Suppose $y=h(u)$ prove$E(Y)=E(h(u))$ Suppose $y$ has a pdf $f(y)$, and $y=h(u)$ for some variable $u$, where $g(u)$ is the pdf for $u$.
Prove that $E(Y)=E(h(U))$
that was to prove
$\int yf(y)dy=\int h(u) g(u)du$.
 A: When you look at this the right way, it becomes so trivial that it's hard to come up even with something to say!  This is a testament to the power of employing the definitions and axioms of probability theory, which have been chosen to make results like this very, very simple.
Rather than invoke those abstractions, I hope I might prevail upon you, gentle readers, by building on intuitive, non-mathematical material introduced in another post.
Recall that $U$ can be thought of as numbers written on each one of a set $\Omega$ of tickets in a box.  $Y$ is determined by applying the function $h$ to the value $h(\omega)$ on each ticket $\omega$ and also writing that on the ticket.  The result we wish to prove asserts you don't actually need to write the numbers down, so long as you know how to compute $h.$
That is so blindingly obvious that you might want to stop reading right here.  The rest just proves the same thing by writing down the argument in a more formal mathematical notation.
When we actually write the new numbers on the tickets, the new value on a ticket $\omega$ is called $Y(\omega).$  When we choose not to write them down, we instead find out the values of $Y$ by applying $h$ to $U(\omega),$ which is written
$$Y(\omega) = h(U(\omega)).$$
The expectation $E(Y)$ is defined to be the average of the values on all the tickets in the box.  The general notation for it is
$$E(Y) = \int_\Omega Y(\omega)\mathrm{d}\mathbb{P}(\omega).$$
This refers to adding the values $Y(\omega),$ weighted by their probabilities $\mathrm{d}\mathbb{P}(\omega),$ and so--despite the sophisticated notation--is identical to definitions in any textbook, regardless of its level.
(The symbol $\int$ is an old-fashioned "S," short for "sum," and $\mathrm{d}\mathbb{P}(\omega)$ is the probability measure: it is a mathematical description of the proportions of tickets in the box.  For any box with a finite number of tickets $n,$ this probability simply is $\mathrm{d}\mathbb{P}(\omega)=1/n$ for every $\omega.$)
The expectation $E(h(U))$ is the same average, which is obvious by making the substitution $Y(\omega) = h(U(\omega))$ for every $\omega\in\Omega$:
$$E(Y) = \int_\Omega Y(\omega)\mathrm{d}\mathbb{P}(\omega) = \int_\Omega h(U(\omega))\mathrm{d}\mathbb{P}(\omega) = E(h(U)),$$
QED.  The left-hand integral uses the values $Y(\omega)$ pre-written on the tickets while the right-hand integral uses the values $U(\omega)$ and computes $h(U(\omega))$ on the fly. That's the only difference between them.

Simple Worked Example
Let the box have four tickets on which the values of $U$ are $1,2,2,3.$  Thus
$$E(U) = \frac{1}{4}\left(1+2+2+3\right)=2.$$
Suppose $$h(1) = h(3) = 0 \text{ and } h(2)=2.$$  Then the values of $Y$ on the tickets are $h(1), h(2),h(2), h(3)$ which  are $0,2,2,0.$  Both $E(Y)$ and $E(h(U))$ equal $1$ because (by definition) both are given by
$$E(Y) = \frac{1}{4}\left(0+2+2+0\right) = \frac{1}{4}\left(h(1)+h(2)+h(2)+h(3)\right) = E(h(U)).$$
A: At least heuristically $f(y) = g(h^{-1}[y])||\text{d}u/\text{d}y|$ means that
$$
\int y f(y) \text{d}y = \int h(u) g(h^{-1}[y])||\text{d}u/\text{d}y| \text{d}y
= \int h(u) g(u)\text{d}u .
$$
