that-ex-mathcalg-x, but I don't get how to comprehend the smallest $\sigma$-field.
Any help would be appreciated!
The random variable $Y:=\mathbb E[X\mid\mathcal G]$ is $\mathcal G$-measurable, which implies that for all $t\in\mathbb R$, the set $A_t:=\{Y\leqslant t\}$ is either the emptyset or $\Omega$. Letting $t_0:=\sup\{t\mid A_t=\emptyset\}$, we can see that $Y$ is constant, equal to $t_0$. Now, in order to determine the value of the constant, we use the definition of conditional expectation $$ \int_GE[X|\mathcal{G}]dP=\int_GXdP $$ with $G=\Omega$. The left hand side is $t_0$ and the right hand side $\mathbb E[X]$.