Convergence of Dirichlet kernel Let $f$ be a monotonic and bounded function defined on [0,1) and $\lim_{x\rightarrow 0^+}f(x)$ exists. Prove
\begin{align}
\lim_{\lambda\rightarrow \infty}\int_0^1 f(x)\frac{\sin\lambda x}{x}dx =  \frac{\pi}{2}f(0^+)
\end{align}
I am thinking of dividing this integral into two parts:
\begin{align}
\int_0^1 f(x)\frac{\sin\lambda x}{x}dx = \int_0^\delta+\int_\delta^1 f(x)\frac{\sin\lambda x}{x}dx
\end{align}
It is easy to show the second part goes to zero, however I don't really know how to deal with the first part. In other words, I need to estimate
\begin{align}
\lim_{\lambda\rightarrow\infty}\int_0^\delta (f(x)-f(0^+))\frac{\sin\lambda x}{x}dx = 0
\end{align}
However $\frac{\sin\lambda x}{x}$ is oscillating and is not absolutely integrable.
 A: 
Inasmuch as $f(x)$ is monotonic and bounded on $[0,1)$, it is almost everywhere continuous there and therefore integrable.


Note that we can write the kernel $\frac{\sin(\lambda x)}{x}$ as the integral
$$\frac{\sin(\lambda x)}{x}=\frac12\int_{-\lambda }^\lambda e^{i x t}\,dt\tag 1$$
Using $(1)$ we can write 
$$\begin{align}
\int_0^1 f(x)\frac{\sin(\lambda x)}{x}\,dx&=\frac12\int_0^1 f(x)\int_{-\lambda }^\lambda e^{i x t}\,dt\,dx\\\\
&=\frac12\int_{-\lambda }^\lambda \int_0^1 f(x)e^{i x t}\,dx\,dt\\\\
&=\frac12\int_{-\lambda }^\lambda \int_{-\infty}^\infty f(x)\xi_{[0,1]}(x)e^{i x t}\,dx\,dt\\\\
\end{align}$$
Letting $\lambda \to\infty$ yields
$$\lim_{\lambda \to \infty}\int_0^1 f(x)\frac{\sin(\lambda x)}{x}\,dx=\frac12\int_{-\infty }^\infty \int_{-\infty}^\infty f(x)\xi_{[0,1]}(x)e^{i x t}\,dx\,dt\tag2$$
Since $f(x)\xi_{[0,1]}(x)\in L^1$, the inner integral on the right-hand side of $(2)$ is the Fourier Transform of $f(x)\xi_{[0,1]}(x)$.  And the outer integral is $2\pi$ times the inverse Fourier Transform of the Fourier Transform, evaluated at $0$. 
Inasmuch as $f(x)\xi_{[0,1]}(x)$ is discontinuous at $0$, we conclude See this on the Fourier Inversion Theorem that 
$$\begin{align}
\lim_{\lambda \to \infty}\int_0^1 f(x)\frac{\sin(\lambda x)}{x}\,dx&=\pi \left(\frac12 (f(0^+)\xi_{[0,1]}(0^+)+\frac12f(0^-)\xi_{[0,1]}(0^-)\right)\\\\
&=\frac\pi2 f(0^+)
\end{align}$$
as was to be shown!
