NETMaths
Log inStart free

Is this true?

|f| Riemann integrable ⇒ f Riemann integrable

No — it is false.

The counterexample

f = 1 on elsewhere

|f| ≡ 1 is integrable; f is discontinuous everywhere.

The kind of mistake this is

Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

Drill statements like this

Watch it happen

Riemann sums closing on the integral

integration

Others that fail the same way

From Integration › Riemann integration and criteria

ShareWhatsAppTelegram