Lebesgue integral, Is the solution right?

Question

I'am trying to understand Lebesgue integration

Compute $\int_{0}^{\pi}$ f(x)dx

Where

$f(x) = \begin{cases} sin x & \text{ if } x \in \mathbb{I} \\ cosx & \text{ if } x \in \mathbb{Q} \end{cases}$

I tried this

(L)$\int_{[0,\pi]} f(x) dx = (L)\int_{[0,\pi] \cap \mathbb{Q}} f(x)dx + (L)\int_{[0,\pi] \cap \mathbb{I}} f(x)dx = (L)\int_{[0,\pi] \cap \mathbb{Q}} cos x dx + (L)\int_{[0,\pi] \cap \mathbb{I}} sinxdx = (R)\int_{[0,\pi]} cos x dx + (R)\int_{[0,\pi]} sinxdx = \int_{0}^{\pi} (sinx+cosx)dx = sin(\pi) - sin(0) + - (cos(\pi) - cos(0)) = 0 - (-2) = 2$

Is my sollution correct? Can you recommend any good materials with emamples of lebesgue integration?

Thank you!

Answer 1

Note that the set $\Bbb Q$ has Lebesgue measure zero. Thus, we outright have

$$\int_{[0,\pi] \cap \Bbb Q} \cos(x)dx = 0$$

since a subset of the rationals will obviously have measure zero. Similarly, $\Bbb I \cap [0,\pi]$ has Lebesgue measure $\pi$ as result since $\Bbb I = \Bbb R \setminus \Bbb Q$. In doing so, we notice that

$$\int_{[0,\pi]} f(x)dx = \int_0^\pi \sin(x)dx$$

as a result. Your answer still ends up being correct, but only by chance: the working is wrong.

Answer 2

The reasoning is not correct, even though the final answer is correct. The Lebesgue integral over $[0,\pi]\cap \mathbb Q$ of $\cos x$ is $0$ becasue $\mathbb Q$ has measure $0$, not because it is equal to the integral of $\cos x$ from $0$ to $\pi$.