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.

  • 4
    $\begingroup$ I believe the $f\in L_{loc}^1(\Omega)$ is just necessary such that $\int_\Omega f\varphi$ is well-defined. $\endgroup$ Dec 24, 2018 at 23:58
  • $\begingroup$ I suppose here we mean $\int_\Omega f\ \mathsf d\varphi$ where $\varphi$ is some measure? $\endgroup$
    – Math1000
    Dec 25, 2018 at 0:03
  • 1
    $\begingroup$ @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 $f\varphi \in L^1(\Omega)$? $\endgroup$ Dec 25, 2018 at 0:11
  • 1
    $\begingroup$ @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. $\endgroup$ Dec 25, 2018 at 1:43
  • 1
    $\begingroup$ @mathworker21 This seems like something for a meta post. math.meta.stackexchange.com/questions/29537/… $\endgroup$ Dec 25, 2018 at 2:31

1 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.

  • $\begingroup$ +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. $\endgroup$ Dec 25, 2018 at 1:40
  • $\begingroup$ Good point, and it is less obvious than you might think, so I edited the post. I think the argument even becomes a bit nicer. $\endgroup$ Dec 25, 2018 at 2:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .