Proof of $P(X < x) = F(x−)$ 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?
 A: The intuition is that the CDF is just "adding probabilities up", a little bit at a time.  Remember, the integral is a sum; that's why $\int$ is an elongated 's'.  
So think of your expression $F(x-\frac 1 n)$ as "adding a little bit more"; as $n$ gets larger, you added in the difference
$$
\int_{-\infty}^{x-\frac 1 {n+1}} f(x) - \int_{-\infty}^{x-\frac 1 n} f(x) = \int_{x-\frac 1 n}^{x-\frac 1 {n+1}} f(x).
$$
Since $f(x) \geq 0$ for all $x$, this difference is also $\geq 0$. The limit is then the sum of the area under the curve $f(x)$ between $x-1/n$ and $x-1/(n+1)$:
$$
\sum_{n=1}^\infty \int_{x-\frac 1 n}^{x-\frac 1 {n+1}} f(x).
$$
Now, if $X$ is a continuous random variable, this sum will converge to $F(x)$. However, if $X$ has "discrete pieces" things can go wrong and the CDF can jump: see the third picture in this diagram.  In general, however, the sum will converge to $\lim_{b\to x^-} F(b)$.
A: EDIT: I rewrite my answer, since it appears it was not clear enough. I will try to go in more details even though I think this is not really interesting.
We define $F(x) := P(X \leq x)$ and we want to prove $F(x-) = P(X < x)$.
Consider a sequence $(a_n)$ such that $\lim_{n \rightarrow \infty} a_n = x$, $a_n < x$ for all $n$. For the sake of simplicity, let's admit $a_n$ is increasing ($a_{n+1} \geq a_n$).
Now, let's consider the sequence $(b_n)$ given by $b_n := F(a_n)$. By definition of $F$, we have $b_n = P(X \leq a_n)$. Thus $(b_n)$ is a sequence in [0,1]. Since $b_{n+1} = P(X \leq a_{n+1}) \geq P(X \leq a_n) = b_n$, we have that the sequence $b_n$ is also increasing. Hence, it converges. Now let's calculate the limit (let's call it $b$).
We know that $P(X \leq a_n) \leq P(X < x)$, so $b_n \leq  P(X < x)$ for all $n$. Thus this should remain true for the limit $b$. You might argue that you are not sure why this should remain true for the limit and why we could not have $b = P(X \leq x)$. Here comes the $\varepsilon$ argument. If  $P(X \leq x) = P(X < x)$ then there is no problem, so suppose $P(X \leq x) \ne P(X < x)$. In that case, let $\varepsilon := \frac{1}{2}\left(P(X \leq x) - P(X < x)\right)$.
By definition of the limit, you should find infinite members of the sequence in $[b- \varepsilon, b + \varepsilon]$. But this can't be, because all the $b_n$ are smaller than $P(X < x)$ and thus at distance at least 2 $\varepsilon$ from the limit. This proves that if $P(X \leq x) \ne P(X < x)$, then $b \ne P(X \leq x)$. 
Ok, so we know $b \leq P(X < x)$. We still need to prove there is an equality here. I will not go into further technical details here, the intuition should tell you that since $a_n \rightarrow x$, you can come as near to $P(X < x)$ as you want with the sequence $b_n$ (consider $\varepsilon = \frac{P(X < x) -b}{2}$ and see what this implies bla-bla-bla).
In the beginning I said we could consider $a_n$ to be increasing. Suppose it is not. Then you can construct a new sequence $c_n$, such that $\lim c_n = x$, $c_n \leq a_n$ for all $n$ and $c_n$ is increasing. Then you compare the sequence $d_n := F(c_n)$ with the sequence $b_n$ and bla-bla-bla. This really becomes technical, but it works. So $F(x-) = P(X < x)$.
by the way, sorry if I can't write my sequences properly, I can't get { } in math mode, even with 1 backslash, 2 backslaches or 3 backslaches before the {. This really annoys me.
A: 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: ${\rm P}(X < x) = \lim _{n \to \infty } {\rm P}(X \le x - \frac{1}{n}) = \lim _{n \to \infty } F(x - \frac{1}{n}) = F(x - )$.
