Sufficient condition for the independence of random variable and the event

Question

I would like to prove the following proposition: Let $X$ be a random variable on some probability space $(\Omega, \mathcal{F}, \mathbb{P})$. If $$ \mathbb{E}(1_{B}\times e^{iuX})=\mathbb{P}(B)\times \mathbb{E}(e^{iuX}) $$ for any $u\in \mathbb{R}$, then random variable $X(\omega)$ and event $B$ are independent.

The idea of the proof is the following:

Step 1: let $f(x)$ be, say, continuously differentiable $2\pi$-periodic function. Then it is well-known that its Fourier series $\sum_{|n|\le N}c_ne^{inx} \to f(x)$ for any $x$ as $n\to \infty$. Using linearity of mathematical expectation, it is easy to show that following is valid $$ \mathbb{E}\left(1_{B}\times \sum_{|n|\le N}c_ne^{inX(\omega)}\right)=\mathbb{P}(B)\times \mathbb{E}\left(\sum_{|n|\le N}c_ne^{inX(\omega)}\right) $$ Then tending $n\to \infty$ in equality above, we see that $$ \mathbb{E}\left(1_{B}\times f(X)\right)=\mathbb{P}(B)\times \mathbb{E}(f(X)) $$ where $f$ is a continuously differentiable $2\pi$-periodic function.

Step 2: By considering function $F(x)=f(\frac{x}{K})$, we can extend it to continuously differentiable $2\pi K$-periodic function for arbitrary $K$.

Step 3: Then we can further extend it to arbitrary continuously differentiable function.

Step 4: Next, we can approximate the indicator of function $1_A(x)$ by a sequence of continuously differentiable functions, which means that $$ \mathbb{E}\left(1_{B}\times 1_A(X)\right)=\mathbb{P}(B)\times \mathbb{P}(X\in A) $$ And it proves that indeed random variable $X(\omega)$ and set $B$ are independent.

I know that my proof is not strictly speaking complete, but I was just wondering if there's any shorter prooof? Or may be this result follows from some other, more well-known , sufficient condition for independence?

Answer 1

1. There is a theorem which says that if $X, Y$ are random variables then ($\perp$ denotes independence) $$ X \perp Y\iff \mathbb{E}e^{iuX+ivY}=\mathbb{E}e^{iuX}\mathbb{E}e^{ivY} \forall_{u,v\in\mathbb{R}}.$$ It follows from the fact that Fourier transform uniquely determines a probability measure.

2. For an event $B$, if $1_B$ is the indicator function of $B$, $$\mathbb{E}e^{iv1_B} = \mathbb{P}(B)e^{iv} + \mathbb{P}(B^c).$$

3. Now, from our assumption, $$\mathbb{E}e^{iuX+iv1_B} = e^{iv}\mathbb{P}(B)\mathbb{E}e^{iuX} + \mathbb{P}(B^c)\mathbb{E}e^{iuX} = \mathbb{E}e^{iv1_B}\mathbb{E}e^{iuX}.$$ Therefore we are allowed to use 1.)