If two continuous functions are equal almost everywhere on $[a,b]$, then they are equal everywhere on $[a,b]$ Suppose $f$ and $g$ are continuous functions on $[a,b]$. Show that if $f=g$ almost everywhere on $[a,b]$, then, in fact, $f=g$ on $[a,b]$. Is a similar assertion true if $[a,b]$ is replaced by a general measurable set $E$?

I have known that the set $A=\{x \mid f(x) \neq g(x)\}$ has measure zero and we want to show that A is empty. Now let's assume $A$ is not empty. I am stuck in getting the contradiction.

Thanks for your hints and answers.
You can do it even more simply:

$f,g$ continuous on $[a,b]$ implies $(f-g)$ continuous.  Thus:

$$(f-g)^{-1}(0, + \infty)\cup(0, -\infty) = A$$

is open, thus $A=\emptyset$ as that is the only open set of measure zero.  QED. No epsilonics required.
If $f = g$ almost everywhere on $[a,b]$ then $|f - g| = 0$ almost everywhere on $[a,b].$ Let $\mu$ denote the Lebesgue measure on $[a,b].$ Then $$\int_{a}^{b} |f-g|\ dx = \int_{[a,b]} |f-g|\ d\mu = 0.$$ Since $f,g \in C[a,b]$ it follows that $|f - g| \in C[a,b]$ and moreover $|f-g| \geq 0$ on $[a,b].$ So $$\int_{a}^{b} |f-g|\ dx = 0 \iff |f-g| \equiv 0 \iff f\equiv g.$$