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

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.

• I believe the $f\in L_{loc}^1(\Omega)$ is just necessary such that $\int_\Omega f\varphi$ is well-defined. Dec 24, 2018 at 23:58
• I suppose here we mean $\int_\Omega f\ \mathsf d\varphi$ where $\varphi$ is some measure? Dec 25, 2018 at 0:03
• @SmileyCraft why is that? I mean $f$ is measurable (presumably) and $\varphi \in C_0^\infty(\Omega)$. So $f\varphi$ is measurable. So we just need to know whether $f\varphi \in L^1(\Omega)$. And it's not clear to me that it isn't. Consider, for example, $\Omega = \{|z| < 1\} \subseteq \mathbb{R}^2$ with $f$ blowing up at the boundary. Then isn't it possible that for all smooth functions $\varphi$ vanishing at the boundary of $\Omega$, that $f\varphi \in L^1(\Omega)$? Dec 25, 2018 at 0:11
• @SmileyCraft may I ask why? I generally think more information is better, and in this case, I think seeing productive math conversations and thought processes is helpful. I'm fine deleting though, if you want. Dec 25, 2018 at 1:43
• @mathworker21 This seems like something for a meta post. math.meta.stackexchange.com/questions/29537/… Dec 25, 2018 at 2:31

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.
• +1 nice! Just one (most likely stupid) question. Is it obvious that [$\forall x$ there exists compact $K \ni x$ s.t. $\int_K |f(y)|dy < \infty$] implies [for all compact $K, \int_K |f(y)|dy < \infty$]. The latter is what I thought the definition of $L^1_{loc}$ is. Dec 25, 2018 at 1:40