A problem about Riemann-Lebesgue lemma Let $f$ be an integrable function over $[a,b]$. Prove that:
$$\lim_{n \rightarrow \infty } \int _a ^b f(x)|\sin(nx)| dx= \frac {2}{\pi} \int _a^b f(x) dx.$$
 A: Hint: Prove the theorem for characteristic functions of intervals, then prove the theorem for step functions. Approximate integrable functions with step functions and apply a convergence theorem.
A: Based on the hints given by Alex Becker, we can obtain a more general result:
If $f$ is integrable on $[a,b]$,which is a bounded closed interval, and $g$ is integrable periodic function with period $\rm T$,
then we have the following:
$$\lim\limits_{n\rightarrow \infty} \int_a^b f(x)g(nx)dx={\rm T}^{-1}\int_0^{\rm T} g(t)dt \int_a^bf(x)dx$$

(1): I show that $$\lim\limits_{n\rightarrow \infty} \int_a^b {\rm g}(nx)dx={\rm T}^{-1}(b-a)\int_0^{\rm T}g(t)dt $$

PROOF: For each $n\in \mathbb{N}$, define $b_n=a+\frac{{\rm T}}{n} \left[\frac{n(b-a)}{{\rm T}} \right]$.
It is not difficult to see that $0\le b-b_n <\frac{{\rm T}}{n}$.
$$\int_a^{{b_n}} g (nx)dx = \sum\limits_{k = 1}^{\left[ {\frac{{n(b - a)}}{{\rm T}}} \right]} {\int_{a + (k - 1)\frac{{\rm T}}{n}}^{a + k\frac{{\rm T}}{n}} g } (nx)dx$$
$$ = \frac{1}{n}\sum\limits_{k = 1}^{\left[ {\frac{{n(b - a)}}{{\rm T}}} \right]} {\int_{na + (k - 1){\rm T}}^{na + k{\rm T}} g } (y)dy = \frac{1}{n}\sum\limits_{k = 1}^{\left[ {\frac{{n(b - a)}}{{\rm T}}} \right]} {\int_0^{\rm T} g } (y)dy$$
(since the integral of a periodic function over a length of its period is the same)
$$\int_a^{b_n} g(nx) dx=\frac{1}{n}\sum_{k=1}^{\left[\frac{n(b-a)}{{\rm T}}\right]}\int_{0}^{{\rm T}} g(y)dy
=\frac{1}{n}\left[\frac{n(b-a)}{{\rm T}}\right]\int_0^{\rm T}g(t)dt$$
Since $$\left|\int_a^bg(nt)dt-\int_a^{b_n}g(nt)dt \right|\le (b-b_n)M \le \frac{{\rm T}}{n} M$$ where $$M=\sup|g(x)|$$
so $$\lim_{x\rightarrow \infty} \int_a^bg(nx)dx=\lim_{x\rightarrow \infty} \int_a^{b_n}g(nx)dx={\rm T}^{-1}(b-a)\int_0^{\rm T} g(t)dt $$

(2): I show $$\lim_{n\rightarrow \infty} \int_a^b f(x)g(nx)dx={\rm T}^{-1}\int_0^{\rm T} g(t)dt \int_a^bf(x)dx$$ for any step function $f$. 

PROOF: This result follows from (1) easily.

(3): I prove $$\lim_{n\rightarrow \infty} \int_a^b f(x)g(nx)dx={\rm T}^{-1}\int_0^{\rm T} g(t)dt \int_a^bf(x)dx$$ for any integrable function.

PROOF: Given $\delta >0$, since $f$ is integrable, we can find a step function $h$ such that $\displaystyle \int_a^b|f-h|<\delta$.
$$\eqalign{
  & \left| {\int_a^b f (x)g(nx)dx - {T^{ - 1}}\int_0^T g (t)dt\int_a^b f (x)dx} \right| \leqslant   \cr 
  & \left| {\int_a^b f (x)g(nx)dx - \int_a^b h (x)g(nx)dx} \right| + \left| {\int_a^b h (x)g(nx)dx - {T^{ - 1}}\int_0^T g (t)dt\int_a^b h (x)dx} \right|  \cr 
  &  + \left| {{T^{ - 1}}\int_0^T g (t)dt\int_a^b f (x)dx - {T^{ - 1}}\int_0^T g (t)dt\int_a^b h (x)dx} \right| < \left( {M + 1 + {T^{ - 1}}\int_0^T g (t)dt} \right)\delta  \cr} $$
for all $n\ge N$,(such $N$ exist by (2)),and since $\delta>0$ is arbitrary so

$$\lim_{n\rightarrow \infty} \int_a^b f(x)g(nx)dx=T^{-1}\int_0^Tg(t)dt \int_a^bf(x)dx$$

