TITLE: Integral of bounded continuous function on $R$
QUESTION [7 upvotes]: Let $f$ be a bounded continuous function on $R$. Prove that
$$\lim_{n \to \infty} \frac{n}{\pi} \int_{ R} \frac {f(t)}{1+n^{2}t^{2}} dt=f(0)$$
I solved this question as follows, but I ran into a problem:
Solution:
Since $f$ is continuous at $0$, given $\epsilon>0$, there is $\delta >0$ such that $|f(x)-f(0)| \leq \epsilon, \forall x \in [-\delta \hspace{2 mm} \delta]$. Now consider $f_{n}(x)=\frac{nf(x)}{1+n^{2}x^{2}}$,$n \in N$, is a sequence of continuous function that converges to $0$ on $R-[-\delta \hspace{2 mm} \delta]$ when $n \to \infty$. Also, according to guestion, $f$ is bounded (i.e.$|f_{n}(x)|<g(x)$). Therefore according to Dominated Convergence Theorem 
 $$\lim_{n \to \infty} \frac{n}{\pi} \int_{ R-[-\delta \hspace{2 mm} \delta]} \frac {f(t)}{1+n^{2}t^{2}} dt=\lim_{n \to \infty} \frac{n}{\pi} \int_{ R-[-\delta \hspace{2 mm} \delta]} 0 \hspace{2 mm} dt=0$$
Now I have a problem to solve the second part. I want to prove   
$\lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(t)}{1+n^{2}t^{2}} dt =  \lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(0)}{1+n^{2}t^{2}} dt $
which 
$\lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(0)}{1+n^{2}t^{2}} dt =\pi f(0)$
If I can prove that, the rest of question is easy, because I can say: 
$\lim_{n \to \infty} n \int_{ R} \frac {f(t)}{1+n^{2}t^{2}} dt=\lim_{n \to \infty} n \int_{ R-[-\delta \hspace{2 mm} \delta]} \frac {f(t)}{1+n^{2}t^{2}} dt+\lim_{n \to \infty} n \int_{ R} \frac {f(t)}{1+n^{2}t^{2}} dt=\lim_{n \to \infty} n \int_{ [-\delta \hspace{2 mm} \delta]} \frac {f(t)}{1+n^{2}t^{2}} dt=\pi f(0)$
Now how can I prove:
$\lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(t)}{1+n^{2}t^{2}} dt =  \lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(0)}{1+n^{2}t^{2}} dt $

REPLY [2 votes]: We have
$$\int_{-\delta}^{\delta}\frac{nf(t)}{1+n^2t^2}\,dt=\int_{-\infty}^{\infty}\frac{f(u/n)}{1+u^2}\chi_{[-n\delta,n\delta]}(u)\,du.$$
Since $f$ is bounded, $|f(t)| \leq M$, and it follows that
$$\left|\frac{f(u/n)}{1+u^2}\chi_{[-n\delta,n\delta]}(u)\right| \leq \frac{M}{1+u^2}$$
Apply the dominated convergence theorem, noting that 
$$\lim_{n\rightarrow \infty}\int_{-\infty}^{\infty}\frac{f(u/n)}{1+u^2}\chi_{[-n\delta,n\delta]}(u)\,du=\int_{-\infty}^{\infty}\lim_{n\rightarrow \infty}\left[\frac{f(u/n)}{1+u^2}\chi_{[-n\delta,n\delta]}(u)\right]\,du\\=\int_{-\infty}^{\infty}\frac{f(0)}{1+u^2}\,du=\pi f(0)$$