continuous almost everywhere versus equal to a continuous function almost everywhere

The concept of almost everywhere can be somewhat tricky to people who are not familiar with it. Let $m$ denote Lebesgue measure. Consider the following two statements about a function $f\colon\mathbb{R}\to\mathbb{R}$ :

•

$f$ is continuous (http://planetmath.org/Continuous) almost everywhere with respect to $m$
•

$f$ is equal to a continuous function almost everywhere with respect to $m$

Although these two statements seem alike, they have quite different meanings. In fact, neither one of these statements implies the other.

Consider the function $\chi_{[0,\infty)}(x)=\begin{cases}1&\text{if }x\geq 0\\ 0&\text{if }x<0.\end{cases}$

This function is not continuous at $0$ , but it is continuous at all other $x\in\mathbb{R}$ . Note that $m(\{0\})=0$ . Thus, $\chi_{[0,\infty)}$ is continuous almost everywhere.

Suppose $\chi_{[0,\infty)}$ is equal to a continuous function almost everywhere. Let $A\subset\mathbb{R}$ be Lebesgue measurable (http://planetmath.org/LebesgueMeasure) with $m(A)=0$ and $g\colon\mathbb{R}\to\mathbb{R}$ such that $\chi_{[0,\infty)}(x)=g(x)$ for all $x\in\mathbb{R}\setminus A$ . Since $\chi_{[0,\infty)}(x)=0$ for all $x<0$ and $m(A\cap(-\infty,0))=0$ , there exists $a<0$ such that $g(a)=0$ . Similarly, there exists $b\geq 0$ such that $g(b)=1$ . Since $g$ is continuous, by the intermediate value theorem, there exists $c\in(a,b)$ with $g(c)=\frac{1}{2}$ . Let $U=(0,1)$ . Since $g$ is continuous, $g^{-1}(U)$ is open. Recall that $c\in g^{-1}(U)$ . Thus, $g^{-1}(U)\neq\emptyset$ . Since $g^{-1}(U)$ is a nonempty open set, $m(g^{-1}(U))>0$ . On the other hand, $g^{-1}(U)\subseteq A$ , yielding that $0<m(g^{-1}(U))\leq m(A)=0$ , a contradiction.

Now consider the function $\chi_{\mathbb{Q}}(x)=\begin{cases}1&\text{if }x\in\mathbb{Q}\\ 0&\text{if }x\notin\mathbb{Q}.\end{cases}$

Note that $m(\mathbb{Q})=0$ . Thus, $\chi_{\mathbb{Q}}=0$ almost everywhere. Since $0$ is continuous, $\chi_{\mathbb{Q}}$ is equal to a continuous function almost everywhere. On the other hand, $\chi_{\mathbb{Q}}$ is not continuous almost everywhere. Actually, $\chi_{\mathbb{Q}}$ is not continuous at any $x\in\mathbb{R}$ . Recall that $\mathbb{Q}$ and $\mathbb{R}\setminus\mathbb{Q}$ are both dense in (http://planetmath.org/Dense) $\mathbb{R}$ . Therefore, for every $x\in\mathbb{R}$ and for every $\delta>0$ , there exist $x_{1}\in(x-\delta,x+\delta)\cap\mathbb{Q}$ and $x_{2}\in(x-\delta,x+\delta)\cap(\mathbb{R}\setminus\mathbb{Q})$ . Since $\chi_{\mathbb{Q}}(x_{1})=1$ and $\chi_{\mathbb{Q}}(x_{2})=0$ , it follows that $\chi_{\mathbb{Q}}$ is not continuous at $x$ . (Choose any $\varepsilon\in(0,1)$ .)

Title	continuous almost everywhere versus equal to a continuous function almost everywhere
Canonical name	ContinuousAlmostEverywhereVersusEqualToAContinuousFunctionAlmostEverywhere
Date of creation	2013-03-22 15:58:47
Last modified on	2013-03-22 15:58:47
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	10
Author	Wkbj79 (1863)
Entry type	Example
Classification	msc 28A12
Classification	msc 60A10