$\phi_{\epsilon} \ast \mu \rightarrow \mu$? Let $\phi$ be a non-negative function on $\mathbb{R}$ with $\int_{\mathbb{R}} \phi = 1$. Define $\phi_{\epsilon}(x)=\epsilon^{-1}\phi(\epsilon^{-1}x)$ for $x \in \mathbb{R}, \epsilon > 0$. 
For $f \in L^1$, $\phi_{\epsilon} \ast f \rightarrow f$ in $L^1$ as $\epsilon \rightarrow 0$. (cf. Theorem 8.14 of Folland's Real Analysis).
Can we replace $f$ by a probability measure $\mu$ to get something like $\phi_{\epsilon} \ast \mu \rightarrow \mu$ weakly as $\epsilon \rightarrow 0$?
If so, can you show me (or point me to a reference containing) the proof?
If $\mu$ has a density function $f \in L^1$, the conjecture is true because $g (\phi_{\epsilon} \ast f) \rightarrow g f$ in $L^1$ for any $g \in L^{\infty}$.
 A: Let $\phi\geq 0$ be a Lebesgue integrable function with $\int_{\mathbb{R}}\phi=1$, and let $\mu$ be a (Borel) probability measure on $\mathbb{R}$. Define $\phi_{\delta}$ as above (I want to reserve the letter $\epsilon$ for later).
Let $g$ be a continuous, bounded function on $\mathbb{R}$. Observe that
$$\int_{\mathbb{R}}\int_{\mathbb{R}}g(x+y)\phi_{\delta}(x)dxd\mu(y)=\int_{\mathbb{R}}\left(\int_{\mathbb{R}}\tilde{g}(-y-x)\phi_{\delta}(x)dx\right)d\mu(y)=\int_{\mathbb{R}}\widetilde{(\tilde{g}\ast\phi_{\delta})}(y)d\mu(y)$$
where $\tilde{g}(x):=g(-x)$.
Recall that for any bounded, continuous function $h:\mathbb{R}\rightarrow\mathbb{C}$, $h\ast\phi_{\delta}\rightarrow h$, as $\delta\rightarrow 0$, uniformly on compact subsets of $\mathbb{R}$. Given $\epsilon>0$, take $N>0$ sufficiently large so that $\mu([-N,N]^{c})<\epsilon$. Let $\delta_{0}>0$ be a sufficiently small so that 
$$\delta\leq\delta_{0}\Longrightarrow\left\|\widetilde{(\tilde{g}\ast\phi_{\delta})}-g\right\|_{L^{\infty}([-N,N])}=\left\|\tilde{g}\ast\phi_{\delta}-\tilde{g}\right\|_{L^{\infty}[-N,N]}<\epsilon$$
By Young's convolution inequality, we have that
$$\left\|\widetilde{(\tilde{g}\ast\phi_{\delta})}\right\|_{L^{\infty}}=\left\|\tilde{g}\ast\phi_{\delta}\right\|_{L^{\infty}}\leq\left\|\tilde{g}\right\|_{{\infty}}\left\|\phi_{\delta}\right\|_{L^{1}(\mathbb{R})}=\left\|g\right\|_{\infty},\quad\forall\delta>0$$
Using these two estimates, we see that for all $\delta\leq\delta_{0}$,
\begin{align*}
\left|\int_{\mathbb{R}}\widetilde{(\tilde{g}\ast\phi_{\delta})}d\mu(y)-\int_{\mathbb{R}}g(y)d\mu(y)\right|&\leq\int_{[-N,N]}\left|\widetilde{(\tilde{g}\ast\phi_{\delta})}(y)-g(y)\right|d\mu(y)\\
&+\int_{[-N,N]^{c}}(\left\|\widetilde{(\tilde{g}\ast\phi_{\delta})}\right\|_{L^{\infty}}+\left\|g\right\|_{L^{\infty}})d\mu(y)\\ & \\
&\leq\epsilon\mu([-N,N])+2\left\|g\right\|_{L^{\infty}}\mu([-N,N]^{c})\\ & \\
&\leq\epsilon+2\left\|g\right\|_{L^{\infty}}\epsilon
\end{align*}
Since $\epsilon>0$ was arbitrary, we conclude that
$$\lim_{\delta\rightarrow 0}\int_{\mathbb{R}}\int_{\mathbb{R}}g(x+y)\phi_{\delta}(x)dxd\mu(y)=\int_{\mathbb{R}}g(y)d\mu(y)$$
I.e. $\phi_{\delta}\ast\mu\rightarrow \mu$ weakly.
