If $f$ is nonnegative and integrable then $F(x) = \int_{-\infty}^x f$ is continuous. I'm learning about measure theory, specifically the Lebesgue integral of nonnegative functions, and need help with the following problem.  

Let $f:\mathbb{R}\to[0,\infty)$ be measurable and $f\in L^1$. Show that $F(x)=\int_{-\infty}^x f$ is continuous.

I know is isn't much but the only thing a could think of is that given $x, y \in \mathbb{R}$ with $x < y$ we note that $F(x) \leq F(y)$, i.e. $F$ is increasing. So maybe we can apply one of the convergence theorems of Lebesgue integration theory.

I was also wondering if this problem can be solved using only Riemann integration theory. 
 A: Let $x \in \mathbb{R}$. 
It suffices to prove that $F(x_n) \to F(x)$ for every $x_n \to x$.
Therefore, let $x_n \to x$.
We have $F(x_n)=\int_{-\infty}^{\infty}f \cdot \chi_{[-\infty,x_n]} $. 
It is easy to see that $f \cdot \chi_{[-\infty,x_n]} \to f \cdot \chi_{[-\infty,x]}$ (except possibly at $x$).
$f \cdot \chi_{[-\infty,x_n]}$ is dominated by an integrable function (namely, $f$). Therefore, we have $F(x_n) \to F(x)$.
A: For $x<y$ since integration (in any reasonable type of integration) is additive on union of disjoint sets,
$$ F(x) + \int_{x}^y f=\int_{(-\infty,x)} f + \int_{(x,y)} f =\int_{(-\infty,x) \cup (x,y)}  f=F(y)$$
$$\implies F(y)-F(x)=\int_{x}^y f$$
$$\implies |F(y)-F(x)|=|\int_{x}^y f| \leq \int_{x}^y |f|$$
LEMMA (A Key lemma, very basic and crucial.): If $f$ is integrable, i.e. in $L^1$, for any $\epsilon >o$ given, there is a $\delta >0$, such that
$$ \forall A, \ \ \ \  |A|<\delta \implies \int_{A} |f| < \epsilon.$$
The important thing is that we don't care what set, and where, $A$ is!
In light of this lemma, if $x$ is $\delta$-close to $y$, we'll have
$$|F(y)-F(x)| < \epsilon .$$
So, we even proved more: $F(x)$ is uniformly continuous, i.e. same $\delta$ works for all points in the definition of continuity.
