What's the use of a complete measure? - MathOverflow

What's the use of a complete measure?

2010-01-12T16:52:23Z

A complete measure space is one in which any subset of a measure-zero set is measurable.

For what reasons would I want a complete measure space? The only reason I can think of is in the context of probability theory: using complete probability spaces forces almost-everywhere equal random variables to generate the same sigma-sub-algebra.

Am I missing some other important technical reasons?

Answer by Anweshi for What's the use of a complete measure?

2010-01-12T17:25:24Z

Wikipedia gives one instance of a situation in which complete measures are needed, for the purpose of defining measures on product spaces.

I suggest that you look into Rudin's "Real and Complex Analysis". There he makes an argument that a completion of an ordinary measure space into a complete measure space is just as fundamental to real analysis, as the completion of the rationals to the reals is.

Many theorems in measure theory, for instance Fubini or Radon-Nikodym, needs completeness to make full sense. Fubini is explained in the wikipedia example. To make the other aspect clear -- quite a few statements in measure theory uses the notion of "almost everywhere" -- for instance the definition of $L^p$ spaces, or Radon-Nikodym.

But this notion of "almost everywhere"(rather, "almost nowhere") becomes better if the measure space is complete. It would look really odd if you declare that some property holds true almost nowhere because it holds only on some set with measure zero, and you so arrange things that some other property holds on a smaller set, and then you are no longer able to make the assertion! The product measure example above is a specific illustration in which the concerned property is simply "being measurable", and the consequences are particularly notable.

Added(Jan 16): There are problems into applications into Ergodic theory, for instance. This definition of ergodic transformation and an ergodic theory built on it will run into all sorts of problems if the underlying measure space is not complete. This is again because you need a proper notion of "almost everywhere" and "almost nowhere".

Answer by The Bridge for What's the use of a complete measure?

2010-01-12T17:33:57Z

Hi Tom E,

For Stochastic Processes, completion of the initial sigma-field of the natural filtration of the process with the negligeable sets of the Probability measure of the limit sigma-field of the filtration (coupled with right continuity of the underlying filtration) is really uselfull.

It allows to find version of processes that are càdlàg (right continuous with left limit) under very general conditions. Càdlàg processes are the main object of study in Stochastic Processes analysis (only my point of view).

As a matter of fact, this is so usefull, that those two conditions are called the "usual conditions" for the probability space and filtration on which process lives (the 3-tuple $(\Omega, (\mathcal{F}_t),P)$ is named a stochastic basis).

If interested, you can have a look at Karatzas and Shreve's book on Brownian motion and Stochastic Calculus, but even if I realise that the matter might be quite far from your day-to-day mathematical activities this is certainly an example which shows how usefull completion of sigma field might be.

Regards

Answer by Joel David Hamkins for What's the use of a complete measure?

2010-01-12T17:51:07Z

Since the existence of non-measurable sets is often seen as undesirable, we naturally want to have as many measurable sets as possible. With Lebesgue measure on the reals, for example, if we were to stop with the collection of Borel sets, we would only have continuum c many measurable sets. But when completing the measure, we gain 2^c many more measurable sets, incomparably more. The newly measurable sets are not just measure zero sets, of course, but all those sets that differ from a previously measurable set by (a subset of) a measure zero set.

But it isn't just about the number of measurable sets. Rather, completing the measure allowed us to increase (or even maximize in a sense) our collection of measurable sets in a way that seems to accord completely with how we wanted to measure sets in the first place. It's a basic part of what we were trying to do with measure to be able to say that something that is less than negligible is also negligible.

Answer by Dmitri Pavlov for What's the use of a complete measure?

2010-01-13T04:16:23Z

From the categorical viewpoint there is no difference, because the category of measurable spaces is equivalent to the category of complete measurable spaces with the equivalence given by the completion functor. Moreover, we are forced to identify objects that are different only on a set of measure 0 or a subset of such a set (otherwise some theorems simply do not make any sense), hence we cannot even see the difference. However, working with complete measurable spaces is technically easier. More precisely, objects of the category of measurable spaces are triples (X,A,N), where X is a set, A is a σ-algebra of measurable subsets of X, N is a σ-ideal of null sets in A. A morphism from (X,A,N) to (Y,B,O) is an equivalence class of maps of sets f: X→Y such that the preimage of every element of B is a union of an element of A and a subset of an element of N and the preimage of every element of O is a subset of an element of N. Two maps are equivalent if they differ on a subset of an element of N. If we restrict our attention to complete measurable spaces, then the definition of morphism becomes significantly simpler: we have to require that the preimage of every element of B is an element of A and likewise for O and N and two maps are equivalent if they differ on an element of N.

