On limit of convolution of function with a summability kernel

Theorem 2.1 Let ##I=(-a,a)## be an interval (finite or infinite). Suppose that ##(K_n)_{n=1}^\infty## is a sequence of real-valued, Riemann-integrable functions defined on ##I##, with the following properties:
1) ##K_n(s)\geq 0##,
2)##\int_{-a}^a K_n(s)ds=1##, and
3) if ##\delta>0##, then ##\lim\limits_{n\to\infty}\int_{\delta<|s|<a} K_n(s)ds=0.##
If ##f:I\to\mathbb{C}## is integrable and bounded on ##I## and continuous for ##s=0##, we then have $$\lim_{n\to\infty}\int_{-a}^a K_n(s)f(s)ds=f(0).$$

Corollary 2.1 If ##(K_n)_{n=1}^\infty## is a positive summability kernel on the interval ##I##, ##s_0## is an interior point of ##I##, and ##f## is continuous at ##s=s_0##, then $$\lim_{n\to\infty}\int_I K_n(s)f(s_0-s)ds=f(s_0).$$

The proof is left as an exercise (do the change of variables ##s_0-s=u##.