Expected Value in measure theory Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and let $X$ be a continuous random variable, i.e., there exists a probability density function $f_X(x)$ such that for every $x\in\mathbb{R}$:
$$\mathbb{P}_X\{(-\infty,x]\}\equiv\mathbb{P}\{\omega\in\Omega:X(\omega)\leq x\}\equiv F_X(x)=\int_{-\infty}^xf_X(z)\  \mathrm{d}z$$
Assume $X$ is bounded and non-negative. Prove by definition:
$$\mathbb{E}[X]=\int_{\mathbb{R}}xf_X(x)\ \mathrm{d}x$$

First I must mention that I'm very new to measure theory and lebesgue integrals, so I apologize if this question is easy and I'm missing something. Anyway, I was able to show by definition that:
$$\mathbb{E}[X]=\int_{\mathbb{R}}x\ \mathrm{d}\mathbb{P}_X(x)$$
As a physicist, I would very much like to just do this:
$$\mathbb{E}[X]=\int_{\mathbb{R}}x\frac{\mathrm{d}\mathbb{P}_X(x)}{\mathrm{d}x}\mathrm{d}x=\int_{\mathbb{R}}xf_X(x)\ \mathrm{d}x$$
But of course this is not rigorous at all. So I have no idea what to do here. The only lebesgue integral we were taught to compute was the lebesgue integral of an idicator of some set $S$ (and in that case, of course, the integral would just be the (probability) measure of $S$), so I thought - maybe I should express $x$ as a sum of simple functions (indicators)? But I couldn't eventually work this through. Another crucial thing is the fact that I was given that $X$ is bounded and non-negative, but I haven't used that yet.
Thanks!
 A: If you cannot use Radon-Nikodym you can always first define a density function $f_X(x)$ and then obtain a measure $\mathbb{P}$ with which compute Lebesgue integrals. Consider $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ and let $f:\mathbb{R} \to \mathbb{R}$ be positive and continuous (thus, Borel measureable). The function
$$\nu(B)=\int_{B}f_X(x)\lambda(dx), \ \ B \in \mathcal{B}(\mathbb{R})$$
is a measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Indeed:
(1). $\nu(\emptyset)=\int\mathbb{I}_\emptyset f_Xd\lambda=0$
(2). Let $(B_n)_{n \in \mathbb{N}} \subset \mathcal{B}(\mathbb{R})$ be pairwise disjoint, then
$$\nu(\cup_nB_n)=\int\mathbb{I}_{\cup_nB_n}(x)f_X(x)\lambda(dx)=\int\sum_{n \in \mathbb{N}}\mathbb{I}_{B_n}(x)f_X(x)\lambda(dx)=$$
$$=\int \sup_{N\in \mathbb{N}}\sum^N_{n=1}\mathbb{I}_{B_n}(x)f_X(x)\lambda(x)\underbrace{=}_{\textrm{Monotone conv. thm.}}\sup_{N\in \mathbb{N}}\sum^N_{n=1}\int\mathbb{I}_{B_n}(x)f_X(x)\lambda(x)=\sum_{n \in \mathbb{N}}\nu(B_n)$$
The measure $\nu$ is a probability measure if we also have that
$$\nu(\mathbb{R})=1$$
in which case we call $\mathbb{P}_X:=\nu$. We then have that
$$\mathbb{P}_X(X \in (-\infty,x])=\int_{(-\infty,x]}f_X(x)\lambda(dx)=F_X(x)$$
and we can also comfortably write
$$\mathbb{E}[X]=\int_{\mathbb{R}}x\mathbb{P}_X(dx)=\int_{\mathbb{R}}xf_X(x)\lambda(dx)$$
A: Let $X_n:=g_n(X)$, where $g_n(x)=2^{-n}\lfloor 2^n x\rfloor$, s.t. $X_n\to X$ as $n\to\infty$. For any $n$,
\begin{align*}
\mathsf{E}X_n&=\mathsf{E}\sum_{i=1}^{M_n} \frac{i}{2^n}1\{X\in [i2^{-n},(i+1)2^{-n})\} \\
&=\sum_{i=1}^{M_n} \frac{i}{2^n}\mathsf{P}(X\in [i2^{-n},(i+1)2^{-n})) \\
&=\sum_{i=1}^{M_n} \frac{i}{2^n}\int_{i2^{-n}}^{(i+1)2^{-n}} f_X(x)\,dx \\
&=\int g_n(x)f_X(x)\,dx,
\end{align*}
where $M_n=\lfloor 2^n\|X\|_{\infty}\rfloor$. Using the monotone convergence theorem,
$$
\mathsf{E}X=\lim_{n\to\infty}\mathsf{E}X_n=\lim_{n\to\infty}\int g_n(x) f_X(x)\,dx=\int xf_X(x)\,dx.
$$ .
