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What is the relationship between the improper Riemann-integral and Lebesgue integrability?

There's a very sloppy section of it in my book with about 4-5 known mistakes found so far, so I fear I have a completely wrong idea of what's what now.

My question is, what must be true for some $u$ and its improper Riemann-integral for it to be Lebesgue-integrable, and what must be true for the Lebesgue-integrability of $u$ for it to have finite or infinite improper Riemann-integrable?

I know for example that $\sin(x) / x$, $x \in (0,\infty)$ is improperly Riemann-integrable but not Lebesgue integrable. What's gone wrong here?

Oversaa
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1 Answers1

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An improper Riemann integral is Lebesgue integrable if it is absolutely convergent. $\int_0^\infty \sin t/t\,dt$ is convergent but not absolutely convergent.

If a Lebesgue integrable function is also Rieman integrable, then its integral is absolutely Riemann convergent.

Julián Aguirre
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  • Don't forget that the same distinction applies to the improper Riemann integral on a bounded interval $[a,b]$. Everyone should know about the existence of a differentiable function $F$ with the property that $\int_0^1 F'(x),dx=F(1)-F(0)$ in the sense of the improper Riemann integral but not in the sense of the Lebesgue integral. – B. S. Thomson Oct 29 '15 at 15:38