Let $\delta\in L^1,\delta\ge 0 $ and $\int_{\mathbb{R}^n} \delta(x)dx = 1$, then $\delta_m := m^n\delta(mx)$ is an approximation for the unity of convolution for all $f\in L^p, 1\le p<\infty$.
We want to show that $\lim_{m\to\infty}\| f * \delta_m - f \|_p=0$. Given is that following inequality holds $$ \|f*\delta_m-f\|_p\le \int_{\mathbb{R}^n} \| \tau_y f-f\|_p\delta_m(y)dy,$$ where $\tau_y$ is the translation operator $(\tau_yf)(x) = f(x-y)$. The integral in the RHS can be split in two parts: one where the integration happens over a ball of radius $\rho$ and one where it happens over the complement of this ball: $$ \int_{|y|\le \rho} \| \tau_y f-f\|_p\delta_m(y)dy + \int_{|y|>\rho}\| \tau_y f-f\|_p\delta_m(y)dy.$$ If we let $\rho\to 0$, then the first term will be arbitrary small, since the translation operator is continuous. How can the second term be bounded using this information?
Note that $\|\tau_yf-f\|_p\leq2\|f\|_p$, and for $m$ sufficiently large, $\int_{|y|>\rho}\delta_m(y)dy$ is going to be small.