Universally measurable set

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In mathematics, a subset $A$ of a Polish space $X$ is universally measurable if it is measurable with respect to every complete probability measure on $X$ that measures all Borel subsets of $X$ . In particular, a universally measurable set of reals is necessarily Lebesgue measurable (see #Finiteness condition) below.

Every analytic set is universally measurable. It follows from projective determinacy, which in turn follows from sufficient large cardinals, that every projective set is universally measurable.

[edit] Finiteness condition

The condition that the measure be a probability measure; that is, that the measure of $X$ itself be 1, is less restrictive than it may appear. For example, Lebesgue measure on the reals is not a probability measure, yet every universally measurable set is Lebesgue measurable. To see this, divide the real line into countably many intervals of length 1; say, N₀=[0,1), N₁=[1,2), N₂=[-1,0), N₃=[2,3), N₄=[-2,-1), and so on. Now letting μ be Lebesgue measure, define a new measure ν by

$\nu(A)=\sum_{i=0}^\infty \frac{1}{2^{n+1}}\mu(A\cap N_i)$

Then easily ν is a probability measure on the reals, and a set is ν-measurable if and only if it is Lebesgue measurable. More generally a universally measurable set must be measurable with respect to every sigma-finite measure that measures all Borel sets.

[edit] Example contrasting with Lebesgue measurability

Suppose $A$ is a subset of Cantor space $2 ω$ ; that is, $A$ is a set of infinite sequences of zeroes and ones. By putting a binary point before such a sequence, the sequence can be viewed as a real number between 0 and 1 (inclusive), with some unimportant ambiguity. Thus we can think of $A$ as a subset of the interval [0,1], and evaluate its Lebesgue measure. That value is sometimes called the coin-flipping measure of $A$ , because it is the probability of producing a sequence of heads and tails that is an element of $A$ , upon flipping a fair coin infinitely many times.

Now it follows from the axiom of choice that there are some such $A$ without a well-defined Lebesgue measure (or coin-flipping measure). That is, for such an $A$ , the probability that the sequence of flips of a fair coin will wind up in $A$ is not well-defined. This is a pathological property of $A$ that says that $A$ is "very complicated" or "ill-behaved".

From such a set $A$ , form a new set $A'$ by performing the following operation on each sequence in $A$ : Intersperse a 0 at every even position in the sequence, moving the other bits to make room. Now $A'$ is intuitively no "simpler" or "better-behaved" than $A$ . However, the probability that the sequence of flips of a fair coin will wind up in $A'$ is well-defined, for the rather silly reason that the probability is zero (to get into $A'$ , the coin must come up tails on every even-numbered flip).

For such a set of sequences to be universally measurable, on the other hand, an arbitrarily biased coin may be used--even one that can "remember" the sequence of flips that has gone before--and the probability that the sequence of its flips ends up in the set, must be well-defined. Thus the $A'$ described above is not universally measurable, because we can test it against a coin that always comes up tails on even-numbered flips, and is fair on odd-numbered flips.

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Categories: Descriptive set theory | Determinacy | Measure theory