Exercise · ZW-RIEMANN-INTEGRAL-005 · ★★★ Advanced
Let \(f=1\) on the rationals and \(f=0\) on the irrationals, over \([0,1]\). Is it Riemann integrable?
Choose one option.
Show answer
- B. No: the lower integral is \(0\) and the upper is \(1\)
Exercise · ZW-RIEMANN-INTEGRAL-005 · ★★★ Advanced
Choose one option.
Every partition has \(L=0\) and \(U=1\), so the supremum of the lower sums (\(0\)) and the infimum of the upper sums (\(1\)) differ. Being \(0\) except on a countable set is not enough.
A bounded function \(f\) on \([a,b]\) is integrable if the supremum \(L(f)\) of its lower sums and the infimum \(U(f)\) of its upper sums, over partitions \(P\), coincide; that common value is its integral.