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A question on locally integrable function on $mathbb{R}^n$

Mathematics Asked by Kevin_H on December 8, 2020

I am currently doing some practice problems for the analysis qual. I have some thoughts on the following problem, and it would be great if someone could see if I am on the right track:

Problem:
Let $Osubset mathbb{R}^n$ be an open set, and let $fin L_{loc}^1(mathbb{R}^n)$ (the set of locally integrable functions on $mathbb{R}^n$). Assume that for all $phi in C_c(O)$ (the set of continuous functions with compact support in $O$), we have $$int_Ofphi dm = 0,$$ where $dm$ stands for Lebesgue measure. Prove that $f(x) = 0$ for a.e. $xin O$.

My thoughts are as follows: I want to show by contradiction that if there exists $f in L_{loc}^1(mathbb{R}^n)$ s.t. there is a Lebesgue measurable set $E$ with $m(E) > 0$ and $|f| > 0$ on $E$, then one can find some $phi in C_c(O)$ s.t. $$int_Ofphi dm ne 0.$$ Since $E$ is with positive measure, then by inner regularity of Lebesgue measure we can find a nonempty compact set $K$ s.t. $K subset U$. Then by Urysohn lemma, there exists a nonnegative function $phi in C_c(O)$ s.t. $phi|_K = 1$ and $phi$ vanishes outside $O$. Now I claim that $phi$ does the job: Note that $$int_O|f|phi = int_E|f|phi ge int_K|f|phi = int_K|f| > 0,$$ where the last inequality is by the fact that $Ksubset E$. But now I don’t find a good way to conclude that $$int_Ofphi dm ne 0.$$
Could anyone help me fill in this gap or point out which place goes wrong here? Much appreciated in advance!

One Answer

Hint:

there are compacts sets $K_n$ such that $K_nsubsetoperatorname{Int}(K_{n+1})$ and $O=bigcup_nK_n$.

You may try to show that $f=0$ in $K_n$. Any measurable subset $E$ of $K_n$ can be approximated in $L_1$ by a sequence of continuous functions with support in $operatorname{Int}(K_{n+1})$ and which are uniformly bounded. Then by dominated convergence $int fmathbb{1}_E=0$.

This implies that $f=0$ in $K_n$ for each $K_n$.

Correct answer by Oliver Diaz on December 8, 2020

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