Q:

$f$ be Lebesgue integrable and $F = m\{f > \alpha\}$, then $F$ is right continuous

The following is a part of a problem 18.2 from from Real Analysis, N. L. Carothers:

Let $f: \mathbb{R} \rightarrow [0, \infty]$ be integrable and define $F: [0, \infty) \rightarrow [0,\infty]$ by $F(\alpha) = m\{f>\alpha\}$. Show that $F$ is right continuous. ($m$ denotes the Lebesgue measure.)

I am able to see that $F$ is decreasing and that if $y>x$, then $F(x) - F(y) = m\{ x<f\le y\}$. If I'm able to make the $m\{x < f\le y\}$ arbitrarily small, I'm done. But I'm stuck here, can you help with a hint or answer?

A:

Hints:

$m$ is continuous from below, i.e. for any sequence $(A_j)_{j \in \mathbb{N}}$ of measurable sets, $A_1 \subseteq A_2 \subseteq A_3 \subseteq \dots$, we have $$m \left( \bigcup_{j \in \mathbb{N}} A_j \right) = \lim_{j \to \infty} m(A_j).$$
Let $(\alpha_j)_{j \in \mathbb{N}}$ be a sequence such that $\alpha_j \downarrow \alpha$. Show that $$A_j := \left\{f>\alpha_j \right\}$$ satisfies all assumptions from step 1 and $$\{f>\alpha\} = \bigcup_{j \in \mathbb{N}} A_j.$$
Conclude

