# Lusin’s Theorem and Continuous Extension

Here we give proofs for two versions of Lusin’s Theorem, one from Exercise 44, Ch2 in Folland’s Real Analysis and the other from the textbook used for my mathematical analysis course in Beijing.  The latter version is a stronger result which in addition discusses the condition for a real-valued function defined on a subset of $\mathbb{R}^n$ to be extended to the whole of $\mathbb{R}^n$. A more general result in topology is the Tietze Extension Theorem.

See the full post here: Lusin’s Theorem and Continuous Extension

Here we let $\mu$ denote the Lebesgue measure on $\mathbb{R}$.

Lusin’s Theorem (Version 1)[Exercise 2.44, Folland]. Suppose $E\subset \mathbb{R}^n$ is Lebesgue  measurable, $f: E\to \mathbb{R}$ is Lebesgue measurable and $\epsilon> 0$, there is a compact set $F\subset E$ such that $\mu(F^c)<\epsilon$ and $f|_F$ is continuous.

Lusin’s Theorem(Version 2)[Huan]. Suppose $E\subset \mathbb{R}^n$ is Lebesgue measurable and $f: E\to \bar{\mathbb{R}}$ is a Lebesgue measurable extended real valued function with $\mu(|f|=+ \infty)=0$, then  $\forall \epsilon >0$, $\exists g\in C(E)$ such that $\mu(f\neq g)<\epsilon$, where $C(E)$ denotes the space of continuous function on $E$

Continuous Extension Theorem[Huan]. Suppose $E\subset \mathbb{R}^n$, then $f$ can be extended to a continuous function on $\mathbb{R}^n$ if and only if $f$ can be extended to a continuous function on the closure $\bar{E}$ of $E$.

Tietze Extension Theorem. Let $X$ be normal and $F \subset X$ be closed and let $f: F \to R$ be continuous. Then there is a map $g: X \to R$ such that
$g(x) = f(x)$ for all $x\in F$. (Note that in topology, by a map we mean a continuous function. )

# The “Dunce Cap” Space Is Contractible

Here is the exercise 6 on P. 50 in the book Topology and Geometry by Glen Bredon. I put it here because I found the drawing of this cap very lovely. Indeed I like that most of the pictures in this book are lovely sketches.

Question. The “dunce cap” space is the quotient of a triangle (and interior) obtained by identifying all three edges in an inconsistent manner. That is, if the vertices of the triangle are $p, q, r$ then we identify the line segment $(p, q)$ with $(q, r)$ and with $(p, r)$ in the
orientation indicated by the order given of the vertices. (See Figure 1-6.) Show that
the dunce cap is contractible.

Following the development in the book, I will use the following theorem that the homotopy type of a mapping cylinder or cone
depends only on the homotopy class of the map [Theorem 14.18, Topology and Geometry by Glen Bredon]. The idea is to identify the dunce cap as a mapping cone.

Theorem 14.18. If $f_0\simeq f_1:X\to Y \text{ then } M_{f_0 } \simeq M_{f_1}\text { rel } X+Y \text{ and }C_{f_0}\simeq C_{f_1}\text{ rel } Y+\mathrm{vertex}.$

Proof of the Qustion. Suppose $f: S^1\to S^1$ is a map from $S^1$ to itself. The cone $C_f=M_f/S^1 \times\{1\}$ for $f$ is obtained by pinching the top of the mapping cylinder to a point. As $M_f$  is the cylinder $S_1\times [0,1]$ with the bottom pasted to $S_1$ by the map $f$, $C_f$ is $D_2$ with $\partial D_2$ pasted to $S_1$ by the map $f$. So the dunce cap is just $C_f \text{ with } f: S^1\to S^1$ defined as

$f(e^{2\pi i t})= \begin{cases} e^{2\pi i (3t)}, 0\leq t\leq 2/3\\ e^{2\pi i(2- 3t)}, 2/3\leq t\leq 1. \end{cases}$

which is homotopic to the identity by a linear homotopy (note that we make the choice of $f$ for an easy definition of the homotopy)

$H(e^{2\pi i t},s)= \begin{cases} e^{2\pi i (3t(1-s)+st}, 0\leq t\leq 2/3\\ e^{2\pi i[(2- 3t)(1-s)+st]}, 2/3\leq t\leq 1. \end{cases}$.

So the dunce cap is homotopic to $C_{id}\simeq D^2$ which is contractible.