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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!

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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 $y_k\gt b$.

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  • $\begingroup$ 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? $\endgroup$ – Michael Maier Jan 8 at 15:24
  • $\begingroup$ And one other question: Is my first approach correct, i.e. it doesn't include any mistakes? Thanks again! $\endgroup$ – Michael Maier Jan 8 at 15:30

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