Let $ g(t) = \int_{0}^{2\pi} f(x)\cdot \sin(tx) dx, $ where $ t \in \mathbb{R}. $ Show that $g$ is uniformly continuous on $ \mathbb{R} $ and that $ \lim_{n \to \infty} g(n) = 0. $
I don't even know if I can assume $f$ to be continuous since not all Riemann integrable functions are continuous.
I'll prove uniform continuity here. First, we see that $h(t)= \sin(tx)$ is a uniformly continuous function with fixed $x$ and varying $t$. This is because $\sin$ itself is uniformly continuous. Hence, for $\epsilon >0$ we can find $\delta >0$ such that: $$ |t_1 - t_2| <\delta \implies |\sin(xt_1) - \sin(xt_2)|<\epsilon $$ Keeping the same $\delta$ distance between $t_1,t_2$, we conclude: $$ |g(t_1) - g(t_2)| = \left|\int_0^{2\pi} f(x)\sin(t_1x)\mathrm{d}x - \int_0^{2\pi} f(x)\sin(t_2x)\mathrm{d}x\right| \leq \int_0^{2\pi}|f(x)||\sin(t_1x)-\sin(t_2x) \mathrm{d}x \ $$ Because $f$ is Riemann integrable, it is bounded. applying this and uniform continuity we conclude that: $$ \int_0^{2\pi}|f(x)||\sin(t_1x)-\sin(t_2x)| \mathrm{d}x \leq \int_0^{2\pi}|f(x)|\mathrm{d}x \leq 2\pi M\epsilon $$
Giving us uniform continuity.