Measure Theory - Expectation and Independence

Question

Can we use $E[f(X)g(Y)]=E[f(X)]E[g(Y)]$ for $f,g$ bounded and measurable to show that $X$ is independent of $\sigma(Y)$?

Answer 1

Let $A\in \sigma(X)$ and $B\in \sigma(Y)$. Then,there exist Borel sets $F$ and $G$ such that $A=(X\in E)$ and $B=(Y\in F)$.

Let $f(x)=1_{E}(x)$ and $g(x)=1_{F}(x)$. Then, by assumption $$\mathbb{P}(A\cap B)=\mathbb{E}(f(X)g(Y))=\mathbb{E}(f(X))\mathbb{E}(g(Y))=\mathbb{P}(A)\mathbb{P}(B)$$ This proves the desired.