This is my problem: Let $F(x, W)$ be a function of $x\in\mathbb{R}$ and a random variable $W$ also taking values in $\mathbb{R}$. Now define $f(x)$ such that
$ f(x) = \int_{\Omega} F(x, W(\omega)) \mathbb{P}(d\omega)$.
How do I prove that F(x,W) is convex in x for any realization of W if f(x) is convex in x???
If $\Omega$ is finite then I can easily prove it by using the fact that the sum of convex functions is convex. If $\Omega$ is finite then $ f(x) = \sum_{\Omega} F(x, W(\omega)) \mathbb{P}(\omega)$. If $f(w)$ is convex, then $F(x,W(\omega))$ is also convex for sample paths with non-zero probability. Otherwise, we have a contradiction (convex + concave = not convex.)
But I'm not sure how to prove that it holds when $\Omega$ is not finite e.g. Gaussian distribution. How do I extend my argument to the more general case???
Edit: Let $\mathbb{E}(W)$ be the expected value of $W$.
How about prove $F(x,\mathbb{E}(W))$ is convex in x if f(x) is convex in x??? Is it incorrect????
It's true, of course. You need to check inequalities like $pf(x)+qf(y)\ge f(px+qy)$. (For all $x$ and $y$, and so on.) These follow by observing that for each $W$, $$pF(x,W)+qF(y,W))-F(px+qy,W)$$ is nonnegative. Now you integrate with respect to $W$: the expectation of a nonnegative random variable is nonnegative.