$$ \lim _{n \to \infty} \int_{-\infty}^{\infty} \frac{e^{-x^{2} / n}}{1+x^{2}} d x=? $$
My opinion is using Monotone Convergence Theorem here.
For every $x \in \mathbb{R}$ the sequence $\left\{e^{-x^{2} / n}\right\}$ monotonically increases and converges to $e^0=1$. Thus by the Lebesgue Monotone Convergence Theorem, $$ \lim _{n \to \infty} \int_{-\infty}^\infty \frac{e^{-x^2/n}}{1+x^2} dx = \int_{-\infty}^\infty \frac{dx}{1+x^2} = \left.\tan^{-1} x \right|_{-\infty}^\infty = \pi $$ I think my answer is right. But I am especially interested in its details.I think applying Lebesgue Classical monotone convergence is enough in this question, not need Beppo Levi version. But i am not sure.What about you?May you give some details ? Thanks for your helps.
$\newcommand{\D}{\,\mathrm{d}}$The notation $$\int_{-\infty}^{\infty} \frac{e^{-x^{2} / n}}{1+x^{2}} \D x$$ is most commonly used to denote the iterated improper Riemann integral $$\lim_{a \to \infty} \lim_{b \to \infty} \int_a^b \frac{e^{-x^{2} / n}}{1+x^{2}} \D x$$ while $$\int_\mathbb{R} \frac{e^{-x^{2} / n}}{1+x^{2}} \D\lambda(x)$$ the analogous Lebesgue integral, where $\lambda$ is the Lebesgue measure. If you want to add more details to your proof then you could make the relation to Lebesgue integral clearer. For that, let $$f_n : \mathbb{R} \to [0, \infty[ : x \mapsto \frac{e^{-x^{2} / n}}{1+x^{2}}$$ and $F$ its pointwise limit as $n \to \infty$. Having in mind that the proper Riemann integral, when it exists, is equal to the corresponding Lebesgue integral, reason as follows \begin{align*} \lim_{n \to \infty} \int_{-\infty}^{\infty} f_n(x) \D x &= \lim_{n \to \infty} \lim_{a \to \infty} \lim_{b \to \infty} \int_a^b f_n(x) \D x \\ &= \lim_{n \to \infty} \lim_{a \to \infty} \lim_{b \to \infty} \int_{[a, b]} f_n \D\lambda \\ &= \lim_{n \to \infty} \lim_{a \to \infty} \lim_{b \to \infty} \int_{\mathbb{R}} f_n \chi_{[a, b]} \D\lambda \\ \end{align*}
Then using the MCT three times, one time for each limit, it follows that
\begin{align*} \lim_{n \to \infty} \lim_{a \to \infty} \lim_{b \to \infty} \int_{\mathbb{R}} f_n \chi_{[a, b]} \D\lambda &= \int_{\mathbb{R}} \lim_{n \to \infty} \lim_{a \to \infty} \lim_{b \to \infty} \left( f_n \chi_{[a, b]} \right) \D\lambda \\ &= \int_{\mathbb{R}} \lim_{n \to \infty} f_n \lim_{a \to \infty} \left( \lim_{b \to \infty} \chi_{[a, b]} \right) \D\lambda \\ &= \int_{\mathbb{R}} F \lim_{a \to \infty} \left( \lim_{b \to \infty} \chi_{[a, b]} \right) \D\lambda \\ \end{align*}
Now use MCT just two more times to get back to the Riemann integral
\begin{align*} \int_{\mathbb{R}} F \lim_{a \to \infty} \left( \lim_{b \to \infty} \chi_{[a, b]} \right) \D\lambda &= \lim_{a \to \infty} \lim_{b \to \infty} \int_{\mathbb{R}} F \chi_{[a, b]} \D\lambda \\ &= \lim_{a \to \infty} \lim_{b \to \infty} \int_{[a, b]} F \D\lambda \\ &= \lim_{a \to \infty} \lim_{b \to \infty} \int_a^b \frac{1}{1+x^2} \D x \\ \end{align*}
which you already computed.
Furthermore you could also argue why all $f_n$ and $F$ are Lebesgue measurable.