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Let $E \subset \mathbb{R}$ be measurable and $f: E \rightarrow \mathbb{R}$ measurable. Define the essential range of $f$ to be

$$ R_{f} = \{ x \in \mathbb{R}: \forall \epsilon > 0, m(f^{-1}(B(x, \epsilon ))) > 0 \} $$

where $B(x, \epsilon)$ is the ball (open interval) of radius $\epsilon$ around $x$.

The first part of the problem is to show that $R_{f}$ is closed, which I didn't have a problem with.

Then I need to show that if $f \in L^{\infty}(E)$, then $R_{f}$ is compact.

My idea is to show that $R_{f}$ is a subset of a compact set, and since a closed subset of a compact set is compact, then $R_{f}$ is compact.

I am thinking that because $f \in L^{\infty}(E)$, it is bounded a.e. on $E$, and hence $f(E)$ must be compact. And then $R_{f} \subset f(E)$... so it's compact?

I feel like I am missing something. Any thoughts would be appreciated.

2 Answers 2

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If $|f| \leq M$ almost everywhere, it should be easy to show that if $|x| > M$, then $x \not\in R_f$. This will prove that $R_f$ is bounded.

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Hint: Show that $R_f$ is bounded and use Heine-Borel. In particular, for sufficiently large $a$, $m(|f(x)| > a) = 0$, so $\{x\mid |f(x)| > a\}$ does not intersect $R_f$.