Proof: distribution function is continuous?

In my textbook we work with the following definition for continuous random variables:

A random variable X is continuous if its distribution function $F_X$ may be written in the form

$F_X(x)=\mathbb P(X\leq x)=\int_{-\infty}^{x}f_X(u)\text{d}u\quad \text{for } x\in \mathbb R,$

for some non-negative function $f_X$.

I'm wondering if $F_X$ is continuous, and if so, how to prove that.

The way I started:

We can write $\int_{-\infty}^{x}f_X(u)\text{ d}u=\lim_{a\to-\infty}\int_{a}^{x}f_X(u)\text{ d}u$. By the fundamental theorem of calculus, we know that $F^{\text{~}}(x)=\int_{a}^{x}f_X(u)\text{ d}u$ is continuous for each $a\in \mathbb R$. But how can I extend this to the case of the limit?

Answer

$F$ is non-decreasing, so for each $x$ the one-sided limits $F(x_+)$ and $F(x_-)$ exist.

$F(x_-)=\int_{(-\infty,x)}f(x)dx$

By monotone convergence theorem:

$F(x_-)=\lim_{n\rightarrow\infty}F(x-1/n)=\lim_{n\rightarrow\infty}\int_{(-\infty,x-1/n]}f(x)=\int_{(-\infty,x)}f(x)dx$

$F(x)=\int_{(-\infty,x]}f(x)dx$

$F$ is right-continuous so $F(x)=F(x_+)$, regardless of its overall continuity:

$F(x)=P(X\le x)=P(\bigcap X\le x+1/n)=\lim_{n\rightarrow\infty}P(X\le x+1/n)=\lim_{n\rightarrow\infty}F(x+1/n)=F(x_+)$

So obviously $F$ is continuous at $x$, since the value of $f$ in one point has no effect on the integral.