Riemann integrable vs Lebesgue integrable

The proof for Riemann could be easier.

Since $\int_a^bf(x)dx>0$, there is a Partition $P$ such that $L(P,f)>0$, where $L(P,f)$ means Darboux lower sum. So there must be a interval, where $\inf f>0$


The theorem is false in the Lebesgue setting. Consider $$f(x) = \begin{cases} 0 & x \in \mathbb{Q} \\ 1 & \textrm{otherwise} \end{cases} .$$ Your very first sentence of your "proof" is wrong.

EDIT: I just realized this was meant to be a proof for Riemann-integrable functions, not Lebesgue.


For Lebesgue integrals, this is not even true up to modification on a set of measure zero. Take for instance the indicator function of a "fat" Cantor set (cf. page 39 of Folland's 'Real Analysis').