Prove $\Bbb E(X)=\int_{\Bbb R}xf(x)\,d\lambda$ in rigorous measure theory

Question

I'm studying theoretical probability theory. Let $(\Omega,\Sigma,P)$ be a probability space. $X:\Omega\to\Bbb R$ be a random variable. Suppose the probability density function $f$ of $X$ exists.

If the definition of expectation is $\Bbb E[X]:=\int_{\Omega}X\,dP$. I want to prove that $\Bbb E(X)=\int_{\Bbb R}xf(x)\,d\lambda$, where $\lambda$ is the Lebesgue measure in $\Bbb R$.

My attempt:

$$\Bbb E[X]:=\int_{\Omega}X\,dP=\int_{\Omega}\text{id}\circ X\,dP=\int_{X(\Omega)}\text{id}\,dP_X,$$ where the last equality came from Lebesgue change-of-variable formula in abstract settings (i.e, $\Omega$ is not necessarily required to be $\Bbb R^n$; an abstract measure space is fine), and $P_X$ is the push-forward measure of $P$.

Next, by Radon-Nikodym Theorem we know that $P_X(A)=\int_Af\,d\lambda$ for every Borel set $A$ in $\Bbb R$. Then how can I run to the next and final step that $$\int_{X(\Omega)}\text{id}\,dP_X=\int_{\Bbb R}xf(x)\,d\lambda?$$

I guess it would be quite easy; however, I haven't studied real analysis for a while, so I was not able to figure out. And on the other hand, is my self-attempted proof correct? Could it be more simpler?

Answer 1

As mentioned in the comment, the integral shouldn't be over $X(\Omega)$. The general change of variables theorem tells you \begin{align} \int_{\Omega}X\,dP&=\int_{X^{-1}(\Bbb{R})}(\text{id}_{\Bbb{R}}\circ X)\,dP=\int_{\Bbb{R}}\text{id}_{\Bbb{R}}\,dP_X. \end{align} The final step is a general exercise in measure theory, that you can 'multiply and divide' by $d\lambda$ to write the last integral as $\int_{\Bbb{R}}\text{id}_{\Bbb{R}}\frac{dP_X}{d\lambda}\,d\lambda=\int_{\Bbb{R}}\text{id}_{\Bbb{R}}\cdot f\,d\lambda$. More precisely, here's a guided exercise/statement which I leave to you to fill in the details of (they really are routine).

Let $(X,\mathcal{A})$ be a measurable space, and $\mu$ a positive measure defined on the $\sigma$-algebra $\mathcal{A}$. Let $f:X\to [0,\infty]$ be a non-negative $\mathcal{A}$-$\mathcal{B}(\overline{\Bbb{R}})$ measurable function, and consider the set function $\nu:\mathcal{A}\to [0,\infty]$ defined as \begin{align} \nu(A)&:=\int_Af\,d\mu. \end{align}

• Prove that $\nu$ is a positive measure (hint: monotone-convergence).
• Prove that for any $\mathcal{A}$-$\mathcal{B}(\overline{\Bbb{R}})$ measurable function $g:X\to [0,\infty]$, we have $\int_{X}g\,d\nu=\int_Xgf\,d\mu$. As an outline, prove this first when $g$ is the indicator of a measurable set, then prove it for non-negative linear combinations (i.e non-negative simple functions), and then use monotone convergence to prove it for all non-negative measurable functions $g$.
• Deduce (by taking positive, negative, real, imaginary parts) that if $g:X\to\Bbb{C}$ is an $\mathcal{A}$-$\mathcal{B}(\Bbb{C})$ measurable function then $g\in L^1(\nu)$ if and only if $gf\in L^1(\mu)$. Furthermore, in this case, $\int_{X}g\,d\nu=\int_Xgf\,d\mu$.

This theorem provides the justification for the notation $d\nu=f\,d\mu$, because it tells us that in order to calculate $\int_Xg\,d\nu$, we just replace $d\nu=f\,d\mu$ to arrive at the final answer of $\int_Xgf\,d\mu$.