Is there any measure on $([0,1], P([0,1]))$ which satisfies: $$ \mu ([0,1]) = 1 $$ $$ \mu(A) \in \{0,1\} $$ for any $A \subset [0,1]$ $$\mu(B) = 0$$ for any $ B \subset [0,1]$ finite.

My "guess" is NO, because of the last measure. If the measure of B is zero, then B should be the empty set. No?

  • No. The measure of the singleton is also zero. Therefore, countable sets have measure zero. – ToposLogos Oct 17 at 12:41
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    It can’t be sigma additive. – Marco Lecci Oct 17 at 12:43
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    There are a lot of non-empty sets with measure zero in the usual measure of $[0,1]$. – Saucy O'Path Oct 17 at 13:13
  • @MeesdeVries: Uhh, why? You'd get a countably complete ultrafilter. How can you get one with just transfinite induction? – Asaf Karagila Oct 17 at 13:46
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    No, you cannot have such a measure. The keywords to search for relevant information here are "measurable cardinals ". – Andrés E. Caicedo Oct 17 at 14:39
up vote 4 down vote accepted

Surprisingly, the answer is no due to serious set-theoretic restrictions. If $X$ is an infinite set such that there is a ($\sigma$-additive) measure on $\mathcal P(X)$ that only takes the values 0 and 1, assigns 1 to $X$ and 0 to singletons, then the cardinality $\kappa=|X|$ of $X$ is much much larger than $\mathfrak c=|\mathbb R|$, the cardinality of the set of reals. (It is immediate that $\kappa$ is uncountable.)

It is so large, in fact, that the standard axioms for set theory do not suffice to prove the existence of sets of that size.

Indeed, if $\kappa$ is the least possible size of such a set $X$, then any such measure $\mu$ is not only $\sigma$-additive, but in fact it is $\kappa$-additive, that is, $\mu\bigl(\bigcup_{i\in I}X_i\bigr)=\sum_i \mu(X_i)$ holds for disjoint unions of subsets of $X$ as long as the index set $I$ has itself size less than $\kappa$. If a set $Y$ admits such a $|Y|$-additive $\{0,1\}$-valued measure on $\mathcal P(Y)$, we say that $|Y|$ is a measurable cardinal.

If $\kappa$ is measurable, then for any $A$ with $|A|<\kappa$, we have that also $|\mathcal P(A)|<\kappa$. In particular, not only is $\kappa$ larger than $\mathfrak c$. It is also larger than $2^{\mathfrak c}$, $2^{2^{\mathfrak c}}$, etc (we say that $\kappa$ is strong limit). Also, if a set of size $\kappa$ is split into fewer than $\kappa$ many pieces, then at least one of these pieces must have size $\kappa$ itself (we say that $\kappa$ is regular). These two properties say that $\kappa$ is a strongly inaccessible cardinal. But more is true: If $\kappa$ is measurable, then there are $\kappa$ strongly inaccessible cardinals smaller than $\kappa$. And being inaccessible is already a very strong property, beyond what the axioms of set theory can show exists.

An excellent reference for these matters is the book

MR1994835 (2004f:03092). Kanamori, Akihiro. The higher infinite. Large cardinals in set theory from their beginnings. Second edition. Springer Monographs in Mathematics. Springer-Verlag, Berlin, 2003. xxii+536 pp. ISBN: 3-540-00384-3

Curiously, the answer is a bit different if you allow the measure to take any values in $[0,1]$ rather than just 0 and 1. The size of a set admitting such a measure is said to be an (atomlessly) real-valued measurable cardinal. Any such cardinal has size at most $\mathfrak c$. The existence of such a cardinal is equivalent to the existence of a measure extending Lebesgue measure and defined on all sets of reals. (Of course, no such extension can be translation invariant.)

If these cardinals exist, then the continuum hypothesis fails very badly. In particular, if $\kappa$ is real-valued measurable, then $\kappa$ is still regular and a limit cardinal, meaning that if $|Y|<\kappa$, then $|Y|^+$, the next infinite size strictly larger than $|Y|$, is also smaller than $\kappa$. (Cantor's theorem says that $|Y|<|\mathcal P(Y)|$, so $|Y|^+\le|\mathcal P(Y)|$, which means that being a strong limit cardinal, as the name suggests, is a stronger requirement than being a limit cardinal.)

Again, we cannot prove that real-valued measurable cardinals exist. However, if it is consistent that they do, then it is consistent that the size of the reals is real-valued measurable.

Suppose you had such a measure $\mu$. Observe that either $\mu([0,1/2])=1$ or $\mu([1/2,1])=1$, since if both were $0$ then $\mu([0,1])$ would be $0$. Let's say that $\mu([0,1/2])=1$. Then by the same reasoning, either $\mu([0,1/4])=1$ or $\mu([1/4,1/2])=1$. Continuing this process, we get a descending sequence of closed intervals $I_0\supset I_1\supset I_2\supset\dots$ such that $\mu(I_n)=1$ for each $n$ and $I_n$ has length $1/2^n$. Since $\mu(I_n)=1$ for each $n$, we must also have $\mu(\bigcap I_n)=1$. But $\bigcap I_n$ is a singleton set, so this is a contradiction.

To relate this to Andrés Caicedo's answer, this argument can be generalized to show that any $\{0,1\}$-valued measure on $P(X)$ which is $\kappa$-additive must also be $2^\kappa$-additive, for any set $X$ and any cardinal $\kappa$. This implies that any measurable cardinal is a strong limit cardinal.

  • One can also build on this argument to show that any non-atomic $[0,1]$-valued measure on $\mathcal P(X)$ cannot be $\mathfrak c^+$-additive, and therefore any real-valued measurable cardinal has size at most $\mathfrak c$. – Andrés E. Caicedo Oct 17 at 20:56

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