I'm struggeling with some small gaps in the reasoning of the problem as stated in the title.
We asume $X: \Omega \rightarrow \mathbb{R}^+$ is a r.v.(random variable). We can easily prove that if this is true that $Y = \log(X + 1)$ also is measurable because $f(\cdot)=\log(\cdot)$ and $g(\cdot)=\cdot + 1$ are both continuous hence the concatentation too and $Y$ is also a r.v.. Yes, this follows from the fact that for continuous functions $h$ open sets in the codomain get mapped to open sets in the domain under the inverse of the continuous function $h^{-1}$. If you now consider the fact that the Borel algebra on the reals $\mathbb{R}^+$ gets generated by open sets $Y$ is also a r.v.
With a little work see that $\int_{\Omega} |Y(\omega)| \,d\mathbb{P}(\omega) \leq 2\int_{\Omega} |X(\omega)| \,d\mathbb{P}(\omega) < \infty$, hence $Y$ is integrable.
Now for the sketchy part, let's consider $h = f \circ g$ we would like to calculate $\mathbb{E}(Y)$ which according to this reference is equal to $\int_{0}^{\infty} (1-F_Y(y)) \,dy$, because $Y$ has a support that is non-negative. Why though? It seems to me that in the proof there is an interchange of integrals and hence this is not so trivial and I want to apreciate why exactly this is true! Furthermore if we consider this reference, the fact that $Y = h(X)$, $y = h(x)$ and that $h$ is monotonically increasing, overmore we have $\frac{\,dh}{\,dx}(x) = \frac{1}{1+x}$ and throwing it all together we get the desired equality $\mathbb{E}(Y) = \int_{0}^{\infty} \frac{1-F_X(x)}{x + 1} \,dx$
Are all my steps legitimate? Should I expand on some steps? How can i make my point watertight?
Thank you so much guys hope to hear from you soon!