Question: 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?

Answer: 
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 

$$\int\_{[0,\pi]}\sin(x)1\_{(\mathbb{R}-\mathbb{Q})}dm=2$$