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 $w_1,...w_n\geq0$ 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.