Ytukyg

In general case, the expectation value is defined like this
$$E[X]=int_Omega X(w)dP(w)$$

and in absolutely continuous or countably case the expectation value is defined:
$$E[X]=int_Bbb R xf(x)dx$$ or $$E[X]=sum xf(x)$$

if $w_1,...w_ngeq0$ and $f_1,...f_n$ are convex, then function $w_1f_1+...w_nf_n$ is also convex

and this can extends to the integrals.

So,My problem is what if today the density function $f(x)$ is doesn't exist

or in general case $E[X]=int_Omega X(w)dP(w)$

Expectation value $$E[X]=int_Omega X(w)dP(w)$$

and Expectation value of convex function $$E[g(X)]=int_Omega g(X(w))dP(w)$$

are convex?

I think you're confusing many things at once.

If $w_1, ldots, w_n ge 0$ and $f_1, ldots, f_n$ are convex functions then yes $sum_i w_i f_i$ is a convex function.

I think you are looking at the sum $E[X] = sum_x x f(x)$ and drawing some sort of connection when there is none. There is only one function $f$ (the PMF of the distribution) and the sum is over values of $x$. The expectation $E[X]$ is a number, not a function.

It does not make sense to ask whether $E[X]$ or $E[g(X)]$ is "convex," since they are not functions.