I'm stuck on the following exercise:
Let $f:\mathbb R\to \mathbb R$ be a nonnegative integrable function ($\int f<\infty$). Show that the function $$F(x)=\int_{-\infty}^xf$$ is continuous.
I was able to prove this very easily by concluding that $\lim_{x\to a} F(x) = F(a)$, but my textbook gave the following hint: "Use Theorem 10" which states:
THM 10:
Let $(f_n)$ be an increasing sequence of nonnegative measurable functions (with respect to the Lebesgue measure in the real line) and let $f=\lim f_n$ almost everywhere. Then $$\int f=\lim\int f_n$$
I just don't see how this theorem is related to this problem. How can we prove this using this theorem?
Using Theorem 10 (Monotone convergence theorem), one can first easily prove that $F$ is left continuous: for every increasing convergent sequence $x_n\to x$, applying the theorem to $f_n:=1_{(-\infty, x_n)}f$, we find $F(x_n)\to F(x)$.
But thanks to the integrability of $f$, we can also prove that $F$ is right continuous: for every decreasing convergent sequence $x_n\to x$, applying the theorem to $f_n:=1_{(x_n,+\infty)}f=f-1_{(-\infty, x_n)}f$, we find $\left(\int f\right)-F(x_n)\to\left(\int f\right)-F(x)$, i.e. again, $F(x_n)\to F(x)$.