If $f$ Lebesgue integrable, then $\lim_{k\to\infty}\int_{\mathbb{R}}\cos(xk)f(x)dx=0$. Let $f$ be a Lebesgue integrable function on $\mathbb{R}$ and let $g$ be defined on $\mathbb{R}$ by $g(y)=\int_{-\infty}^{\infty} \cos(xy)f(x)dx$.
Prove that $\lim_{k\rightarrow\infty}g(k)=0$.
 A: Your $g$ is the imaginary part of the Fourier transform of a real function $L^1(\mathbb R)$. It is one of the basic properties of the Fourier transform that it maps $L^1$ into the space of continuous function decaying for $y\to\pm\infty$, this is the Riemann-Lebesgue lemma.
A: Step 1:
Suppose that $f = \chi_{[a,b]}$, where $\chi_E$ is the indicator function defined by
$$ \chi_E(x) = \begin{cases} 1 & \text{if}\ x\in E, \\ 0 & \text{otherwise}.
\end{cases}$$
Step 2:
Suppose that $f$ is a step function; that is, $f=\sum_{i=1}^nc_i\chi_{[a_i,b_i]}$.
Step 3:
Use the fact that the step functions are dense in the space of Lebesgue integrable functions on $\mathbb{R}$.

To expand on step 3, let $f$ be a Lebesgue integrable function and $\varepsilon>0$. The density of step functions in the space of Lebesgue integrable functions yields a step function $h$ such that $\int |h(x)-f(x)|dx < \varepsilon/2$. Then
$$ \left|\int \cos(xn)f(x)dx\right|
\leq \int |cos(xn)||f(x)-h(x)|dx + \left|\int\cos(xn)h(x)dx\right|.
$$
Can you take it from here?
