# 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,

$$\int_{[0,1]} f \, d\mu = \sum_{k=1}^\infty \int_{2^{-k}}^{2^{-k+1}} \sin kx \, dx \\ = \sum_{k=1}^\infty \frac{\cos \frac{k}{2^{k}} - \cos \frac{k}{2^{k-1}}}{k} \\ \underbrace{\approx 0.605359}_{\text{WolframAlpha}}$$