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Given that $E[g(X)] = int_-infty^infty g(x)f(x) dx$, how to prove $E[g(X)] = int_0^infty g'(x)S(x)dx$, where $S(x) = 1-F(x)$?

By integration by parts, I can get the following:
beginalign
E[g(X)] &= int_-infty^infty g(x)f(x) dx \
&= int_-infty^infty g(x)F'(x) dx \
&= g(x)F(x)|_-infty^infty - int_-infty^infty g'(x)F(x) dx \
&= g(x)F(x)|_-infty^infty + int_-infty^infty g'(x)[1-S(x)] dx \
&= g(x)F(x)|_-infty^infty - int_-infty^infty g'(x)dx +int_-infty^infty g'(x)S(x) dx\
endalign

Then I am stuck.

I think what you are trying to prove is not exactly correct. I will work through the problem and show the correct answer at the end.

You need to split the integral into two, one for the region $xge 0$ and one for $xle 0$.

In the positive region, instead of writing $f(x)=F'(x)$ and integrating by parts, you need to use $f=(F(x)-1)'$. This is the only choice which ensures that the resulting integral actually exists; note that $F(x)-1$ goes to zero as $xto+infty$, but $F(x)$ does not.

beginalign
int_0^infty g(x)f(x),dx
&=int_0^infty g(x)[F(x)-1]',dx
\&=g(x)[F(x)-1]Big|^infty_0-int_0^infty g'(s)[F(x)-1],dx
\&=underbraceBig(lim_xtoinftyg(x)[F(x)-1]Big)_=0-g(0)[F(0)-1]+int_0^infty g'(x)S(x),dx
endalign
It takes some doing, but you can show that that limit actually is zero, as long as $E[g(X)]$ is finite. See Is it true that $limlimits_xtoinftyx·P[X>x]=0$? for some inspiration. Edit: Actually, I am not quite sure this is true without some additional assumptions on $g.$ It is true as long as $g$ is either bounded or monotonic.

For the negative region, you do want to use $F(x)$ as the antiderivate of $f(x)$, because $lim_xto-inftyF(x)=0$.
beginalign
int_-infty^0 g(x)f(x),dx
&=int_-infty^0 g(x)F(x)',dx
\&=g(x)F(x)Big|^0_-infty-int_-infty^0 g'(s)F(x),dx
\&=g(0)F(0)-underbraceBig(lim_xto-inftyg(x)F(x)Big)_=0-int_-infty^0 g'(x)F(x),dx
endalign
Putting this all together, we get
$$
bbox[7px,border:2px solid red]int_-infty^infty g(x)f(x),dx=g(0)+int_0^infty g'(x)S(x),dx-int_-infty^0g'(x)F(x),dx
$$
This is correct for any function $g$ such that $E[g(X)]$ is finite. Edit: Well, as long as $g$ is either bounded or monotonic.

If we make further assumptions about $g$, we can write this more nicely. Assuming $g(-infty):=lim_xto-inftyg(x)$ exists, you would have
$$
g(0)=g(-infty)+int_-infty^0 g'(x),dx
$$
so
$$
bbox[7px,border:2px solid green]int_-infty^infty g(x)f(x),dx=g(-infty)+int_-infty^infty g'(x)S(x),dx
$$
You cannot in general avoid the presence of some constant $g(a)$. This is because the $int g'(x)S(x),dx$ part of the formula does not change when $g$ is shifted by a constant (this does not affect $g'$), but $E[g(x)]$ does change when $g$ is shifted by a constant.

Write $1-F(x)=int_(x,infty) dmu(y)$ where $mu$ is the measure corresponding to $F$. Now apply Fubini/Tonelli Theorem to the right hand side.