use indicator function to prove independence

Question

I would like to know why for $\forall A \in \mathcal{F}$ $$E[1_{A}e^{itX}] = P(A)E[e^{itX}]$$ will imply the independence between $X$ and $\mathcal{F}$. Thanks in advance!

Answer 1

Use simple function approximation to prove that $Ee^{itY}e^{itX}=Ee^{itY}Ee^{itX}$ for all $t$ for any $\mathcal F$ measurable $Y$. This proves that $X$ and $Y$ are independent for any $\mathcal F$ measurable $Y$. QED.

Answer 2

Hint: let $\Omega$ denote the probability space and $F$ the cdf of $X$. Since $$\int_A \exp itx dF(x)\int_\Omega dF(y)=\int_A dF(x)\int_\Omega\exp ity dF(y)$$(where I've inserted a factor of $1$ on the left-hand side),$$\int_\Omega dF(y)\int_A dF(x) [\exp itx-\exp ity]=0.$$