For a $$f \in L^2(\mathbb{R})=L^2(\mathbb{R})^* \subset D^*$$ so we can regard it as a distribution.
So we have support set of f as a distribution. $$supp f=\{x\in \mathbb{R}: \text{for any open neighborhood U of x , restrictions of f on U is not }0\}.$$
Actually $f \in L^2$ as a function we have $$Supp f=\overline{\{x\in \mathbb{R}:f(x) \neq 0\}}$$ But they are different. For example ,$f(x)=1$,when x is rational,otherwise $f=0$. For f as a distribution we have $supp f$ is empty,but $Supp f$ is the $\mathbb{R}$.
Here is my question: For a nonnegative function $f\in L^2$ , do we have $f>0 \quad a.e$ on its support set $supp f$.
Any suggestion will be appreciated.
Look at $$f=\sum_{n\ge 1}\sum_{k =0}^{2^n} 1_{\displaystyle [\frac{k}{2^n}-\frac1{2^{2n+2}},\frac{k}{2^n}+\frac1{2^{2n+2}}]} \in L^2[0,1]$$
$f\ge 1$ on a neighborhood of each $k/2^n$ so $$supp(f)= [0,1]$$ but $f$ vanishes on the set with non-zero measure $$[0,1] - \bigcup_{k,n}(\frac{k}{2^n}-\frac1{2^{2n+2}},\frac{k}{2^n}+\frac1{2^{2n+2}})$$
($f>0$ a.e. on its support means that $\mu(f^{-1}(0)\cap supp(f)) = 0$ right ?)