Let $K\in L^1(\mathbb{R})$ satisfy $\int_\mathbb{R} K(x)dx=1$, and denote $K_a(x)=\frac1aK(\frac{x}{a})$ for $a>0$. We get trivially that $\int_\mathbb{R}K_a(x)\, dx=1$.
Let $y\in\mathbb{R}$ and $f\in C_0(\mathbb{R})$, the set of continuous function that tends to $0$ at $\pm\infty$. I want to show that $$\lim_{a\rightarrow 0^+}f\ast K_a(x)=f(x),$$
here $f \ast K_a$ is the convolution.
By simple manipulations, this is equivalent to $$\lim_{a\rightarrow 0^+}\int_\mathbb{R}(f(x-y)-f(x))K_a(y)dy=0.$$
How can we show this last statement?
The trick is to split the integral into small $y$ and large $y$.
Given $\epsilon>0$, choose $\delta>0$ so that $|f(x-y)-f(x)|<\epsilon$ when $|y|<\delta$ (this can even be done uniformly in $x$, because functions in $C_0$ are uniformly continuous). Hence
$$\left|\int_{|y|<\delta} (f(x-y)-f(x))K_a(y)\,dy\right|<\epsilon$$ Also, since $$\lim_{a\rightarrow 0^+}\int_{|y|>\delta} K_a(y)\,dy =\lim_{a\rightarrow 0^+}\int_{|x|>\delta/a} K (x)\,dx =0$$ and $f$ is bounded, it follows that $$\lim_{a\rightarrow 0^+}\int_{|y|>\delta}(f(x-y)-f(x))K_a(y)\,dy =0$$