Universally measurable set

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.

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, "N"₀= [0,1), "N"₁= [1,2), "N"₂= [-1,0), "N"₃= [2,3), "N"₄= [-2,-1), and so on. Now letting μ be Lebesgue measure, define a new measure ν by: $u(A)=sum\_\{i=0\}^infty\; frac\{1\}\{2^\{n+1mu(Acap\; N\_i)$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.

Example contrasting with Lebesgue measurability

Suppose $A$ is a subset of Cantor space $2^omega$; 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 $A\text{'}$ 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 $A\text{'}$ 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 $A\text{'}$ "is" well-defined, for the rather silly reason that the probability is zero (to get into $A\text{'}$, 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 $A\text{'}$ 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.