Problem with almost everywhere equality

In several aspects of mathematics dealing with measurability (such as measure theory, descriptive set theory, stochastic processes, ergodic theory, study of L p -spaces, etc.) the idea of identifying functions which are equal almost everywhere is quite natural. The experience gained from real-valued functions may lead to an oversight that the set on which two measurable functions (taking values in an arbitrary topological space) coincide is always measurable. It is well-known and quite easy to prove that this happens when functions take values in a space with countable base or, if they have separable images lying in a metrizable space. However, it is not true in general. The reason for this is that B(Y ) ⊗ B(Y ) differs (in general) from B(Y × Y ) where B(Y ) is the σ-algebra of all Borel subsets of a topological space Y . Let us say that a topological space Y has almost everywhere equality property

The aim of this short note is to prove that a metrizable space has (AEEP) iff card(Y ) 2 ℵ 0 . Thus, among metrizable spaces only those whose topological weight is no greater than 2 ℵ 0 have (AEEP). It may seem surprising that not only separable spaces appear in this characterization.

Nonseparable metric spaces are widely investigated in functional analysis and operator theory. In fact, the Banach algebra of all bounded linear operators acting on a separable Banach space is usually nonseparable. Also infinite-dimensional von Neumann algebras are nonseparable. A special and a very important in theory of geometry of Banach spaces example of them is the Banach space l ∞ of all bounded sequences. So, our result may find applications in investigations of these spaces.

Notation and terminology. For every topological space Y , B(Y ) stands for the σ-algebra of all Borel subsets of Y . That is, B(Y ) is the smallest σ-algebra containing all open sets. Whenever (Ω, M) and (Λ, N) are two measurable spaces, a function f

and card(E) is the cardinality of E.

We begin with a simple 2.1. Lemma. For a topological space Y the following conditions are equivalent:

Proof. If f, g : (Ω, M) → Y are two measurable functions, then the function h

This shows that (i) follows from (ii). To see the converse, notice that the natural projections

) are measurable and that ∆(p 1 , p 2 ) = ∆ Y which finishes the proof.

2.2. Lemma. For an arbitrary topological space Y , every member F of B(Y ) ⊗ B(Y ) may be written in the form (2-2)

Now let F consists of all sets F which may be written in the form (2-1)

To this end, put Λ = [0, 1] N (here 0 / ∈ N) and observe that

which finishes the proof since there is a bijection between Λ and [0, 1].

As an immediate consequence of the above results, we obtain (x,t)) = (u(x), t) and u(ω X ) = ω Y is continuous as well.

An important example of a topological cone is the so-called hedgehog space ([1, Example 4.1.5]). The hedgehog J(m) of spininess m ℵ 0 is the topological cone over a discrete space of cardinality m. Its importance is justified by the following result of Kowalsky [2] (see also [1,Theorem 4.4.9]; it is already known that [J(m)] ℵ 0 with infinite m is homeomorphic to the Hilbert space of Hilbert space dimension m, see [4,5] or Remark in Exercise 4.4.K of [1]).

As a colorrary of Theorem 2.5, we obtain the following result, which may be interesting in itself.

2.6. Proposition. Every metrizable space Y such that card(Y ) 2 ℵ 0 admits a continuous one-to-one function of Y into a separable metrizable space.

Proof. Let D = [0, 1] and T = [0, 1] be equipped with, respectively, the discrete and the natural topology. By Lemma 2.4, the function C(D) ∋ z → z ∈ C(T ) is continuous. This means that there is a one-to-one continuous function u : J(2 ℵ 0 ) → W where W is a separable metrizable space. But then the function

continuous and one-to-one as well. Now it suffices to apply Theorem 2.5.

We are now able to prove the main result of the paper. Assume (iii) is fulfilled. We infer from Proposition 2.6 that there is a separable metrizable space X and a continuous one-to-one function u : Y → X. Then v = u×u : (Y ×Y, B(Y )⊗B(Y )) → (X ×X, B(X)⊗ B(X)) is measurable (v(x, y) = (u(x), u(y)) for x, y ∈ Y ). Since X is separable, B(X) ⊗ B(X) = B(X × X) and therefore ∆ Y = v -1 (∆ X ) ∈ B(Y ) ⊗ B(Y ) and we are done.

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