Let $f$ be a function defined over $[0,1]$, $$f(x) = \begin{cases}1 , & x\in \Bbb R-\Bbb Q \\ 0, & x\in \Bbb Q\end{cases}$$
Show that $f$ is Lebesgue-Integrable but not Riemann-Integrable.
I know that $f$ is Lebesgue-Integrable iff $|f|$ is Lebesgue-Integrable. But not sure how this can be proved by the function. I'm also stuck on how to prove it is not Riemann-Integrable. Could someone help to provide a complete explanation please? Thanks.
Best Answer
Observe that $\mathbb{R} \setminus \mathbb{Q}$ is measurable because $\mathbb{Q}$, being a countable union of singletons, is measurable. Hence $$ f = \chi_{\mathbb{R} \setminus \mathbb{Q}} + 0\chi_{\mathbb{Q}} $$ $f$ is simple (it is in fact only a characteristic function), and thus integrable.
$f$ is not Riemann integrable on $[0,1]$ because its discontinuity set has positive Lebesgue measure. This can be proved by noticing the fact that a rational number and an irrational number exist inbetween every two real numbers.