Q:

Independence via conditional expectation and characteristic function

I have encountered the following statement but can't think of a proper argument to explain/prove it:
If for every $A \in \mathcal F$ s.t. $P(A>0)$ we have $ E\left[ \exp (i \lambda X) | A \right] = E \left[\exp(i \lambda X)\right]$, then $X$ is independent from $\mathcal F$.
I know that the characteristic function defines the distribution completely and independence is somewhat a distribution property, but I am still not very convinced. In particular, only $E\left[ X | A \right] = E \left[X \right] \Rightarrow$ $X$ and $\mathcal F$ are independent, is false, isn't it?

A:

The statement is true.  Proving it takes a fair amount of work, depending on what results you already know (about Fourier transforms and characteristic functions in particular).  However, for some very rough intuition, the idea is that any bounded Borel function $f(x)$ can be approximated in some sense by "trigonometric polynomials" of the form $g(x)=a_1 e^{i \lambda_1 x} + \dots + a_n e^{i \lambda_n x}$.  Now since $E[g(X) \mid A] = E[g(X)]$ for each such $g$, with some limiting arguments you conclude that $E[f(X) \mid A] = E[f(X)]$ for every bounded Borel function $f$.  In particular, taking $f = 1_B$, you have $P(X \in B \mid A) = E[1_B(X) \mid A] = E[1_B(X)] = P(X \in B)$.  This shows that the event $\{X \in B\}$ is independent of $A$, and since every event in $\sigma(X)$ is of the form $\{X \in B\}$, you are done.
The best way I know to execute the "limiting argument" is the multiplicative system theorem, which is to measurable functions what the Stone-Weierstrass theorem is to continuous functions (and indeed the proof of the former uses the latter, or ideas from its proof).

In particular, only $E\left[ X | A \right] = E \left[X \right] \Rightarrow$ $X$ and $\mathcal F$ are independent, is false, isn't it?

Yes, it's false.  But notice that here you have this relation not just for the random variable $X$ itself, but for infinitely many random variables from $\sigma(X)$.  That makes it sound a little more plausible, hopefully.

