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• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeMar 2nd 2014

New page: Henstock integral.

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeMar 2nd 2014

Is it the first theorem that explains why the Henstock integral behaves “better” than the Lebesgue integral? From memory, it’s absolutely continuous functions which have Lebesgue integrable derivatives (almost everywhere).

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeMar 2nd 2014

Yeah, I think so. Although Hake’s theorem may not be true in quite as strong a way for the Lebesgue integral either.

• CommentRowNumber4.
• CommentAuthorTobyBartels
• CommentTimeMar 5th 2014

Right, a function can be improperly Lebesgue integrable (and therefore Henstock integrable) without being Lebesgue integrable. (Such a function must alternate sign a lot.) Wikipedia gives the example of

$\int_{-1}^1 \frac{\sin (1/x^3)}{x} = \frac{1}{3} \pi - \frac{2}{3} Si(1) \approx 0.416476 .$

(Well, Wikipedia didn't specify the endpoints, so I did the integral on Wolfram Alpha.)

• CommentRowNumber5.
• CommentAuthorDavidRoberts
• CommentTimeFeb 20th 2015

I added the example that Toby gave, but as an integral from 0 to $x$ - as the integrand is even, one can easily figure out what happens for intervals with negative endpoint(s). I too am trusting WolframAlpha, but I would prefer a reference (or perhaps we can do it as an exercise!) that does it by hand.