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.
Best Answer
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.