TITLE: Proof of $P(X < x) = F(x−)$
QUESTION [2 upvotes]: The definition of a Cumulative Distribution Function $(CDF)$ says that
$$P(X \le x) = F(x)$$
This is all good.
Then my text book gives the following theorem without proof:
$$P(X \lt x) = F(x-)$$
The book says that the proof is easy, but with my rusted calculus skills, I have trouble to even intuitively understand this theorem. 
I tried to sketch a proof, to improve my understanding:
$$F(x-) = \lim_{n \to \infty} F(x - \frac{1}{n}) = \lim_{n \to \infty} P(X < x - \frac{1}{n}) = P(X < x).$$
I assume we can't say $P(X \le x - \frac{1}{n})$, because then $X(\omega) \ge x$, for some $\omega \in \Omega$.
Is this the correct reasoning?

REPLY [2 votes]: For $x$ fixed, define events $A_n$ by $A_1 = { X \leq x-1 }$ and, for $n \geq 2$, $A_n = { x-\frac{1}{{n - 1}} < X \leq x-\frac{1}{n} }$. Since the disjoint union $A_1 \cup A_2 \cup \ldots$ gives the event ${ X < x }$, we have ${\rm P}(X < x)= {\rm P}(A_1 \cup A_2 \cup \ldots) = \sum\nolimits_{k = 1}^\infty {{\rm P}(A_k )}$. Noting that $\sum\nolimits_{k = 1}^n {{\rm P}(A_k )} = {\rm P}(X \leq x - \frac{1}{n})$, we thus obtain the desired result: ${\rm P}(X < x) = \lim _{n \to \infty } {\rm P}(X \le x - \frac{1}{n}) = \lim _{n \to \infty } F(x - \frac{1}{n}) = F(x - )$.
{"set_name": "stack_exchange", "score": 2, "question_id": 8550}