What is the proof for compact essential range of a measurable function?

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Here is this week's POTW:

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Let $(X,\mathcal{M},\mu)$ be a positive measure space. The essential range of a measurable function $f : X \to \Bbb C$ consists of all complex numbers $c$ such that for every $\epsilon > 0$, $\mu(\{x\in X : \lvert f(x) - c\rvert < \epsilon\}) > 0$. Prove that $f$ has compact essential range if $f\in \mathcal{L}^\infty(X,\mu)$.

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This week's problem was solved correctly by GJA. You can read his solution below.

Spoiler

We will make use of the Heine-Borel Theorem by demonstrating that the essential range of $f$ is closed and bounded.

Notation:

• The essential range of $f$ will be denoted $\text{essrng}(f)$.
• An open ball of radius $\varepsilon$ centered at $a$ will be given by $B(a,\varepsilon)$.
• The closed ball will be denoted $\overline{B}(a,\varepsilon)$.

The Essential Range is Bounded

This will follow from the fact that $f\in L^{\infty}(X,\mu)$. Specifically, we will demonstrate that $\text{essrng}(f)\subseteq \overline{B}(0,\|f\|_{\infty}).$

Indeed, suppose $\lvert c\rvert >\|f\|_{\infty}$. Let $M\in\mathbb{R}$ be chosen so that $\|f\|_{\infty}<M<\lvert c\rvert$, and consider the closed ball $\overline{B}(0,M)$. Since $c\notin\overline{B}(0,M)$ there is $\varepsilon_{c} >0$ such that $B(c,\varepsilon_{c})\cap \overline{B}(0,M)=\emptyset$. Now observe that, since $B(c,\varepsilon_{c})\subseteq\mathbb{C}\backslash\overline{B}(0,M)$,

$$\mu\left(\left\{x\in X: \lvert f(x)-c\rvert <\varepsilon_{c} \right\}\right)\leq\mu\left(\left\{x\in X: \lvert f(x)\rvert >M \right\}\right)=0,$$

where the zero is the result of $M>\|f\|_{\infty}.$ Hence, $c\notin\text{essrng}(f)$ and we have $\text{essrng}(f)\subseteq\overline{B}(0,\|f\|_{\infty})$.

The Essential Range is Closed

We will show that the essential range contains all its limit points. Indeed, let $c$ be a limit point of $\text{essrng}(f)$, $\{c_{n}\}$ a sequence in $\text{essrng}(f)$ converging to $c$, and $\varepsilon >0$ be given. Choose $N$ large enough so that $c_{N}\in B(c,\varepsilon /2).$ Consider $B(c_{N},\varepsilon /2)$ and note that $B(c_{N},\varepsilon /2)\subseteq B(c,\varepsilon)$. Now observe that, since $B(c_{N},\varepsilon /2)\subseteq B(c,\varepsilon)$, we have

$$\mu\left(\left\{x\in X: \lvert f(x)-c\rvert <\varepsilon \right\}\right)\geq\mu\left(\left\{x\in X: \lvert f(x)-c_{N}\rvert <\varepsilon /2\right\}\right)>0,$$

where the strict inequality is the result of $c_{N}\in\text{essrng}(f)$. Thus $c\in\text{essrng}(f)$ and therefore $\text{essrng}(f)$ is closed.

By the Heine-Borel Theorem, it follows that $\text{essrng}(f)$ is a compact set.