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.

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  • B. No: the lower integral is \(0\) and the upper is \(1\)

Explanation

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.

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Review the concept: Riemann integral

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.

\[L(f)=\sup_P L(f,P)=\inf_P U(f,P)=U(f)=\int_a^b f(x)\,dx\]

More practice problems: riemann integral

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