# Is there any $F \in \mathscr{F}$ such that $\mu(F)=x$?

Let $(\Omega,\mathscr{F},\mu)$ be a probability space such that $\mu$ is non-atomic, and fix $x \in [0,1]$. Is it true that one can find $F \in \mathscr{F}$ for which $\mu(F)=x$? And what if $\mu$ is only finitely additive?

Ps. Here we recall that $\mu$ is non-atomic if $\mu(X)>0$ for some $X \in \mathscr{F}$ implies that there exists $Y \in \mathscr{F}$ for which $0<\mu(Y)<\mu(X)$.

• If $\mathscr{F}=\{\Omega,\varnothing\}$, then $\mu(F)=x$ has a solution only for $x=0$ and $x=1$. – Blackbird Aug 15 '15 at 22:10
• It is not non-atomic. Seeing your comment, I add the definition in the text – Paolo Leonetti Aug 15 '15 at 22:12
• Erm... I had a different definition in mind, sorry! Then, the answer is yes (for a countably additive measure). Wikipedia calls this Sierpinski's theorem: en.wikipedia.org/wiki/Atom_%28measure_theory%29 – Blackbird Aug 15 '15 at 22:23
• I see; the problem is that if $\lim \mu(A_n)$ is not necessarily equal to $\mu(\lim_n A_n)$ if $\mu$ is not countably additive, and provided that limits are meaningful.. – Paolo Leonetti Aug 15 '15 at 23:22
• References for the sigma-additive case are provided on MO: mathoverflow.net/questions/222583. As for the additive case, Sierpiński's main theorem in: Sur les fonctions d'ensemble additives et continues, Fund. Math. 3 (1922), No. 1, 240-246 (in French), provides an affirmative answer when $\Omega = {\bf R}^n$. I don't know what happens in general and haven't checked 6005's answer below (AFAICS, he claims a counterexample to the surjectivity of $\mu$ if $\mu$ isn't more than additive and non-atomic). – Salvo Tringali Nov 24 '15 at 13:39

Here is a proof of the first question using Zorn's lemma, and a counterexample to the second question using an ultrafilter. (So both cases used some form of the axiom of choice!)

Theorem (Sierpinski): For a non-atomic probability space $(\Omega, \mathcal{F}, \mu)$, $\mu$ is surjective onto $[0,1]$.

Proof: Let $x \in [0,1]$, and let $$\mathcal{G} = \{F \in \mathcal{F} \;:\; \mu(F) \ge x\},$$ and consider the ordering of $\mathcal{G}$ under inclusion up to measure-zero sets. Explicitly, say that $F_1 \le F_2$ whenever $\mu(F_1 \setminus F_2) = 0$. Note that $\mathcal{G}$ is nonempty, as $\Omega \in \mathcal{G}$.

This isn't a partial order because if $F_1$ and $F_2$ are equivalent up to a measure zero set, then $F_1 \le F_2$ and $F_2 \le F_1$. However, it is a pre-order, and Zorn's lemma works for pre-orders.

So, to this end, consider a totally ordered subset (chain) $\mathcal{T}$ of $\mathcal{G}$, under $\le$. For any $y \in \mathbb{R}$, $\mathcal{T}$ contains at most one set $F$ with $\mu(F) = y$; if it contained $F_1$ and $F_2$ with equal measure, with say $F_1 \le F_2$, then $F_2 \le F_1$ as well, and we would not have a total order. Therefore, $\mathcal{T}$ is order-isomorphic to a subset of the nonnegative real line. If this subset of $\mathbb{R}$ has a minimum element, it corresponds to a lower bound, and we are done; else we can find a strictly decreasing sequence $T_1, T_2, T_3 \ldots$, with $T_i \in \mathcal{T}$, such that a lower bound of $(T_i)$ is a lower bound of $\mathcal{T}$. Now, $T_i$ are measurable, with $\mu(T_1) < \infty$, and $\mu(T_{i+1} \setminus T_i) = 0$ for all $i$, so by a consequence of countable additivity $$\mu\left( \bigcap_{i=1}^\infty T_i \right) = \lim_{i \to \infty} \mu(T_i) \ge x$$ since $\mu(T_i) \ge x$ for all $x$. Hence the intersection $\bigcap_{i=1}^\infty T_i$ is the desired lower bound of the chain $\mathcal{T}$.

Therefore by Zorn's lemma $\mathcal{G}$ has a minimal element, say $E$. We have $\mu(E) \ge x$. If $\mu(E) > x$, then since the probability space is non-atomic, there must be a subset $E'$ of $E$ with arbitrarily small measure, and we can find $E'$ such that $\mu(E) > \mu(E \setminus E') > x$, contradicting the minimality of $E$. So $\mu(E) = x$.

Counterexample: There exists a non-atomic $(\Omega, \mathcal{F}, \mu)$ where $\mu$ is only finitely additive, $\mu(\Omega) = 1$, and $\mu$ is not surjective onto $[0,1]$.

