Math Problems Vault

Problem 5.7.1

(a)(b)(c)

In Lusin's Theorem, we may replace the condition that $f$ has range contained in $\mathbb {R}$ with the more general assumption that $f$ is finite $\mu$ -a.e.on $A$ .

Lusin's Theorem (the result to be generalized here): Consider the space $(\mathbb {R}^{k}, \mathcal{F}, \mu )$ , where $\mathcal{F}$ contains $\mathcal{B}^{k}$ , and where $\mu$ is such that (i) $\mu (A) = \inf \left\{ \mu (G) : A \subseteq G, G \text{ open}\right\}$ for every $A \in \mathcal{F}$ , and (ii) $\mu (A) = \sup \left\{ \mu (K) : K \subseteq A, K \text{ compact}\right\}$ for every $A \in \mathcal{F}$ . Let $A \in \mathcal{F}$ be such that $\mu (A)<+\infty$ , and suppose that $f : \mathbb {R}^{k} \rightarrow \mathbb {R}$ is $\mathcal{F}/\mathfrak {B}^{*}$ -measurable with $f(\omega )=0$ for each $\omega \notin A$ . Then for any $\epsilon >0$ there exists a function $g : \mathbb {R}^{k} \rightarrow \mathbb {R}$ with the following properties:

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(a)

$g$ is continuous and has compact support.

(b)

$\mu (\left\{ \omega \in \mathbb {R}^{k} : f(\omega ) \neq g(\omega )\right\} ) < \epsilon$ .

(c)

$\sup_{\omega \in \mathbb {R}^{k}} \left|g(\omega )\right| \leqslant \sup_{\omega \in \mathbb {R}^{k}} \left|f(\omega )\right|$ .

Problem 5.7.2

(a)(b)(c)

If $\left|f\right| \leqslant 1$ in Lusin's Theorem, there exists a sequence $\left\{ g_{n}\right\}_{n=1}^{\infty }$ of $\mathcal{F}/\mathfrak {B}^{*}$ -measurable functions such that

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(a)

$\left|g_{n}\right| \leqslant 1$ for all $n \in \mathbb {N}$ ,

(b)

$g_{n}$ is continuous and has compact support for each $n \in \mathbb {N}$ , and

(c)

$\lim g_{n} = f$ $\mu$ -a.e.

Remark.

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Lusin's Theorem is stated in problem 5.7.1's task.

Problem 5.7.3

Let $F \subseteq G \subseteq \mathbb {R}^{k}$ , where $F$ is closed and $G$ is open. Without using Urysohn's Lemma (Claim 2 below), quickly construct a lower semicontinuous $\phi : \mathbb {R}^{k} \rightarrow [0,1]$ such that $\phi (\boldsymbol {x})=1$ if $\boldsymbol {x} \in F$ and $\phi (\boldsymbol {x})=0$ if $\boldsymbol {x} \notin G$ . Then quickly construct an upper semicontinuous $\phi$ with the exact same properties.

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Remark.

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Take $\phi =\; \mathbb {1}_{F}$ and $\phi =\; \mathbb {1}_{G}$ . The point of Urysohn's Lemma is that we can state something much stronger, namely, that there exists a continuous $\phi$ with these properties. Claim 2 (Urysohn's Lemma, special case): Let $\varnothing \subsetneq F \subseteq G \subseteq \mathbb {R}^{k}$ , where $F$ is closed and $G$ is open. Then there exists a continuous function $\phi : \mathbb {R}^{k} \rightarrow \mathbb {R}$ such that (i) $0 \leqslant \phi (\boldsymbol {x}) \leqslant 1$ for each $\boldsymbol {x} \in \mathbb {R}^{k}$ , (ii) $\phi (\boldsymbol {x})=1$ for each $\boldsymbol {x} \in F$ , (iii) $\phi (\boldsymbol {x})=0$ for each $\boldsymbol {x} \in G^{c}$ , and (iv) $0 < \phi (\boldsymbol {x}) < 1$ for each $\boldsymbol {x} \in G - F$ .

Problem 5.7.4

We defined the support of a function $g : \mathbb {R}^{k} \rightarrow \overline{\mathbb {R}}$ to be the closure of the set $\left\{ \boldsymbol {x} \in \mathbb {R}^{k} : g(\boldsymbol {x}) \neq 0\right\}$ , i.e. $\operatorname {supp}\left(g\right) = \overline{\left\{ \boldsymbol {x} \in \mathbb {R}^{k} : g(\boldsymbol {x}) \neq 0\right\} }$ . This is equivalent to defining the support of $g$ to be the complement of the interior of the set $\left\{ \boldsymbol {x} \in \mathbb {R}^{k} : g(\boldsymbol {x}) = 0\right\}$ .

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Problem 5.7.5

Is it the case that every compact subset of $\mathbb {R}^{k}$ is the support of some continuous function $f$ defined on $\mathbb {R}^{k}$ ?

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Problem 5.7.6

Consider the measure space $(\mathbb {R}^{k}, \mathcal{F}, \mu )$ , where $\mu$ assigns finite measure to bounded $\mathcal{B}^{k}$ -sets and $\mathcal{F} \supseteq \mathcal{B}^{k}$ . Let $f : \mathbb {R}^{k} \rightarrow \overline{\mathbb {R}}$ be given. If for every $\epsilon >0$ there exists a closed set $K$ such that (i) $\mu (K^{c})<\epsilon$ and (ii) the restriction of $f$ to $K$ is continuous, then $f$ is $\mathcal{F}/\mathfrak {B}^{*}$ -measurable.

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Problem 5.7.7

Consider the measure space in problem 5.7.6. Let $A \in \mathcal{F}-\left\{ \varnothing \right\}$ be such that $\mu (A)<+\infty$ . Let $\left\{ f_{n}\right\}_{n=1}^{\infty }$ denote a sequence of $\mathcal{F}/\mathfrak {B}^{*}$ -measurable functions that are finite $\mu$ -a.e., and suppose that there exists an $\mathcal{F}/\mathfrak {B}^{*}$ -measurable function $f : A \rightarrow \mathbb {R}$ with $\lim f_{n}=f$ $\mu$ -a.e. Then, for any $\epsilon >0$ , there exists $B \in \mathcal{F}-\left\{ \varnothing \right\}$ with $B \subseteq A$ , $\mu (A-B)<\epsilon$ , and $f_{n} \rightarrow f$ uniformly on $B$ .

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