Talk:Almost everywhere

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I've always used "null set" to refer to a set contained within a set of measure zero, and "set of zero measure" when the set is actually measurable. (Most of the measures I work with are Borel, not Lebesgue.) Am I just wrong, or can we change the definitions to reflect this? -- Anon

I do the same, FWIW. -- Toby 20:11 Feb 12, 2003 (UTC)

I am no mathematician, but it would be useful to see in this article what distinction is being made between "almost everywhere" and "everywhere". Does the latter apply to neighborhoods rather than elements? --FOo 20:47, 14 Aug 2004 (UTC)

The only way I see to construe the first sentence of this article as failing to answer this question is to think that "null set" means "empty set", which is how it is defined in many books, so the sentence is confusing. I've corrected it. Michael Hardy 00:43, 15 Aug 2004 (UTC)

[edit] Fixing Dino's damage

The following true theorems were removed by User:Dino:

• If f : R → R is Lebesgue measurable and ∫_a^b|f(x)|dx<∞ for every real numbers a<b then there exists a null set E (depending on f) such that, if x is not in E, the Lebesgue mean 1/(2e)∫_x-e^x+ef(t)dt converges to f(x) as e decreases to zero. In other words, the Lebesgue mean of f converges to f almost everywhere. The set E is called the Lebesgue set of f.
• If f(x,y) is Borel measurable on R² then for almost every x, the function y→f(x,y) is Borel measurable.

The following false theorem was added:

• A bounded function f : [a, b] -> R is Riemann integrable if and only if it is continuous almost everywhere.

Counter-example: the indicating function of the rationals in [0,1] is constant outside the rationals (hence continuous there.) Since the rationals have measure zero, this function is a.e. continuous. However this is the canonical example of a non-Riemann-integrable function (all lower sums are zero, all upper sums are 1.)

Be more careful next time.

Loisel 17:42, 16 Sep 2004 (UTC)

The above marked "false theorem" is actually true. I'm currently working out a proof of it (it was assigned by my teacher, Paul J. Sally @ U of C.

The "counter-example" above is actually not a counter-example because, the indicator function is discontinuous everywhere. Proof: Given any x in [0,1], let ɛ = 1/2, for all ʛ>0 there is an x' in (x-ʛ,x+ʛ) such that |x-x'|=1>1/2=ɛ. That x' is in (x-r,x+r) follows from the fact that in any open interval there is an irrational and a rational. This shows that the function cannot be continuous at x because there is no delta for this ɛ. QED

The above theorem should be added back immediately. I'm not sure if I'm the person to do that...??

[edit] almost nowhere?

is this said of a property only true on a set of measure zero? MotherFunctor 06:01, 15 May 2006 (UTC)