Calculate the Lebesgue integral
$$ \int_{[0,\pi]}^{} \sin x\: \epsilon[\mathbb{R}\setminus\mathbb{Q}] d\mu,$$
where is $\epsilon $ is the characteristic function.
I tried this way $ \int_{\mathsf ([0,\pi ] \cap I) }^{} \sin x \: d\mu$...
$\mathsf ([0,\pi ] \cap I)$ = $[0,\pi ]$..
$ \int^{\pi}_0 \sin x \: d\mu=-\cos(\pi)+\cos(0)=2$
Is this correct?
You are not far, just use that $m([0,\pi])=m((R-Q) \cap [0,\pi])$
And the fact that if $m(E)=0$ then $\int_Ef=0$ for some measurable $E$ and $f$ integrable.(where $f$ can also be negative)
$\int_{\mathbb{Q \cap [0,\pi]}}\sin{x}=0$ because $m( \mathbb{Q} \cap[0,\pi])=0$
So $$\int_{(\mathbb{R}-\mathbb{Q}) \cap [0,\pi]} \sin{x}=\int_0^{\pi}\sin{x}-\int_{\mathbb{Q \cap [0,\pi]}}\sin{x}=\int_0^{\pi}\sin{x}=2$$
Let $1_{E}$ be the charecteristic function and $m$ be the Lebesgue measure (Note that $m$ is not a variable). Then $$\int_{[0,\pi]}\sin(x)1_{(\mathbb{R}-\mathbb{Q})}dm=\int_{[0,\pi]}\sin(x) dm-\int_{[0,\pi]}\sin(x)1_{\mathbb{Q}}dm$$
Since $\sin(x)$ is positive on the interval $[0,\pi]$ and bounded above by $1$ we see that:
$$0\leq\int_{[0,\pi]}\sin(x)1_{\mathbb{Q}}dm\leq \int_{[0,\pi]}1_\mathbb{Q}dm=m(\mathbb{Q}\cap [0,\pi])=0$$
So $$\int_{[0,\pi]}\sin(x)1_{(\mathbb{R}-\mathbb{Q})}dm=\int_{[0,\pi]}\sin(x) dm$$
In the case of riemann integratable functions on a closed inteval like $\sin(x)$ the Lebesgue integral just becomes the riemann integral:
$$\int_{[0,\pi]}\sin(x)dm=\int_{[0,\pi]}\sin(x)dx=-\cos(\pi)+\cos(0)=2$$
so
$$\int_{[0,\pi]}\sin(x)1_{(\mathbb{R}-\mathbb{Q})}dm=2$$
