If the cdf of $X$ is $I_{[a,+\infty)}$ then $P(X=a)=1.$

Question

Let $F$ be the probability distribution function on $\mathbb{R}$ for some random variable $X:\Omega \rightarrow \mathbb{R}$, such that $F(x)$ is defined as the indicator function $I_{[a,+\infty)}(x)$, i.e. $0$ for $x<a$ and $1$ for $x\geq a$ (where $a\in \mathbb{R}$ constant). Prove that $X$ is degenerate and, in fact, $P(X=a)=1$.

Some thoughts. Of course, the converse holds. I am aware that if two cdf's $F_X,~F_Y$ coincide, then there is no need for $X,~Y$ to be ae equal (eg $[0,1],$ with the Lebesgue measure and $X(\omega)=\omega$, $Y=1-X$). But it seems that the converse holds in this special case. Still don't know how to prove it though.

Thanks a lot in advance!

Answer 1

Hint: $\mathsf P(X=a) ~=~ \mathsf P(X\leq a)-\mathsf P(X<a)$

Where $\mathsf P(X\leq a) = F(a)$ by definition of CDF, and $\mathsf P(X<a)=\lim\limits_{h\nearrow a} F(h)$ by right continuity of the CDF.