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

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!

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 $$y_k\gt b$$.
• Thanks, I just figured that out a few minutes ago. I have now found an alternative approach by splitting $f_n$ in a negative and positive part and then using MCT. However, I struggle to show that $f_n$ is indeed a non-decreasing sequence, do you have any hints on that? – Michael Maier Jan 8 at 15:24