Smoothing measure by convolution

Question

I'm reading Optimal Transport for Applied Mathematicians by Fillipo Santambrogio. At one point he says that taking the convolution of a measure with a smooth mollifier, i.e $\eta_\varepsilon \in C^\infty_c (X)$ such that $$ \int_{\mathbb{R}^d} \eta_\varepsilon = 1$$ then considering $$ \mu * \eta_\varepsilon $$ we obtain an absolutely continuous (with respect to the d dimensional lebesgue measure) smoothing of $\mu$. However, I am somewhat struggling to understand how this is the case. For example, if I consider the dirac measure $\delta_0$ at 0 on $\mathbb{R}$, then take $\eta_\varepsilon * \delta_0$, I should have $$ \eta_\varepsilon * \delta_0 (\{0\}) = \int_{\{0\}} 1 d\eta_\varepsilon * \delta_0 = \int_{\{0\}} 1 * \eta_\varepsilon d(\delta_0) = \int_{\{0\}} \left( \int_\mathbb{R} 1 \cdot \eta_\varepsilon(z) dz \right) d(\delta_0) = \int_{\{0\}} 1 d(\delta_0) = 1$$ which would seem to imply that $\delta_0 * \eta_\varepsilon$ gives mass to a set of lebesgue measure zero, which obviously means it cannot be absolutely continuous. What am I missing here?

Answer 1

Since $\delta_0$ is the neutral element for convolution, you have $\eta_{\epsilon}\star \delta_0=\eta_{\epsilon}$ which is smooth. Your computation went wrong, you should have written, since $\delta_0$ is a distribution, $$(\delta_0\star\eta_{\epsilon})(x)=\left<\delta_0,\eta_\epsilon(x-\cdot)\right>=\eta_\epsilon(x-0)=\eta_\epsilon(x).$$