Let $X$ be a random variable and $\mathcal{G}$ a sigma-algebra. I'm asked to prove that $X$ is independent of $\mathcal{G}$ if and only if
$E(g(X)|\mathcal{G}) = E(g(X))$,
for all borelian functions $g:\mathbb{R} \to \mathbb{R}$ such that $E(|g(X)|)< \infty$.
The definition I'm using for the independence of X regarding $\mathcal{G}$ is that $X$ and $I_{G}$ are independent for all $G \in \mathcal{G}$.
I was able to prove the first part, showing that independence implies the desired equality. However, I'm stuck trying to prove the converse.
For me, it sounds reasonable that it's enough to start with $g(x) =x$. It makes sense for me that if $E(X|\mathcal{G}) = E(X)$, then $X$ and $\mathcal{G}$ are independent. But I'm not sure of this claim. Anyway, I tried to exploit the fact, in order to prove the result in the other direction, I can use any borelian $g$, but I had no success.
Any ideas? Hints? Thanks a lot in advance!
By definition of conditional expectation, equality $$\mathbb E\left[\mathbb E\left[g(X)\mid\mathcal G\right]\mathbf 1_A\right] =\mathbb E\left[ g(X)\mathbf 1_A\right] $$ holds for any $A\in\mathcal G$ and any measurable function $g\colon\mathbb R$ such that $\mathbb E\left|g(X)\right|$ is finite. Using The assumption, we derive that $$\mathbb E\left[ g(X)\mathbf 1_A\right]=\mathbb E\left[g(X)\right]\mathbb P\left(A \right).$$ Now, for a Borel set $B$, choose $g(x)=1$ if $x\in B$ and $0$ otherwise.