I'm trying to prove the following claim: Let $f$ be a nonnegative measurable function. We define $F(x)=\int_{-\infty}^x {f(u)du}$. Show that $F$ is continuous by using the Monotone Convergence Theorem.
I started by writing the integral as:
$$F(x_0)=\int_{-\infty}^{x_0}{f(u)du} = \int{f(u)\cdot \chi_{(-\infty,x_0)}(u)du}$$
and tried to prove the claim using Heine's criterion: $F(x)$ is continuous at $x_0$ if and only if for each sequence $\{x_n\}_{n=1}^\infty$ such that $x_n\rightarrow x_0$, $F(x_n)\rightarrow F(x_0)$.
I've managed to prove this for monotone sequences, i.e. $x_n\uparrow x_0$ and $x_n\downarrow x_0$, using the Monotone Convergence Theorem, however I got stuck when trying to prove this for a general sequence.
For $f$ just non-negative and measurable the monotone convergence theorem for decreasing sequences doesn't hold in general, for this case we will need to assume other hypothesis, by example that $f$ is essentially bounded or that is dominated for an integrable function.
Assuming that the monotone convergence theorem holds for decreasing sequences then you can use the limit superior and limit inferior of a real-valued sequence to show the convergence for an arbitrary convergent sequence $(x_n)$, that is, if $\lim_{n\to \infty }x_n=x_0$ then $$ \begin{align*} \limsup_{n\to \infty }x_n=\lim_{n\to \infty }\sup_{k\geqslant n}x_n=x_0\tag1\\ \liminf_{n\to \infty }x_n=\lim_{n\to \infty }\inf_{k\geqslant n}x_n=x_0\tag2 \end{align*} $$ Now setting $s_n:=\sup_{k\geqslant n}x_n$ and $i_n:=\inf_{k\geqslant n}x_n$ we have that $(s_n)\downarrow x_0$ and $(i_n)\uparrow x_0$ and that $i_n\leqslant x_n\leqslant s_n$ for all $n\in \Bbb N $, hence $$ \int_{-\infty }^{i_n}f(u)\,\mathrm d u\leqslant \int _{-\infty }^{x_n}f(u)\,\mathrm d u\leqslant \int_{-\infty }^{s_n}f(u)\,\mathrm d u\tag3 $$ Then taking limits in $\rm(3)$ you are done.