Example of non-locally integrable $f$ that $\int_\Omega f\varphi=0$.

Question

In real analysis we know that if $f\in L_{loc}^1(\Omega)$ and $\int_\Omega f\varphi=0$ for any $\varphi\in C_0^\infty(\Omega)$, then $f=0$ a.e. I am thinking if it is possible to find an example to show that the condition $f\in L_{loc}^1(\Omega)$ is necessary.

Answer 1

The condition $f\in L_{loc}^1(\Omega)$ is necessary for the statement to even make sense. If $f\varphi$ is integrable for all $\varphi\in C_0^\infty(\Omega)$, then $f\in L_{loc}^1(\Omega)$. Since we are considering the properties of locally integrableness and smoothness, I assume we are considering open sets $\Omega\subseteq\mathbb{R}^n$.

Proof of claim: Let $\Omega\subseteq\mathbb{R}^n$ open and $f:\Omega\to\mathbb{R}$ such that $f\varphi$ is integrable for all $\varphi\in C_0^\infty(\Omega)$. Let $K\subseteq\Omega$ compact. We can find (*) $U\subseteq\Omega$ open and $L\subseteq\Omega$ compact such that $K\subseteq U\subseteq L$. By smooth Urysohn we find $\varphi\in C^\infty$ such that $\varphi|_{K}\equiv1$ and $\mbox{supp}(\varphi)\subseteq U$. Because $U\subseteq L\subseteq\Omega$ we get $\varphi\in C_0^\infty(\Omega)$. By hypothesis, $f\varphi$ is integrable on $\Omega$, and hence also on $K$. Since $(f\varphi)|_K=f|_K$ we find that $f$ is integrable on $K$. Hence $f\in L_{loc}^1(\Omega)$.

(*) This is a bit tedious. Since $K$ is compact, it is contained in some ball $B_R(0)$. Then $S=(\Omega\cap B_R(0))^c$ is closed and disjoint from $K$. Because $\mathbb{R}^2$ is normal, we find disjoint open sets $U,V\subseteq\mathbb{R}^2$ such that $K\subseteq U$ and $S\subseteq V$. Choosing $L=V^c$ we get the desired properties.