Show that $\lim_{n\to\infty}\int_{\mathbb{R}}f_nd\mu$ exists, where $\mu=\sum_{k=1}^M \delta_{y_k}$

Question

I want to show that $\lim_{n\to\infty}\int_{\mathbb{R}}f_nd\mu$ exists and calculate it, with $\mu=\sum_{k=1}^M \delta_{y_k}$ and $\delta_{y_k}$ being the dirac measure. Additionally, $f_n$ is defined from $\mathbb{R}\to\mathbb{R}$ where $$x \to arctan(n(x-a))-arctan(n(x-b))$$ with $a,b\in\mathbb{R}, a\lt b, n\in\mathbb{N}$ and $y_k\in\mathbb{R}$ is given. What I tried: $$\lim_{n\to\infty}\int_{\mathbb{R}}f_nd\mu=\lim_{n\to\infty}\left(\sum_{k=1}^M\int_{\mathbb{R}}f_n\delta_{y_k}\right)$$ Is this correct? Then, $$\lim_{n\to\infty}\left(\sum_{k=1}^M\int_{\mathbb{R}}f_n\delta_{y_k}\right)=\lim_{n\to\infty}\left(\sum_{k=1}^Mf_n(y_k)\right)=\sum_{k=1}^M\left(\lim_{n\to\infty}f_n(y_k)\right)$$ If I can now show that $f_n$ converges pointwise to a limit function $f$, I should be done. However, I struggle with that and am completely unsure if this is anyhow correct. Thanks for help!

Answer 1

The only thing you have to know on the $\arctan$ function is that $$ \lim_{x\to -\infty}\arctan x=-\frac{\pi}2\mbox{ and }\lim_{x\to +\infty}\arctan x= \frac{\pi}2. $$ Therefore, the value of $\lim_{n\to +\infty}f_n\left(y_k\right)$ depends whether $y_k\lt a$, $y_k=a$, $a\lt y_k\lt b$, $y_k=b$ or $y_k\gt b$.