I want to verify somehow (by direct computation if possible) the easy fact that for any simple, non-negative, Lebesgue-measurable function $f :[a,b] \to \mathbb{R}$ where $f = \sum_{i=1}^{k}m_i \chi_{E_i}$ (for measurable subsets $E_i \subset [a,b]$)
$$\lim_{n \to \infty} \int_{[a,b]} f(x) \sin(nx) \,d \mu(x) =0$$
I know this is meant to be the easiest part of the proof of the Riemann-Lebesgue lemma (if one takes this kind of approach to the proof) but I'm getting lost past this point:
$$ \int_{[a,b]} f(x) \sin(nx) \, d \mu(x) = \int_{E_1} f(x) \sin(nx) \, d \mu(x) + \int_{E_2} f(x) \sin(nx) \, d \mu(x) + \cdots + \int_{E_k} f(x) \sin(nx) \, d \mu(x)$$
because $E_1,\ldots,E_k$ are just Lebesgue measurable subsets of $[a,b]$ and not necessarily intervals in their own right (so I can't antidifferentiate $\sin$ directly....)
Am I just not seeing something trivial?
If $E_i = [c,d]$ is some interval, it is simple to determine the Fourier-transform of $f= 1_{E_i}$. (And it vanishes for $n \rightarrow \infty$.) Now, by regularity of the Borel-measure on $\mathbb{R}$, we can approximate any $E$ by open sets $V$. Since $V$ is a countable union of intervals $(a_i,b_i)$, we may use the continuity of the measure, to find $$\mathcal{F}(1_E) = \sum_{k=1}^n \mathcal{F}(1_{(a_k,b_k)})+\mathcal{O}(\varepsilon).$$ This gives the vanishing at infinity of the Fourier-transform of $1_E$. Now, it is easy to extend this property for any function $f \in L^1([a,b])$ by approximation.
However, this proof is more inconvenient than the usual proof by approxmating $f \in L^1[a,b])$ by smooth compactly supported functions.