If $f(x)=x^2$ for all $x\not\in\mathbb{Q}\cap[0,1]$ and $f(x)=0$ for all $x\in\mathbb{Q}\cap[0,1]$, then $f$ is Lebesgue integrable in $[0,1]$?
I am learning about Lebesgue integration, but I have trouble in determining if that kind of functions are Lebesgue integrable or not. I know that $f$ is not Riemann integrable in that interval, since $f$ is discontinuous in all $x\in(0,1]$. What argument can I use to prove or disprove the Lebesgue integrability of $f$?
Note that it is sufficient to prove that $f$ is measurable. Enumerate the rationals in [0,1] as $\{q_1,q_2,....\}$, then define functions $f_n : [0,1] \to [0,1]$ with $f_n(x) = x^2$ for $x \neq q_n$, $f_n(q_n)=0$. It is clear each $f_n$ is measurable as it is even piecewise continuous. Now your $f$ is the infimum of the countably many $f_n$, and hence measurable.