I've read the following definition:
Let $u \in \mathcal{D}^\prime(\mathbb{R}^n)$. We say that $y_0 \notin \mathrm{sing} \ \mathrm{supp} \ u$ if there exists $\phi \in \mathcal{C}_c^\infty(\mathbb{R}^n)$ such that $\phi(y_0) \neq 0$ and $\phi u \in \mathcal{C}^\infty(\mathbb{R}^n)$.
However, after thinking for a while, I don't understand it very well.
Firstly, in general, the support of a function is a subset of its domain, isn't it? In this case, I guess the singular support is just a type of support -that of singularities-, but the domain of an element in $\mathcal{D}^\prime(\mathbb{R}^n)$ is not $\mathbb{R}^n$ but $\mathcal{C}_c^\infty(\mathbb{R}^n)$. However, $y_0 \in \mathbb{R}^n$. What if $u$ is not a distribution induced by some $L^1_{\text{loc}}(\mathbb{R}^n)$ function?
On the other hand, what's the meaning of $\phi u \in \mathcal{C}^\infty(\mathbb{R}^n)$? I mean, by definition, given $u \in \mathcal{D}^\prime(\mathbb{R}^n)$, we have that for every $\phi \in \mathcal{C}_c^\infty(\mathbb{R}^n)$ and $j = 1, \dots, n$ $ \partial_j u (\phi) = - u(\partial_j \phi). $ So, it doesn't follow that $u$ is infinitely differentiable?
Thanks a lot for your help!