Prove Lebesgue integrability for an (almost) trigonometric function Let $X = [0,1]$, $\mathfrak{M}$ - is a $\sigma$ algebra of Lebesgue measurable subsets of $X$, $\mu$ - Lebesgue measure on $\mathfrak{M}$ Function $f:X\to\mathbb{R} $ is defined as: $f(x) = \sin{nx}$ if $x \in (\frac{1}{2^n},\frac{1}{2^{n-1}}], n \in \mathbb{N}, f(0) = 0$
Prove that $f$ is Lebesgue integrable and calculate $\int\limits_{X}fd\mu$
As far as I understand, we could use the Lebesgue dominated convergence theorem in this case showing that $\sin{nx}$ converges pointwise on each interval and is bounded by a constant. Is that correct? 
As for integration, using the sigma-additive property of the integral, we can express the integral this way:
$\int\limits_{X}fd\mu = \sum\limits_{n=1}^{\infty}\int\limits_{A_n}fd\mu, n \in \mathbb{N}$
Unfortunately, I don't exactly understand how to calculate this integral. 

The sequence
$$f_n(x) = \sum_{k=1}^n \sin kx \, \mathbf{1}_{\left(2^{-k},2^{-k+1} \right]}(x)$$
converges pointwise to $f(x)$ on $[0,1]$.
Since $0 < \sin kx < 1$ for $x \in \left(2^{-k},2^{-k+1} \right]$ we have $|f_n(x)| \leqslant 1$ for all $x \in [0,1]$.  By the LDCT, $f$ is integrable and 
$$\int_{[0,1]} f \, d\mu= \lim_{n\to \infty}\int_{[0,1]} f_n \, d\mu = \lim_{n\to \infty}\sum_{k=1}^n \int_{(2^{-k},2^{-k+1}]} \sin kx \, d\mu \\ = \sum_{k=1}^\infty \int_{(2^{-k},2^{-k+1}]} \sin kx \, d\mu $$
The integrals on the right-hand side can be evaluated as Riemann integrals (since Riemann integrable functions are Lebesgue integrable) and, thus, 
$$\int_{[0,1]} f \, d\mu = \sum_{k=1}^\infty \int_{2^{-k}}^{2^{-k+1}} \sin kx \, dx \\ =  \sum_{k=1}^\infty \frac{\cos \frac{k}{2^{k}} - \cos \frac{k}{2^{k-1}}}{k} \\ \underbrace{\approx 0.605359}_{\text{WolframAlpha}}$$
