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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 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 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 $\mathbb{R}$. 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 $\varepsilon\in(0,1)$.)
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