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If $(X,\mathcal{B},\mu)$ is a (non-necessarily complete) measure space, we can give two different notions of a property $P(x)$ that is true almost-everywhere :

(D1) There is a measurable set $A$ such that $\mu(X\backslash A)=0$ and such that for all $x\in A$, $P(x)$ is true.

(D2) The subset of all $x\in X$ such that $P(x)$ holds is measurable and its complement has zero measure.

(D1) seems the one commonly used, and is obviously weaker than (D2). Of course, (D1) and (D2) are equivalent when the measure is complete.

My question is : if we take (D2) as the definition of "almost-everywhere", do we "lose" anything fundamental in the theory of measure and integration ? It seems to me (but I'm not sure) that everything is still true, for instance, the integral of a positive measurable function is zero implies that the function is zero almost everywhere in the sense of (D2). The Fubini theorem seems again to hold with (D2) too. The completeness of $L^p$ spaces (with the equivalence relation associated to almost everywhere equality corresponding to (D2)) seems to hold again (but I'm not completely sure).

I ask this question because even if (D2) is a little more annoying to use when proving that something holds almost-everywhere (we have to check measurability of the set on which the property holds), in the end it gives statements that are more precise (because precisely mesurability is implied). Therefore, I am wondering if I'm not missing anything by using (D2) only.

Andrés E. Caicedo
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Jon-S
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    My main objection to (D2) would be that it loses the monotonicity, which is quite a natural quality for a notion of "almost everywhere" – Pietro Majer Dec 17 '16 at 17:21
  • Yes indeed, and my question in this case becomes : does monotonicity play a crucial role in the theory (or is it just a convenient property to have). – Jon-S Dec 17 '16 at 17:26
  • (D1) and (D2) are equivalent when $P$ is a measurable property. So, the distinction becomes important only in the not-so-common cases where $P$ is not measurable. When $P$ is not measurable, (D1) still conveys some information, while (D2) is useless. – Algernon Dec 17 '16 at 18:11
  • Yes, and therefore, my question is : are those not-so-common cases where P is not measurable actually happen sometimes ? I could not find one example where this happen "naturally". – Jon-S Dec 17 '16 at 18:14
  • As for the ``precision of (D2)'' that you mention in the last paragraph, wouldn't it be less confusing to break (D2) into its two independent ingredients, namely that $P$ is measurable (lemma 1) and (D1) holds (lemma 2)? – Algernon Dec 17 '16 at 18:19
  • I can't think any natural scenario in which $P$ is not measurable, but I suppose it can happen in big spaces, where there is more than one natural $\sigma$-algebra. – Algernon Dec 17 '16 at 18:33
  • @Algernon : Yes, you are right, one way to do it is to separate both properties, in a way I wanted to put those two properties in one unique "compact" terminology since all examples I could find systematically imply both properties together. But maybe in some settings, one has to separate them. – Jon-S Dec 17 '16 at 18:41
  • @Jon-S: you might get more detailed answers if you add a "set theory" tag. I think a lot is known about how much ordinary mathematics can be done in models in which all sets of reals are measurable. Set theorists could have something to say here. – Nik Weaver Dec 18 '16 at 00:19
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    I would imagine the key advantage of the first definition is not having to think about whether the set is measurable. – Will Sawin Dec 18 '16 at 11:33
  • @NikWeaver : I tried, but it was removed. – Jon-S Dec 18 '16 at 15:16
  • Hmm. Probably the connection to set theory wasn't clear. Try asking a separate question about how much measure theory goes through in models where all sets of reals are measurable (set-theory, measure-theory)? – Nik Weaver Dec 18 '16 at 15:38

2 Answers2

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There are relatively concrete examples of a Borel set $E$ in the plane $\mathbb R^2$ such that the projection $$ p_1(E) = \{ x \in \mathbb R | \exists y, (x,y) \in E\} $$ is not a Borel set. (But it is a Lebesgue set.) LINK
So, as far as Borel sets in the line are concerned, condition $p_1(E)$ is a non-measurable condition. There is a Borel set $Q \supset p_1(E)$ such that $Q \setminus p_1(E)$ is a null set. And then:

$\qquad$for almost all $x \in Q$ there exists $y$ such that $(x,y) \in E$

is true in sense (D1) but not in sense (D2).

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Gerald Edgar
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With the (D2) definition of "almost everywhere", the corresponding notion of almost everywhere convergence of measurable sequences of is pathological. You can have a sequence of functions $(f_n)$ that converges to $f$ almost everywhere with a subsequence $(f_{n_k})$ that doesn't converge to $f$ almost everywhere in the sense of (D2), simply because the set of points at which $(f_{n_k})$ converges to $f$ need not be measurable.

Cameron Zwarich
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    Maybe I didn't understand well, but isn't the set of points for which $(f_n)$ converge measurable, and also the set of points $B$ for which $(f_{n_k})$ converge ? We can then define a measurable function $g$ equal to $\lim f_{n_k}$ on $B$ and zero elsewhere. Then the set of points for which $(f_{n_k})$ converge to $f$ is $B\cap C$ where $C$ is the measurable set on which $f$ and $g$ coincide. – Jon-S Dec 18 '16 at 15:43
  • Here is an example of the above situation. Consider the Borel sets on the reals, let $C$ be the Cantor set and $\Omega\subseteq C$ a non-measurable set. Further define $f=\chi_\Omega$, $f_{2n} = 0$ and $f_{2n+1} = \chi_C$. With (D1): $(f_n)$ and every subsequence converge almost everywhere to f. With (D2): $(f_n(x))$ converges to f(x) iff $x\in C$, hence almost everywhere. But $(f_2n(x))$ converges to $f(x)$, iff $x\notin\Omega$, i.e. on a nonmeasurable set. – gsa Dec 18 '16 at 21:33
  • @gsa : I'm not sure I understand, don't we have $f_n(x)=(1+(-1)^{n+1})/2$ if $x\in C$ with your definitions ? In that case, it doesn't converge to $f(x)$ iff $x\in C$. – Jon-S Dec 19 '16 at 02:01
  • @Jon-S sorry, I wanted to write: $(f_n(x))$ converges to $f(x)$ iff $x\notin C$. – gsa Dec 19 '16 at 06:25
  • I see, but I was understanding the answer of CameronZwarich with $f$ measurable. This hypothesis seems natural, since a sequence of measurable functions that converge almost everywhere, always converge almost everywhere to a measurable function. Modifying $f$ artificially to a non-measurable function indeed creates the pathology, but this feels really forced. My question can be interpreted as : when is there a situation where "naturally" (meaning starting only with measurable functions) we end up having to use (with no choice about that) a non measurable function or property? – Jon-S Dec 19 '16 at 06:56