If the function $f:G \to \mathbb{R}$ with $G$ a domain in $\mathbb{R}^n$,and $f$ is continuous.
Prove if $f = 0$ almost everywhere(In Lebesgue measure) then $f = 0$ everywhere.
My attempt: w.l.o.g assume $f(x)>0$ for some $x$,since $f$ is continuous ,there exist a neighborhood of $x$ with all $f(y)>0$ on the neighborhood,and the neighborhood is not measure zero.So we have the result.
Is my proof correct?
Best Answer
Yes, your proof is correct.
You can make it even more precise (but maybe cumbersome) using a $\delta$-$\epsilon$ argument. Since $f$ is continuous at $x$, for every $\varepsilon>0$ there exists $\delta>0$ such that $||x-y||<\delta$ implies $|f(x)-f(y)|<\varepsilon$. Since $f(x)>0$, say $f(x)=l$, we can take $\varepsilon=\frac{l}{2}$ and so $f(y)>\varepsilon>0$. This is true for all $y\in B(x,\delta)$ -the open ball of center $x$ and radius $\delta$-, whose measure is positive (you can even look at the volume of the ball if you want to check that). So we have the contradiction.