If $X$ is a random variable with distribution $\mu$, prove $\int \limits_{\Omega} X(\omega) \, dP(\omega) = \int \limits_{\Bbb R} x \, \mu(dx)$. I'm trying to prove $\int \limits_{\Omega} X(\omega) \, dP(\omega) = \int \limits_{\Bbb R} x \,\mu(dx)$ if $X$ is a random variable defined on $(\Omega, \mathcal{F}, P)$.
I understand how to prove this for indicator functions.  I'm not sure how to prove it for simple functions.  Specifically, let $s(\omega) = \sum \limits_{i = 1}^{n} \alpha_{i} \chi_{A_{i}}(\omega)$, where $\alpha_{i} \neq 0$ and $P(A_{i}) < \infty$ for each $i$ and the $A_{i}$'s are pairwise disjoint.  Then we have:
$\int \limits_{\Omega} s(\omega) \,dP(\omega) = \int \limits_{\Omega} \sum \limits_{i = 1}^{n} \alpha_{i} \chi_{A_{i}}(\omega) \,dP(\omega) =  \sum \limits_{i = 1}^{n}  \alpha_{i} \int \limits_{\Omega} \chi_{A_{i}}(\omega) \,dP(\omega) = \sum \limits_{i = 1}^{n}  \alpha_{i} \int \limits_{\Bbb R} x \, \mu_{i}(dx)$ (where $\mu_{i}$ is the distribution measure of $\chi_{A_{i}}(\omega)$).
Now what?  How can I say that this integral equals $\int \limits_{\Bbb R} x \,\mu(dx)$, where $\mu$ is the distribution of $s(\omega)$?  Is the distribution of $s(\omega)$ equal to the "sum" of distributions of its indicator functions?
Also, how would you then prove this for a non-negative measurable function?  I know you would use a sequence of simple functions increasing up to the function, but I'm not sure how to move the limit from out of the integral back into it at the end.
 A: The definition of $\mu$ is that $\mu(B)=P(X^{-1}(B))$, for $B\subset\mathbb{R}$ and measurable. You've got a sum of indicators, and you've proven the result for single indicators. So when looking at $\int_\mathbb{R}xd\mu(x)$, just split $\mathbb{R}$ into sets $B_1,B_2,\cdots$, where $B_i=X(A_i)$. You can always rewrite any simple function into one that's a sum over disjoint sets, so this will imply $B_i$ are disjoint too, allowing you to decompose the integral into a sum.
A: The problem you run into with the $X\ge0$ part of your proof is when you convert to $EX_n$ to $\int x\mu_n(dx)$, you can't use MCT on the right side (here, $\mu_n$ is the measure induced by $X_n$, for simple $X_n$ approaching $X$).  I think it is easier to prove the more general fact that
$$
Ef(X)=\int f(x)\,\mu(dx)
$$
for any measurable $f$. The proof method is the same, except you start with simple $f$, then $f\ge 0$, then all $f$, instead of doing the same for $X$. To get your result, set $f(x)=x$.
Addendum: Suppose $f$ is an indicator function, $f(x)=1_{x\in A}$. Then
$$Ef(X)=E1_{X\in A}=P(X\in A)=\mu(A)=\int 1_{x\in A}\,\mu(dx)$$
The third equality is the definition of $\mu$. When $f=\sum c_i1_{A_i}$ is simple, then
\begin{align*}
Ef(X)=E\sum c_i1_{X\in A_i}=\sum c_iE1_{X\in A_i}=\sum c_i\int 1_{x\in A_i}\,\mu(dx)
&=\int \sum c_i1_{x\in A_i} \,d\mu(dx)\\
&=\int f(x)\,\mu(dx)
\end{align*}
Now, suppose $f\ge 0$. Let $f_n$ be a sequence of simple functions so that $f_n\to f$ pointwise, and $f_n\ge f_{n-1}$. For example, you could pick $f_n=\min(\lfloor 2^nf \rfloor/2^n,n)$. Then by the Monotone convergence theorem,
$$
Ef(X)=E\lim_n f_n(X)=\lim_n E f_n(X)=\lim_n \int f_n(x)\,\mu(dx)= \int \lim_n f_n(x)\,\mu(dx)= \int f(x)\,\mu(dx)
$$
A: Here is what I think is a more clear proof for a beginner (though aside from notation, the ideas are pretty much the same as one of the other answers):
Claim:  Suppose $X$ is a random variable with distribution $\mu$.  Then $\int \limits_{\Omega} X(\omega) \,dP(\omega) = \int \limits_{\Bbb R} x \,\mu(dx)$.
To prove this claim, we will prove a stronger statement, which is the following theorem:

Theorem. Suppose $X$ is a random variable with distribution $\mu$.  Let $f : \Bbb R \to \Bbb R$ be any measurable function such that either $f \geq 0$ or $E[|f(X(\omega))|] < \infty$.  Then $\int \limits_{\Omega} f(X(\omega)) \,dP(\omega) = \int \limits_{\Bbb R} f(x) \,\mu(dx)$.