This definition is too general to be useful for measure theory. Once we restrict ourselves to the subcategory of localizable measurable spaces (all major theorems of measure theorem such as Riesz representation theorem and Radon-Nikodym theorem imply the property of localizability) the resulting category becomes contravariantly equivalent to the category of commutative von Neumann algebras, also known as W*-algebras. In my opinion this constitutes the best possible definition for the main category of measure theory, both in terms of conceptuality and effectiveness, just as the best way to define the category of affine schemes is to make it equal to the opposite category of the category of commutative rings. Such a viewpoint is unfortunately highly unlikely to be adopted by analysts (especially hard analysts) considering their unwillingness to study even the most elementary notions of category theory.

Answer by Matus Telgarsky for What's the use of a complete measure?

2010-01-13T04:53:45Z

In light of the comments here, I'm going to show why completeness can be a pain. In exercise 9 of section 2.1 of Folland, he develops a function $g: [0,1] \to [0,2]$ by $g(x) = f(x) + x$ where $f : [0,1] \to [0,1]$ is the Cantor function. In that exercise it is established that $g$ is a (monotonic increasing) bijection, and that its inverse $h = g^{-1}$ is continuous from $[0,2]$ to $[0,1]$.

Since $h$ is continuous, it is Borel measurable. On the other hand, $h$ is not $(\mathcal{L}, \mathcal{L})$-measurable!! In particular, let $C$ be the Cantor set; $m(g(C)) = 1$, but this means there is a subset $A \subseteq g(C)$ which is not Lebesgue measurable. On the other hand $B := g^{-1}(A) \subseteq C$ whereas $m(C) = 0$; thus this preimage $B$ is Lebesgue measurable (with measure zero). But therefore $h^{-1}(B) = A$ is not Lebesgue measurable, meaning $h$ is not $(\mathcal{L}, \mathcal{L})$-measurable.

On one hand, this function is contrived. On the other hand, it shows that completing measures can mess things up. The typical definition of "measurable function" is a Borel measurable function, and I suppose reasons like the above led to this convention. I do not know the material Bridge references above, and so can't say what breaks when completeness is dropped. Although it seems mathematically convenient to throw in completeness, I don't know any examples in basic probability theory where it helps. For instance, Fubini-Tonelli can be formulated just fine without completeness. Your statement of the theorem only need mention completeness if your measures happen to be complete!

EDIT I corrected the nonsense in the second paragraph; also I meant to talk about $(\mathcal L, \mathcal L)$-measurable functions, which I accidentally refered to as Lebesgue measurable (which means $(\mathcal L, \mathcal B)$-measurable). My whole point is that if you take completion in $\sigma$-algebra of the range space, the extra sets you added could map back to basically anything. IE it is somewhat nonsensical to add in all sorts of null sets, but not all sorts of finite measure sets. Sometimes completion gives you something you want, but sometimes it does not, as I showed here--the function is better behaved wrt the non-completed measure.

Answer by Kaleb for What's the use of a complete measure?

2011-08-05T01:16:35Z

What's missing here is the Caratheodory extension process creates a complete measure space. Hence, to have a product measure that is not complete requires one use a different method to create it. If this is the case, then completeness is not required in Fubini or Tonelli, e.g. (R, A, mu), (R, B, nu) Borel measure spaces and product measure defined as mu X nu for sigma-algebra A X B. However, this is not the traditional way we construct a measure and in particular would not construct the Lebesgue measure on R2. However, if we take two measure spaces which are both not complete and create their product measure using the Caratheodory extension of mu X nu (a complete measure), the equality in the conclusion of Fubini and Tonelli would not necessarily hold.

Hence, completeness of one or both measure spaces in the hypotheses is only important (required) if we want to construct our product measures using the Caratheodory extension process or if our product measure is complete.

Answer by heinrichvw for What's the use of a complete measure?

2012-07-15T01:40:26Z

Hi Tom E,

Here is another reason: Let $E$ be a Borel set in Euclidean space. Then its image under a continuous map is always Lebesgue-measurable but in general not Borel measurable. Results like this make the completion useful; The theory behind this is the theory of analytic sets or the Souslin operation.