Expected value of a function, continous So my query is about the following proof
If $Y$ is a contiouous random variable with pdf $f_Y(y)$ and if $g(Y)$ is a continouous function, then the expected value of then random variable $g(Y)$ is 
$$E(g(Y))=\int_{-\infty}^{\infty} g(y)f_Y(y) dy $$
Now I am a bit stuck onthe proof but have the following:
Let $W=g(Y)$ and assume $g(Y)$ monotone (or else we split the interval up to make it montone) 
$$E(g(Y))=E(W)=\int_{-\infty}^{\infty} wf_W(w) dw $$
And there I am stuck..
 A: You can prove this easily using measure theory like so:
Let X be a real valued stochastic variable with density f with regards to the lesbegue measure and g be a continous function. Then
$E(g(X))=\int_\Omega g\circ X dP = \int_{\mathbb{R}} g(x) dX(P)(x) = \int_{\mathbb{R}} g(x) d(f \cdot m)(x) = \int_{\mathbb{R}} g(x)f(x) dm(x) = \int_{-\infty}^{\infty} g(x)f(x) dx$
Where the first equality is the definition of expected value, the second is the “abstract change of variables” formula, the third follows from the fact that X has density f with regards to the lesbegue measure, fourth is a basic result for measures with density and fifth is a theorem that allows you to switch between lesbegue and Riemann integrals whenever they are both finite and welldefined.
It is sometimes known as the “law of the unconscious statistician” or LOTUS bechause people mistakenly thinks it follows from the definition of expected value.
I know there are more elementary proofs of this theorem without measure theory. You can probably find some by searching for law of the unconscious statistician on google.
