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.
 A: Can I prove like this:
Assume that $A\neq \emptyset$. There exists $x_0\in[a,b]$ such that $f(x)-g(x)\neq 0$, WLOG, say $f(x)-g(x)=a>0$. Since $f,g$ are continuous, $f-g$ is also continuous. By the property of continuous function, for any $\epsilon$ there exists $\delta$ such that $|x-x_0|<\delta$ then $|(f-g)(x)- a|=|(f-g)(x) -(f-g)(x_0)|<\epsilon.$ By choosing $\epsilon = \frac{a}{2}$, $|(f-g)(x)-a|<\frac{a}{2}\to 0<\frac{a}{2}<(f-g)(x)< \frac{3a}{2}$. That means $f\neq g$ for all $x\in \{x:|x-x_0|<\delta\}\subset A$. Since $m \{x:|x-x_0|<\delta\}=2\delta>0$, we have a contradiction.
A: A nonempty open subset of $[a,b]$ always has positive measure. Does that help?
More Hints: The inverse image of an open set under a continuous function is _ _ _ _? A nonempty open set always contains an open interval.
A: 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.
A: Hint: Think about topological properties of $A$ and you'll get your contradiction.
A: 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 $$\int_{a}^{b} |f-g|\ dx = 0 \iff |f-g| \equiv 0 \iff f\equiv g.$$
