Self-independent random variable Let $X$ be a self-independent random variable. Show that $X$ is almost sure constant.
My proof (by condradiction): Assume that there are two disjoint Borel set $A$,$B$ such that:
$\Pr(X \in A)>0$ and $\Pr(X \in B)>0$
Since $X$ is self-independent, there must be $\Pr(X \in A)=1$ and $\Pr(X \in B)=1$.
It's a contradiction, because then $\Pr(X \in A \cup B)>1$.
Is everything ok?
 A: $$\mathbb{E}\left[\left(X-\mathbb{E}\left[X\right]\right)^2\right]
=\mathbb{E}\left[X^2\right]-\mathbb{E}\left[X\right]^2
=\mathbb{E}\left[X\right]^2-\mathbb{E}\left[X\right]^2=0$$
thus $X-\mathbb{E}\left[X\right]=0$ a.e. .
IMPORTANT EDIT :
This requires $X$ to be square integrable.
A: A direct proof:
For any measurable set $A$, applying independence of the events $\{X\in A\}$ and $\{X\in A\}$ gives
$$
\mathbb P(X\in A)=\mathbb P(X\in A)\mathbb P(X\in A)\implies \mathbb P(X\in A)\in \{0,1\}.
$$
Hence $\mathbb P$ is a dirac measure. Therefore $X$ is a.s. constant.

Edit, 4 years later.
From a certain perspective, the more interesting part of the answer is the passage from "$\mathbb P(X\in A)\in \{0,1\}$" to "there exists $x\in\mathbb R$ such that $\mathbb P(X=x)=1$" (the latter is the more common definition of a dirac measure), which my original answer did not address. In fact, it is a special case of a more abstract result, with $\mathbb R$ replaced by any separable metric space.
Lemma. Let $\mu$ be a Borel probability measure on a separable metric space $M$ such that $\mu(A)\in \{0,1\}$ for all Borel subsets $A\subseteq M$. Then there exists $x\in M$ such that $\mu(\{x\})=1$.
Proof. In a separable metric space, every open cover has a countable subcover. Consider the cover of $M$ by open balls of radius $\epsilon$ around every point, and extract a countable subcover. There must exist a ball $B_\epsilon$ in the countable subcover with $\mu(B_\epsilon)=1$, otherwise all balls would be null sets and thus $M$ would be null as well, a contradiction. Observe that the family of sets with measure $1$ is closed under finite (and therefore countable) intersections, from which it follows that $$\mu(B)=1,\qquad B=\bigcap_{n\in\mathbb N}B_{1/n}.$$
Finally, diam$(B)=0$ and thus $B=\{x\}$, yielding the claim.