Proof. Let $\lambda$ denote Lebesgue measure. Let $\Omega = [0,1]$, and let $\mathcal{F}$ be the $\sigma$-algebra of Lebesgue-measurable sets. Let $$\mathcal{A} = \{F \in \mathcal{F} \; : \; \lambda(F) = 1\}.$$ Note that $\mathcal{A}$ is a filter on $\Omega$. Moreover, $\mathcal{A}$ is free and proper. Therefore by the Ultrafilter lemma, there is a free ultrafilter $\mathcal{U} \supset \mathcal{A}$. Now define the following function $\nu: \mathcal{F} \to [0,1]$ corresponding to the ultrafilter $\mathcal{U}$, $$\nu(A) = \begin{cases} 1 &\text{if } A \in \mathcal{U} \\ 0 &\text{if } A \not \in \mathcal{U} \end{cases}$$ Check that $\nu$ is finitely additive by properties of an ultrafilter, and $\nu([0,1]) = 1$. Finally, let $$\mu(A) = \frac{1}{2}\left( \lambda(A) + \nu(A) \right)$$ for all $A \in \mathcal{F}$.

We have that $\mu([0,1]) = 1$, and $\mu$ is finitely additive as both $\lambda$ and $\nu$ are. We need to show that $\mu$ is non-atomic, but does not achieve every value between $0$ and $1$.

First we show there is no set $A \in \mathcal{F}$ such that $\mu(A) = \frac12$. This is proved by casework on $\nu(A)$ :

• If $\nu(A) = 0$, then $A \not \in \mathcal{U}$, so $A \not \in \mathcal{A}$, so $\lambda(A) < 1$, so $\mu(A) < \frac12$.

• If $\nu(A) = 1$, then $\nu(A^c) = 0$, so by the above reasoning $\mu(A^c) < \frac12$, so $\mu(A) > \frac12$.

Lastly, we need to show that $\mu$ is non-atomic: if $A$ is a set with $\mu(A) > 0$, then because it is impossible for $\nu(A) = 1$ and $\lambda(A) = 0$, $\lambda(A) > 0$. Therefore, since $\lambda$ is non-atomic, split $A$ into disjoint subsets $A_1, A_2$ such that $\lambda(A_1) > 0$ and $\lambda(A_2) > 0$; then at least one of $A_1$ and $A_2$ is not in $\mathcal{U}$, so if e.g. it is $A_1$ then $$0 < \mu(A_1) = \frac12 \lambda(A_1) < \frac12 \lambda(A) \le \mu(A).$$

Thus, $(\Omega, \mathcal{F}, \mu)$ is non-atomic and finitely additive, but $\mu$ is not surjective onto $[0,1]$ as there is no set with measure $\frac12$.

Notes on filters:

• A filter $\mathcal{A}$ on $X$ is a subset of $\mathcal{P}(X)$ such that (1) $X$ is nonempty; (2) if $A \in \mathcal{A}$ and $B \supseteq A$ then $B \in \mathcal{A}$; (3) if $A, B \in \mathcal{A}$, $A \cap B \in \mathcal{A}$.

• A filter $\mathcal{A}$ on $X$ is called proper if $\varnothing \not \in \mathcal{A}$, or equivalently $\mathcal{A} \subsetneq \mathcal{P}(X)$. (Sometimes, proper is included in the definition of a filter.) $\mathcal{A}$ is called free if the intersection $\bigcap \mathcal{A} = \varnothing$, or equivalently for every $x \in X$ there is $A \in \mathcal{A}$ such that $x \not \in A$. Free filters only exist on infinite sets; the smallest free filter is the set of cofinite sets, called the Fréchet filter; a filter is free if and only if it contains the Fréchet filter.

• An ultrafilter $\mathcal{U}$ on $X$ is a filter that is "maximal" in the following sense: (4) for any $A \subseteq X$, exactly one of $A, X \setminus A$ is in $\mathcal{U}$.

• The ultrafilter lemma says that every proper filter can be extended to an ultrafilter. Of course, if the original filter is free, then so is the ultrafilter. (Free ultrafilters are essentially the only interesting ones; non-free ultrafilters are called principal and are all equal to $\{A \subseteq X \,:\, x \in A\}$ for some element $x \in X$.)

• Neat! Only one question re non-atomicness: if $\mu(A)>0$, how do you know $A$ can be split into disjoint subsets $A_1,A_2$ with $\lambda(A_1),\lambda(A_2)>0$? Couldn't there be a measurable set $A$ with $\mu(A)>0$ but $\lambda(A)=0$? – Blackbird Aug 16 '15 at 11:42
• Could you please also mention your definitions of filter and ultrafilter? – Blackbird Aug 16 '15 at 11:48
• @Blackbird The non-atomic thing was a consequence of Lebesgue measure being non-atomic and the fact that if $\lambda(A) = 0$ then $\nu(A) = 0$ I clarified this somewhat in the answer. And I have dumped all my filter definitions at the bottom of the answer, at the risk of including too much. – 6005 Aug 16 '15 at 16:57
• Thanks, but I still didn't get why $\lambda(A)=0$ implies $\nu(A)=0$. – Blackbird Aug 16 '15 at 20:40
• @Blackbird If $\lambda(A) = 0$, then $\lambda(A^c) = 1$, which means $A^c \in \mathcal{A}$ (by definition), therefore $A^c \in \mathcal{U}$ (since $\mathcal{U} \supseteq \mathcal{A}$), so $A \not \in \mathcal{U}$ (property of an ultrafilter), so $\nu(A) = 0$. – 6005 Aug 16 '15 at 20:43