If $Eh(X) = \int h d \mu$, then show that $Eh(X) = \int_\mathbb{R} h(x)f(x) \ d x$

Question

I have shown that if $h$ is an integrable function with respect to the measure $\mu$, and $X$ is a random variable on the usual probability space, then $Eh(X) = \int h \ d\mu$. I am wondering how this extends to: if $X$ has density $f$, then $$Eh(X) = \int_\mathbb{R} h(x)f(x) \ d x$$

which is the usual high school definition of expectation. I'm not really sure how to show this using the definition of density

Answer 1

The probability density $f$ is just the Radon-Nikodym derivative of $\mu$ with respect to the relevant measure (e.g. Lebesgue measure for a random variable taking values in $\mathbb{R}$).

Answer 2

The equation $\mathbb Eh(X)=\int hd\mu$ holds if measure $\mu$ is prescribed by $B\mapsto P(X\in B)$ where $P$ denotes the probability measure on the space that serves as domain of $X$.

This measure $\mu$ on $\langle\mathbb R,\mathcal B\rangle$ is often denoted as $P_X$.

It is the probability measure induced by $X$ on $\langle\mathbb R,\mathcal B\rangle$ and is called the distribution of $X$.

It can be endowed with a density $f$ with respect to the Lebesguemeasure on $\langle\mathbb R,\mathcal B\rangle$.

Then $P_X(B)=\int_Bfd\lambda$ for $B\in\mathcal B$ where $\lambda$ denotes the Lebesguemeasure.

In that case: $$\mathbb Eh(X)=\int hd\mu=\int hdP_X=\int hfd\lambda$$

The RHS is also written as $\int h(x)f(x)dx$