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 $f:ℝ\to ℝ$:

• $f$ 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

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

Suppose ${\chi }_{[0,\mathrm{\infty })}$ is equal to a continuous function almost everywhere. Let $A\subset ℝ$ be Lebesgue measurable (http://planetmath.org/LebesgueMeasure) with $m(A)=0$ and $g:ℝ\to ℝ$ such that ${\chi }_{[0,\mathrm{\infty })}(x)=g(x)$ for all $x\in ℝ\setminus A$. Since ${\chi }_{[0,\mathrm{\infty })}(x)=0$ for all and $m(A\cap (-\mathrm{\infty },0))=0$, there exists such that $g(a)=0$. Similarly, there exists $b\ge 0$ 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, $g^{-1}(U)\ne \mathrm{\varnothing }$. 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 , a contradiction.

Now consider the function

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 $x\in ℝ$. Recall that $\mathbb{Q}$ and $ℝ\setminus \mathbb{Q}$ are both dense in (http://planetmath.org/Dense) $ℝ$. Therefore, for every $x\in ℝ$ and for every $\delta >0$, there exist $x_1\in (x-\delta ,x+\delta )\cap \mathbb{Q}$ and $x_2\in (x-\delta ,x+\delta )\cap (ℝ\setminus \mathbb{Q})$. 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 $\epsilon \in (0,1)$.)