Not every distribution is a locally integrable function

Suppose $u_f={\delta }_0$ with $f\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1({\mathbb{R}}^n)$. On the open domain $U={\mathbb{R}}^n\setminus \{0\}$, all tests evaluate to zero at the origin, so F2 gives $u_{f\mathrm{\mid }U}=0$. F1 applied on this entire open domain, without selecting pointwise neighborhoods, gives $f=0$ almost everywhere on $U$. By F4 the omitted singleton is null, so $f=0$ almost everywhere on ${\mathbb{R}}^n$.

Consequently $u_f=0$, but F3 supplies $\phi$ with $\phi (0)=1$, and F2 gives ${\delta }_0(\phi )=1$. This contradiction proves that the witness is not regular and refutes the proposed universal statement. The zero function does represent the zero distribution; the failure is the nonzero point mass, not a failure of the regular-distribution construction. Dimension zero is excluded, since its singleton has mass one in the library convention. $\mathrm{\square }$