Is a bounded real function defined on a singleton Riemann Integrable?

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I know to be Riemann integral a real function $f$ must bounded be defined on a closed bounded interval. A singleton is technically a closed bounded interval so is this permitted?

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Yes. And note that a function from a singleton into $\Bbb R$ is automatically bounded.

On the other hand, the integral of such a function will alwyas be equal to $0$.