I have a question about my Mathematics Stack Exchange post: How to prove that the surface of sphere is measure zero.

I posted the question with context, but it is said that additional context is needed. What do you need?

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    $\begingroup$ Did you click on the "please provide context" link so you could read what it says at math.meta.stackexchange.com/questions/9959/… Eggman? $\endgroup$ Nov 8, 2022 at 12:29
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    $\begingroup$ @GerryMyerson Yes, I clicked. What context do you need?The definition of sphere?The definition of Riemann integrability? $\endgroup$
    – user1114152
    Nov 8, 2022 at 16:21
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    $\begingroup$ The context needed ,essentially, is why the fact that the sphere has Lebesgue measure zero helps in finding the volume of the sphere via a Riemann integral. It's the clash of the notions of "Riemann" integrability contrasted with the "Lebesgue" measure that is causing the confusion, at least as far as I'm concerned. If the question was merely restricted to "why does the boundary have Lebesgue measure zero?", then I think this confusion could have been avoided (although other context might have been required / the question may have been a duplicate). $\endgroup$ Nov 8, 2022 at 17:43

1 Answer 1


To integrate on the sphere, I need to prove the Riemann integrability,

Riemann integrability of what?

so I want to prove that the boundary of the sphere is measure zero.

Which measure?

It's also not at all clear what role the boundary being measure zero plays in your volume calculation.

Honestly, I couldn't make much sense out of your question.

  • $\begingroup$ I think this level of detail can easily be inferred without ambiguity. 1. Riemann integrability of a function which I want to integrate on the sphere. 2. Is there any possibility other than Lebesgue measure in this context? You seem to be unaware of the relationship between Riemannian integrability and the set of discontinuous points having measure zero. $\endgroup$
    – user1114152
    Nov 8, 2022 at 16:49
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    $\begingroup$ You asked for details about how people reacted to your question. I told you how I reacted to your question. You corrected me. You win; I quit. $\endgroup$
    – JonathanZ
    Nov 8, 2022 at 17:04
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    $\begingroup$ @Eggman: It is easy to misjudge how well "this level of detail" can be "inferred without ambiguity" by Readers who lack knowledge of what you are trying to do. You edited the post slightly soon after posting to introduce "measure zero" in place of zero volume, so I tried to grasp your purpose. At the outset you stated you want to calculate the volume $V=\pi r^3/3$ enclosed by a sphere of radius $r$. It's unclear what discontinuities on the sphere appear in your calculation. $\endgroup$
    – hardmath
    Nov 8, 2022 at 17:06
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    $\begingroup$ They still are keeping secret the exact nature of the function they wish to integrate. @Eggman, making the readers do all this extra work of figuring out what you are trying to do is exactly the kind of thing that gets questions closed. $\endgroup$
    – JonathanZ
    Nov 8, 2022 at 17:17
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    $\begingroup$ I have an idea of what kind of context is needed. Thanks. $\endgroup$
    – user1114152
    Nov 8, 2022 at 17:40
  • $\begingroup$ Apologies for omitting a factor of $4$ in my comment. I'm posting from my phone. $\endgroup$
    – hardmath
    Nov 8, 2022 at 17:53
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    $\begingroup$ @hard, your phone has a thing about factors of $4$? $\endgroup$ Nov 8, 2022 at 21:31

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