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 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 $\chi_{[0, \infty )}(x)=\begin{cases} 1 & {if } x \ge 0 \\ 0 & {if } x<0. \end{cases}$ 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 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 $b \ge 0$ 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, $g^{-1}(U) \neq \emptyset$ 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 $0<m(g^{-1}(U)) \le m(A)=0$ a contradiction.

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