Does $f(x)$ is continuous and $f = 0$ a.e. imply $f=0$ everywhere? I wanna prove that 
"if $f: \mathbb{R}^n \to \mathbb{R}$ is continuous and satisfies $f=0$ almost everywhere (in the sense of Lebesgue measure), then, $f=0$ everywhere."
I am confident that the statement is true, but stuck with the proof. Also, is the statement true if the domain $\mathbb{R}^n$ is restricted to $\Omega \subseteq \mathbb{R}^n$ that contains a neighborhood of the origin "$0$"?

A set of measure zero has dense complement.  So if a continuous function zero on a set of full measure, it is identically zero.
