Non-atomic probability takes all values between $0$ and $1$

Posted 2020-11-12 xiongruimath

I got to know the following problem yesterday given by professor Feder Patrov. 

Let $mu$ be a probability over $X$ (i.e. a measure with $mu(X)=1$). If $mu$ is non-atomic, then for any $Cin [0,1]$, there exists $Ysubseteq X$ such that $mu(Y)=C$.

Where a measure is said to be non-atomic if every measurable $E$ such that $mu(E)>0$ contains a measurable subset $A$ such that $0< mu(A)< mu(E)$.

The intuition is to approximate the number. But after a peroid of attemption, it turns to be not that easy to ensure the approximation works. Actually, I read a proof from Rudin‘s textbook functional analysis Page 120. 

We can define a functional $$Lambda : L^infty(sigma)longrightarrow mathbb{R} qquad glongmapsto int gmathrm{d}mu. $$ Consider the ``unit cube‘‘ of $L^infty(sigma)$, $$K={gin L^infty(sigma): 0leq gleq 1}$$ then $K$ is weak-$*$ compact (by Banach-Alaogu theorem) and convex (trivially). Hence $Lambda(K)$ is a compact convex subset of $mathbb{R}^n$.

We will prove that $Lambda(K)$ is the desired range. The main idea is that the extreme points of a finite dimensional cube (resp. unit cube in $L^infty(sigma)$) is the $0,1$-tuple (resp. characteristic function). 

For any $pin Lambda(K)$, denotes its inverse image by $K_g=Lambda^{-1}(p)$. If $g_0in K_p$ is not a characteristic function, then some measurable $E$ with $sigma(E)>0$ and $rleq g_0leq 1-r$ on $E$. Since $sigma$ is not atomic, $chi_E cdot L^infty(sigma)$, the space of functions living on $E$, is of infinite dimension, so some nonzero element gets killed by $Lambda$, say $Lambda g=0$. We can assume $-r<g<r$, then $g_0+g$ and $g_0-g$ are both in $K_p$, so $g_0$ turns not to be an extreme point. %In conclusion, $g_0$ is not extreme point. By Krein-Milman Theorem, there always exists extreme point.

The image is as follow. 

 

 

 There are some remarks on this proof. 

• It requires axiom of choice, of course (since Alaogu Theorem and Krein-Milman Theorem use). 

• It is equivalent to the following assertion

  "Suppose $mu_1,ldots,mu_n$ are non-atomic measures. Then the set consisting all
  $$igg(,mu_1(E),ldots ,mu_n(E),igg)in mathbb{R}^n$$
  is a compact convex subset of $mathbb{R}^n$."

  Because a subset of $mathbb{R}^n$ is convex iff for any linear function $f$, the image under $f$ is convex. Or modify the above proof a little to get it.

• In Rudin‘s book, the theorem is more general for real-valued measure. The proof is similar by considering the total variation measure.
• The usage of extreme points seems to only work for $infty$-norm and $1$-norm, because in that case the unit ball is polygon (``straight‘‘). 

After that, I considered whether there exists an elementary proof? Follow the intuition of approximation the number, the gap is that we cannot approximate it well. For example, $c=1/2$, you may find a series of measurable sets of measures $frac{3}{4}+frac{1}{4},frac{3}{4}+frac{1}{8},frac{3}{4}+frac{1}{16},cdots$. They get close to $1/2$ but not approach $1/2$. Now you should choose the intersection of them to get a measurable set of measure $3/4$, and repeat such procedure, but it may be $frac{5}{8}+frac{1}{8},frac{5}{8}+frac{1}{16},cdots$. If we want to make correct induction, we must use some tool in set theory. 

We use ordinal number to prove the assertion. We will define a function $f$ from the first uncountable ordinal number $aleph_1$ to the sigma-algebra of measurable sets $mathfrak{M}$. Such that $alphain eta$ implies $f(alpha)subseteq f(eta)$ or $f(eta)subseteq f(alpha)$. %$0<mu(f(eta))-c<mu(f(alpha))-c$.

• define initially $f(varnothing)=X$.
• assume that for $alphain mathbf{Ord}$, elements less than $alpha$ is defined. Denote
  $$U_{alpha}=igcapleft{f(eta):egin{array}{l}mu(f(eta))>c\\ eta<alphaend{array} ight}
  qquad L_{alpha}=igcupleft{f(eta):egin{array}{l}mu(f(eta))<c\\ eta<alphaend{array} ight}$$
  Now $U_{alpha}setminus L_{alpha}$ is measurable (only at most countable many) and also of measure nonzero (otherwise, the conclusion follows). So we can find some $A$ with the property with respect to $U_{alpha}setminus L_{alpha}$, then we define $f(alpha)=L_{alpha}cup A$.

Such function exists by transfinite induction(which is equivalent to axiom of choice). But now we have uncountable many subsets with distinct measures, but $mathbb{R}$ is so crowed to accept uncountable many nonintersetion open intevals 
$${ extrm{real numbers between $mu(f(alpha))$ and $mu(f(alpha+1))$}}$$
(This is a pure topology argument by picking one rational number in each interval).

 Of course, this is a natural proof but not an elementary proof.