Here $\xi$ is some random variable on some probability space $(\Omega,\mathcal{F},P)$, and $\mathcal{A}\subset\mathcal{F}$ is a sub $\sigma$-algebra. Apparently, two conditions are sufficient and necessary to each other, but one direction is trivial. My idea is based on following result:
$\xi$ is independent with $\mathcal{A}$ $\iff$ $E(g(\xi)|\mathcal{A})=E(g(\xi))$ for all bounded measurable $g$ on $\mathbb{R}$.
Certainly, the condition "bounded measurable" can be loosened, but the hard part is how to get such a family from the approximation of $\{e^{it(\cdot)}\}$.
So far, I've come up with:
- Inverse Fourier transform together with some Fubini trick. However, Fourier inversion theorem requires $\xi$ to possess good properties, at least $\xi$ must have density function. That's not the case here.
- Fourier series. However, Fourier series only converges inside a compact interval. To be more precise, a Fourier series $s_n$ that consists of terms in $\{e^{ix(\cdot)}\}$ satisfies $s_n\to g$ pointwise or $L^2$ on supp($g$). We hope that $E(s_n(\xi)|\mathcal{A})=E(s_n(\xi))$ would give us $E(g(\xi)|\mathcal{A})=E(g(\xi))$, which involves taking limit under the integral. However, we only have $s_n\to g$ on supp($g$), not on $\mathbb{R}$.
Do you have other ideas to prove this one?