If $X < a$, $EX < a$? If a r.v. $X < a$, does it imply $EX < a$?


*

*If not, why is it different from what I know: If a r.v. $X \leq a$,
it  implies $EX \leq a$, proved by replacing $X$ with $a$ as the integrand.

*Note that $a \in \mathbb R$. If it allows that $a= \infty$, the
answer is no: if $X$ has a Pareto distribution with $α=1$, then $X <
\infty$, but $EX = \infty$, from http://en.wikipedia.org/wiki/Pareto_distribution and https://stats.stackexchange.com/a/91515/1005.
How can we explain that difference?


Thanks.
 A: There exists $b<a$ such that $P(X\leq b)>0$. Hence if $X<a$, $EX<a$.
A: Let $(\Omega, \mathcal F,\mathbb P)$ be a probability space and $X:\Omega\to\mathbb R$ a random variable. By "$X<a$" I'm assuming you mean $X(\omega)<a$ for all $\omega\in\Omega$. This is true, but it's sufficient that $X<a$ almost surely, i.e. $X(\omega)<a$ for all $\omega\in\Omega\setminus E$ where $\mathbb P(E)=0$. This follows from linearity of the Lebesgue integral (and the finiteness of $\mathbb P$):
$$ \mathbb E[X] = \int_\Omega X\,\mathrm d\mathbb P = \int_{\Omega\setminus E} X\,\mathrm d\mathbb P + \int_E X\,\mathrm d\mathbb P < \int_{\Omega\setminus E} a\,\mathrm d\mathbb P + 0 = a\mathbb P(\Omega\setminus E) = a.$$
(Note that the integral of any function over a null set is zero.)
A: If $\alpha$ is a real number then it's true that if $X<\alpha$ then $EX<\alpha$. If we allow $\alpha$ to be an extended real number ($\infty$ or $-\infty$) then we aren't really saying much about X in terms of bounding, X can still grow towards $\infty$ without ever actually reaching $\infty$. Think of how the sequence $x_n=n$ goes to $\infty$ as $n\rightarrow \infty$ but no individual term is ever itself $\infty$.
