<s>[INST] Two Lebesgue's integrals I need to calculate these two integrals:
$$\lim_{n\rightarrow\infty}\int_\mathbb{R}\frac{nx^2+1}{nx^4+n^2}d\lambda(x)$$
$$\int_0^1\sum_{n=1}^\infty n\cdot\chi_{(\frac{1}{n+1},\frac{1}{n})}d\lambda$$
I did the second one like this:
$$\int_0^1\sum_{n=1}^\infty n\cdot\chi_{(\frac{1}{n+1},\frac{1}{n})}d\lambda=\int_0^11\cdot\chi_{(\frac{1}{2},1)}+2\cdot\chi_{(\frac{1}{3},\frac{1}{2})}+\dots d\lambda=$$
$$=\int_0^1\chi_{(\frac{1}{2},1)}d\lambda+2\cdot\int_0^1\chi_{(\frac{1}{3},\frac{1}{2})}d\lambda+3\cdot\int_0^1\chi_{(\frac{1}{4},\frac{1}{3})}d\lambda+\dots$$
First integral if equal to the area of rectangle $(\frac{1}{2},1)\times [0,1]$ which is $\frac{1}{2}$. Every next integral is the area of rectangle $(\frac{1}{n+1},\frac{1}{n})\times [0,1]$ times $n$. From that we get:
$$\frac{1}{2}+2\cdot \frac{1}{6}+3\cdot \frac{1}{12}+\dots=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots=\sum_{n=1}^\infty\frac{1}{n+1}=\infty.$$
For the first one I think I can use one of the theorem of Lebesgue which allow me to rewrite it as
$$\lim_{n\rightarrow\infty}\int_\mathbb{R}\frac{nx^2+1}{nx^4+n^2}d\lambda(x)=\int_\mathbb{R}\lim_{n\rightarrow\infty} \frac{nx^2+1}{nx^4+n^2}d\lambda(x)$$
I'm not sure how will that limit behave in that case though. I would be grateful for any hint on how to complete this.
 [/INST] You have for $\vert x \vert \ge 1$
$$\begin{aligned}
\left\vert \frac{nx^2+1}{nx^4+n^2} \right\vert &\le \frac{x^2+1}{x^4+n} \le \frac{x^2+1}{x^4+1}
\end{aligned}$$
and for $\vert x \vert \le 1$
$$\left\vert \frac{nx^2+1}{nx^4+n^2} \right\vert \le \frac{n+1}{n^2} \le 2.$$
Therefore $f_n(x) = \frac{nx^2+1}{nx^4+n^2} $ is integrable for all $n \ge 1$, and bounded by
$$g(x) = \begin{cases}
\frac{x^2+1}{x^4+1} & \vert x \vert \ge 1\\
2 & \vert x \vert \lt 1
\end{cases}$$ which is integrable. As $\{f_n\}$ converges pointwise to zero, you can apply Lebesgue dominated convergence theorem to conclude that
$$\lim_{n\rightarrow\infty}\int_\mathbb{R}\frac{nx^2+1}{nx^4+n^2}d\lambda(x) = 0$$
</s>