Monotone Convergence theorem Application 
$$
\lim _{n \to \infty} \int_{-\infty}^{\infty} \frac{e^{-x^{2} / n}}{1+x^{2}} d x=?
$$

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.
 A: $\newcommand{\D}{\,\mathrm{d}}$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.
