What it says on the tin. Using the Borel $\sigma$-algebra on the reals instead of the Lebesgue $\sigma$-algebra has the advantage that it allows a broader class of measures, many of which are quite natural: For example the "uniform" measure on the Cantor set is defined on the Borel $\sigma$-algebra, but cannot be defined on the Lebesgue algebra. So why don't we just use the Borel $\sigma$-algebra for everything? What advantage does the Lebesgue $\sigma$-algebra have?

I mean, it has more measurable sets, but sets that are Lebesgue-measurable but not Borel-measurable (or for that matter, sets that are not Borel-measurable, period) are extremely pathological, not explicitly constructible, and (as far as I can tell) never show up naturally. And it's complete, but I have no idea what makes that a useful property.

  • $\begingroup$ If you have choice, I think it is not so hard to come up with an absolutely continuous function whose derivative fails to exist on a Lebesgue measurable, Borel nonmeasurable, measure zero set. Thus the fundamental theorem of calculus does not hold the way we would like in the Borel setting. By the way, some sets that are Lebesgue measurable and Borel nonmeasurable are explicitly constructible, it's just that the nonconstructive proof of their existence (using a Lebesgue nonmeasurable set as a starting point) is easier. $\endgroup$
    – Ian
    Feb 3, 2016 at 2:38
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    $\begingroup$ Lebesgue is complete. $\endgroup$
    – Aloizio Macedo
    Feb 3, 2016 at 2:38
  • $\begingroup$ The Borel $\sigma$-algebra doesn't even include all Jordan measurable sets, so you would end up with something weaker than the Riemann integral. $\endgroup$
    – user310283
    Feb 3, 2016 at 2:41
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    $\begingroup$ @user310283 It's not strictly weaker, since the inclusion doesn't go either way. Still a good point. $\endgroup$
    – Ian
    Feb 3, 2016 at 2:57
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    $\begingroup$ @Ian In fact for a completely arbitrary function $F$ the set of points at which the derivative $F'$ exists is a Borel set. (Due, I believe, to O. Hájek (1957).) So one does have to go in some other direction to find natural examples of non Borel sets that arise in analysis. $\endgroup$ Feb 3, 2016 at 20:18

1 Answer 1


You might be interested in what Borel thought about this. Every modern student learns that Lebesgue's measure is the completion of Borel's measure and that this is rather obvious. Take the Cantor set of measure zero (i.e., Borel measure zero). All subsets are Lebesgue measurable but not all subsets are Borel sets. It is clearly a "bad" thing to have sets (and functions) around that your theory has to avoid, so of course Lebesgue's measure is clearly more useful than Borel's.

But Borel didn't buy that. He was a bit of a constructionist. Not like you sometimes find among people that don't believe in infinite sets or even infinite decimal expansions. He didn't accept that any of these Lebesgue measurable sets that are not also Borel sets can be constructed in any acceptable way. He had given a procedure (countable but transfinite) that constructed all the Borel sets and Lebesgue had no demonstration that there were any other sets that you could actually encounter.

Borel and Lebesgue were the best of friends---until they weren't. It was this issue that drove them apart. Borel was a bit older and had supervised Lebesgue's dissertation. But he quite resented the acclaim that Lebesgue was getting for his measure and his integral when the original ideas were all due to Borel. If you fully believe that non Borel sets don't truly exist then it appears Lebesgue has stolen the glory and with no justification. Priority disputes among mathematicians are fairly rare, but they can be as bitter as such disputes in other fields.

I am not enough of an historian to tell much more of this story. (Of course that wouldn't stop me from telling such stories in lectures.) But I would say that, at least formally, this dispute couldn't have been settled until around 1914. That is when a young Russian mathematician (Suslin) showed that the projection of a two-dimensional Borel set onto one-dimension need not be a Borel set, but did have to be Lebesgue measurable.

I hope that this would have settled the issue in Borel's mind but, if so, it did not restore their friendship. But Borel might have enjoyed one aspect: Suslin made his discovery by finding a rather gross error in a paper of Lebesgue's, a paper that claimed the projection of a Borel set would be a Borel set. The mistake Lebesgue made was an embarrassingly simple one.


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