Let $f$ be a uniformly continuous function with at most polynomial growth at infinity, and $m$ a finite (non-negative) measure, both defined on $X=\mathbb R^n$. By $\bullet_\varepsilon$ we denote convolution with the standard mollifier $\varphi_\varepsilon$ on $X$, that is the density of the normal distribution with mean $0$ and variance $\varepsilon>0$.
Statement: For any $\eta>0$ there is $\varepsilon>0$ such that, for all $z\in X$, \begin{equation} |(f \cdot m_\varepsilon - (f\cdot m)_\varepsilon)(z)| \le \eta \cdot m_\varepsilon(z). \end{equation}
How to prove it?