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To my understanding the definition of the McShane integral is identical to the definition of the Henstock–Kurzweil integral with the exception that each tag does not have to be contained in the subinterval for which it is assigned to. However, the gauge function in general will force the corresponding subinterval to be relatively close to the tag. Still, the tag can be picked a bit outside of the subinterval. If I am not mistakes this implies that a function is McShane integrable if and only if it is absolutely integrable. This is not the case for the Henstock–Kurzweil integral. I do not understand how such a subtle modification of the definition of the Henstock–Kurzweil integral makes every McShane integrable function absolutely McShane integrable. By functions I am simply refering to functions on $\mathbb{R}$ for the sake of simplicity. Is there an intuitive way to view the differences between the aforementioned integrals and in particular why McShane integrable functions are absolutely integrable?

J. W. Tanner
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David
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    Intuitively it is because if you have a (free) tagged partition that 'highlights' the negative values of the function, and one that 'hightlights' the positive values, and both chosen as in the definition of integral such that all refinements are close to the integral, then for the common refinement of the two partitions, you can choose the tags using the previous tags either 'highlighting' the positive values of the function or the negative values of the function. But both choices must give sums that are close, by definition of the integral. – MoonLightSyzygy Dec 22 '19 at 15:32
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    In Swartz's Introduction of Gauge Integrals you can find a proof that the McShane integral is an absolute integral. – MoonLightSyzygy Dec 22 '19 at 15:33
  • Thank you, it seems a bit more clear to me now. – David Dec 22 '19 at 20:21

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