I think you mean “The Lebesgue measure of the complement is zero” or something of the sort, not “The Lebesgue measure is non-zero”. A set with nonzero measure has to be “fairly big” but it certainly doesn’t amount to “almost all”.
You can’t say “almost all natural numbers are even” unless you have already picked a nonprincipal ultrafilter and it happens to have {even numbers} in it. Since there is no construction of a nonprincipal ultrafilter, what one always actually does is to say “Let U be a nonprincipal ultrafilter” and then proceed using just the properties that any nonprincipal ultrafilter has.
There’s also “the complement is countable” when you’re talking about real numbers, or more generally “the complement is of strictly smaller cardinality than the whole set”. (When put like that, of course it’s the same as “cofinite” for integers. Speaking of which, your heading for that one has “not finite” but again you mean (and say later on) “cofinite”.)
One proves the existence of nonprincipal ultrafilters with the axiom of choice. (You can show “every filter is contained in some ultrafilter” pretty directly using Zorn’s lemma.) You don’t actually need the full strength of AC—I think “every infinite set has a nonprincipal ultrafilter” is weaker than countable choice, for instance—but you definitely need some choice, and in particular “there are no nonprincipal ultrafilters” is known to be consistent with ZF.
I think you mean “The Lebesgue measure of the complement is zero” or something of the sort, not “The Lebesgue measure is non-zero”. A set with nonzero measure has to be “fairly big” but it certainly doesn’t amount to “almost all”.
You can’t say “almost all natural numbers are even” unless you have already picked a nonprincipal ultrafilter and it happens to have {even numbers} in it. Since there is no construction of a nonprincipal ultrafilter, what one always actually does is to say “Let U be a nonprincipal ultrafilter” and then proceed using just the properties that any nonprincipal ultrafilter has.
There’s also “the complement is countable” when you’re talking about real numbers, or more generally “the complement is of strictly smaller cardinality than the whole set”. (When put like that, of course it’s the same as “cofinite” for integers. Speaking of which, your heading for that one has “not finite” but again you mean (and say later on) “cofinite”.)
Thanks for the advice! Still learning how to phrase things correctly & effectively.
I wasn’t aware that you can’t actually explicitly construct a nonprincipal ultrafilter—this is interesting and nonintuitive to me!
One proves the existence of nonprincipal ultrafilters with the axiom of choice. (You can show “every filter is contained in some ultrafilter” pretty directly using Zorn’s lemma.) You don’t actually need the full strength of AC—I think “every infinite set has a nonprincipal ultrafilter” is weaker than countable choice, for instance—but you definitely need some choice, and in particular “there are no nonprincipal ultrafilters” is known to be consistent with ZF.