Proof.  We will first prove this for $f$ a characteristic function, then $f$ a simple function, followed by $f$ a non-negative measurable function, and finally $f$ an integrable function.
For characteristic functions:

Let $B \in \Bbb B(\Bbb R)$ and suppose $f = \Bbb 1_{B}$.  Then $\int \limits_{\Omega} \Bbb 1_{B}(X(\omega)) \,dP(\omega) =\int \limits_{X^{-1}(B)} 1 \,dP(\omega) + \int \limits_{\Omega - X^{-1}(B)} 0 \,dP(\omega) = \int \limits_{X^{-1}(B)}  1 \,dP(\omega) = P(X^{-1}(B)) = \mu(B) = \int \limits_{\Bbb R} \Bbb 1_{B}(x) \,\mu(dx)$.

Now for simple functions:

Let $s(x) = \sum \limits_{i =1}^{n} \alpha_{i} \chi_{A_{i}}(x)$.  Then $\int \limits_{\Omega} s(X(\omega)) \,dP(\omega) =\int \limits_{\Omega} \sum \limits_{i =1}^{n} \alpha_{i} \chi_{A_{i}}(X(\omega)) \,dP(\omega) =  \sum \limits_{i =1}^{n} \alpha_{i} \int \limits_{\Omega}\chi_{A_{i}}(X(\omega)) \,dP(\omega) $
$= \sum \limits_{i =1}^{n} \alpha_{i} \int \limits_{\Bbb R}\chi_{A_{i}}(x) \,\mu(dx) = \int \limits_{\Bbb R} \sum \limits_{i =1}^{n} \alpha_{i} \chi_{A_{i}}(x) \,\mu(dx) = \int \limits_{\Bbb R} s(x) \,\mu(dx)$.

Now if $f \geq 0$ is measurable, then find a monotonically increasing sequence of simple functions $s_{n}$ such that $s_{n} \to f$, and we have:

$\int \limits_{\Omega} f(X(\omega)) \,dP(\omega) = \int \limits_{\Omega} \lim \limits_{n \to \infty} s_{n}(X(\omega)) \,dP(\omega) \underbrace{=}_{\text{by MCT}}  \lim \limits_{n \to \infty} \int \limits_{\Omega} s_{n}(X(\omega)) \,dP(\omega)$
$ = \lim \limits_{n \to \infty} \int \limits_{\Bbb R} s_{n}(x) \,\mu(dx) \underbrace{=}_{\text{by MCT}}  \int \limits_{\Bbb R} \lim \limits_{n \to \infty} s_{n}(x) \,\mu(dx) = \int \limits_{\Bbb R}  f(x) \,\mu(dx)$

Up to now, we haven't needed the assumption that $E[|f(X(\omega))|] < \infty$, but we will use to now.
Final case: $f$ satisfies the integrability condition in the previous sentence:

Decompose $f$ into $f^{+} - f^{-}$.  Then $E[|f(X(\omega))|] < \infty$ implies  $E[f^{+}(X(\omega))] < \infty$ and $E[f^{-}(X(\omega))] < \infty$, so:
$\int \limits_{\Omega} f(X(\omega)) \,dP(\omega) = \int \limits_{\Omega} f^{+}(X(\omega)) - f^{-}(X(\omega)) \,dP(\omega) $
$= \int \limits_{\Omega} f^{+}(X(\omega))\,dP(\omega) - \int \limits_{\Omega} f^{-}(X(\omega)) \,dP(\omega) = \int \limits_{\Bbb R} f^{+}(x)\,\mu(dx) - \int \limits_{\Bbb R} f^{-}(x) \,\mu(dx)$
$ = \int \limits_{\Bbb R} f^{+}(x) - f^{-}(x) \,\mu(dx) = \int \limits_{\Bbb R} f(x) \,\mu(dx)$

