Suppose $f$ and $g$ are continuous functions on $[a,b]$. Show that if $f=g$ a.e. 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 $E$?
OK. So here's my thought:
Set $A = \{ x \in [a,b]: f(x) \neq g(x)\}$, we're going to show that $A = \emptyset$.
We have $A = \{ x \in [a,b]: f(x) – g(x) \neq 0\} = [a,b] – \{ x \in [a,b]: f(x)-g(x)=0 \}=[a,b]-B$.
Since $f$ and $g$ both are continuous on $[a,b]$ the function $(f-g)(x)=f(x)-g(x)$ is well-defined on $[a,b]$ and is continuous. Therefore, since $B = (f-g)^{-1}(\{0\})$ and singletons are closed in a metric space, we conclude that $B$ is a closed set in $[a,b]$ and therefore $A$ is an open set in $[a,b]$.
The non-empty open sets in $[a,b]$ are one of these forms: $[a,x)$, $(x,b]$, $(x,y)$ or $[a,b]$ itself for $x<y \in (a,b)$ or we can have a union of these sets. The measure of all these sets is positive. Hence, contradiction. So, $A$ must be equal to the empty set. Q.E.D.
For the second part, if my proof is correct, then since I never used the fact that $[a,b]$ is closed, compact or any special set, I think we can generalize the proof to general measurable sets as long as they contain an open set. Am I right?
For example, if $E$ is the Cantor set, we can't generalize our proof to $E$.
Am I right?
I'm using the Lebesgue measure on $\mathbb{R}$. Whatever I say here is supposed to be true when we're using the Lebesgue measure on $\mathbb{R}$, not other possible measures.
Best Answer
Your proof goes wrong here "The non-empty open sets in [a,b] are one of these forms: [a,x), (x,b], (x,y) or [a,b] itself..."
That statement about open sets is just wrong. For instance, the union of any two such sets is also open. What IS true is that every open set contains a set of one of those three types. And if it contains $[a, x)$, it also contains $( \frac{a+x}{2}, x)$. So it alwasy contains one of the third type, whose measure is positive.
For your second part: what can you say about a measurable set that contains an open set? Also: would the statement be true if your set, was, say, the Cantor set $E$ together with the interval $(0, 0.1)$? Or do you need something MORE about open sets in relation to your set?