Let $X$ be a locally compact Hausdorff space, and $\mu$ be a regular positive Borel measure on $X$. For any $p\in \mathbb{R}$, we denote by $L^p(X)$, the space $L^p(X, \mu)$.
My question is: If a function $f\in L^p(X)$ vanishes almost everywhere except on a compact set $A$ in $X$, then without loss of generality, can we assume $f$ to be a continuous function? Is there a proof or a counter-example? (We may assume that $1<p<2$, if necessary).
I saw this argument somewhere, and it seemed new. I understand that in general, there will always be a $g\in C_c(X)$ such that $||f-g||_p<\epsilon$. But I do not see immediately how the compactness of the essential support of $f$ leads to the above assumption.
In my understanding, the question boils down to: Does there exist a continuous function in the equivalence class of $f$ in $L^p(X)$?
Any help is greatly appreciated.
The answer is no. Let $f(x)=0$ for $x <0$, $1$ for $0\leq x <1$ and $0$ for $x \geq 1$. Then $f$ vanishes outside the compact set $[0,1]$. If there is a continuous function $g$ such that $f=g$ almost everywhere then, since the complement of a set of measure $0$ is dense, it follows that $g(x)=0$ for all $x <0$ and $g(x)=1$ for all $x\in [0,1]$. This violates continuity of $g$ at $0$.
[ We have $g(x)=0$ almost everwhere on $(-\infty,0)$. Suppose $g(x) \neq 0$ for some $x <0$. Then, by continuity, $g(y) \neq 0$ for all $y$ is some interval $(x-r,x+r)$. But this interval has positive measure , so $g(y) \neq 0$ on a set of positive measure. This contradiction shows that $g(x)$ must be $0$ for all $x <0$. A similar argument shows that $g(x)=1$ for $0 <x<1$.