Simplification of the Expected Value via CDF: Does it work for ALL Probability Distributions? If a random variable $X$ has a density $f$, then the expected value can be simplified:
$$\mathbb{E}[X]=a+∫_{a}^{b}(1-F(x))dx,$$
where $F$ is the cumulative distribution function, $F(x)=\Pr(X≤x)$. 
My question is: Does this simplification should work for all probability distributions, even for those that are not absolutely continuous with respect to Lebesgue-measure on $[a,b]$?
If $X$ is any real-valued random variable with support $[a,b]$ and $F(x)=\Pr(x≤X)$, is it always true that
$$ \mathbb{E}[X]=a+∫_{a}^{b}(1-F(x))dx$$
The answer to this question would be simple if one could generally extend integration-by-parts to Lebesgue-Stieltjes integration. However, this is not possible; see Wikipedia. 
 A: Let's use $X$ for the random variable, keeping $x$ for the variable of integration.
In general, we have a probability measure $\mu$ on $I = [a,b]$ and 
$$\eqalign{E[X] &= \int_I x\ d\mu(x) = a + \int_I (x-a) d\mu(x)\cr
    &= a + \int_I \int_a^x 1 \ dt \ d\mu(x) = a + \int_a^b \int_{[t,b]}  1 \ d\mu(x)\ dt \cr
    &= a + \int_a^b (1 - F(t)) \ dt\cr}$$
(note that $\int_{[t,b]} 1 \ d\mu(x) = 1 - F(t-)$, but that is $1 - F(t)$ (Lebesgue) almost everywhere)
A: "The answer to this question would be simple if one could generally extend integration-by-parts to Lebesgue-Stieltjes integration."
This comment moves me to add another answer: Riemann--Stieltjes integration is enough for this.  Look it up and/or make it an exercise to prove it.  I.e. suppose you have a probability distribution on the set of Borel-measurable subsets of $\mathbb{R}$.  Let $X$ be a random variable so distributed and let $F$ be the c.d.f., so $F(x)=\Pr(X\le x)$.  Then
$$
\mathbb{E}(X) = \int_{-\infty}^\infty x \, dF(x),
$$
where that is a Riemann--Stieltjes integral.  If the support of the distribution has a finite a lower bound, then applying integration by parts to the Riemann--Stieltjes integral does it.
