Lebesgue integrable function in rationals

Question

Function $f : [0,1] \to \mathbb{R}$ defined as $ f(x) = \begin{cases} 1 & x\notin\mathbb{Q}\\ 0 & x\in\mathbb{Q} \end{cases} . $ As is well known $f$ is not integrable in the Riemann sense. Is it Lebesgue integrable?

Answer 1

$\lambda\left([0,1]\cap\mathbb{Q}\right)=0$ since $[0,1]\cap\mathbb{Q}$ is countable. So, $\lambda\left([0,1]\setminus\mathbb{Q}\right)=\lambda([0,1]) - \lambda\left([0,1]\cap\mathbb{Q}\right) = 1.$ Hence $\int f d\lambda = 0 \cdot\lambda\left([0,1]\cap\mathbb{Q}\right) + 1\cdot\lambda\left([0,1]\setminus\mathbb{Q}\right) = 1$.