Consider the following expression: $$s_n(x) - s = \frac{1}{2\pi} \int_{-\pi}^{\pi} h(t) \exp(i\frac{1}{2}t)\exp(int) \ dt - \frac{1}{2\pi} \int_{- \pi}^\pi h(t) \exp(-i\frac{1}{2}t)\exp(-int) \ dt $$ where $$ h(t) = \begin{cases} \frac{ f(x+t) + f(x-t) - 2s}{2i \sin (\frac{1}{2}t)} & t \in (0,\pi] \\ 0 & t \in [-\pi, 0] \\ \end{cases} $$ I wish to show that $s_n(x) - s$ converges to $0$ as $t$ tends to $0$ from above. ($s_n(x)$ is a partial sum and $s$ is the sum to which I wish to show convergence). I am told that this follows from Riemann's lemma if $f$ is piecwiese continuous (it is) and if $h(t)$ has a limit as $t$tends to 0 from above (this is an assumption we are allowed to make)...
... but why does this prove that $s_n(x) - s$ tends to zero?
The book gives does not give much more information.
The Riemann-Lebesgue Lemma says that if $g\colon[a,b]\to\mathbb{R}$ is Riemann integrable, then $$ \lim_{\lambda\to\infty}\int_a^bg(x)\cos(\lambda\,x)\,dx=\lim_{\lambda\to\infty}\int_a^bg(x)\sin(\lambda\,x)\,dx=0. $$ Under the conditions on $f$, the function $h$ is Riemann integrable on $[-\pi,\pi]$. This implies that $$ \lim_{n\to\infty}\int_{-\pi}^\pi h(t)\exp\Bigl(i\,\frac{1}{2}\,t\Bigr)\exp(i\,n\,t)\,dt=\lim_{n\to\infty}\int_{-\pi}^\pi h(t)\exp\Bigl(i\,\frac{2\,n+1}{2}\,t\Bigr)\,dt=0. $$