# A Borel measure on $\mathbb{R}$ whose nullsets are exactly the countable sets

When learning measure theory for the first time, students often think that all (Lebesgue) nullsets in $$\mathbb{R}$$ are countable - but then have those hopes dashed by counterexamples like the Cantor set. What I'm wondering is the following: is there a "non-boring" measure $$\mu$$, defined on all Borel subsets of $$\mathbb{R}$$ (under its standard topology), such that, if $$A \subseteq \mathbb{R}$$ is Borel, then $$\mu(A) = 0 \iff A \text{ is countable}?$$

If not, why?

By "non-boring", I mean that there is some set $$B \subseteq \mathbb{R}$$ measurable with $$0 < \mu(B) < \infty$$. Otherwise, you could let $$\mu$$ be the measure which sends all countable sets to 0 and all uncountable sets to $$\infty$$, which would be a pretty boring measure.

• No, not possible. Hint: Given a Borel measure with $\mu(\{t\})=0$ for all $t$ and $0<\mu(E) < \infty$ for some Borel set $E$, construct a Cantor-like set $C \subseteq E$ with zero measure which is uncountable. Apr 15 at 0:38

We claim that any measure $$\mu$$ of your desired form must be the unique measure that maps every countable set to $$0$$ and every uncountable Borel set to $$\infty$$.

Consider the sets \begin{align*}C_1&=\{0\le x<1:\textrm{the decimal expansion of x has only digits 0 or 1}\}\\C_2&=\{0\le x<1:\textrm{the decimal expansion of x has only digits 0 or 2}\}\end{align*} Standard "Cantor-set" arguments show that

• $$C_1,C_2$$ are compact subsets of $$[0,1)$$ which both are uncountable;
• If $$x\in C_1+C_2$$, then there exists a unique pair $$(c_1,c_2)\in C_1\times C_2$$ such that $$x=c_1+c_2$$; (Hint: Consider the decimal expansion of such an $$x$$.)
• The last item implies $${+}:C_1\times C_2\to (C_1+C_2)$$ is a bijection; in particular, $$C_1+C_2=\bigsqcup_{c\in C_2}(C_1+c)$$ is a partition of $$C_1+C_2$$ into uncountably many translated copies of $$C_1$$;
• The last items also imply that $$D:=C_1+C_2$$ is homeomorphic to the standard Cantor set $$C$$, and that under this homeomorphism $$C$$ can be partitioned into uncountably many homeomorphic copies of $$C_1$$.

Now let $$B$$ be any uncountable Borel set. We'll show that $$\mu(B)=\infty$$.

We use the so-called perfect set theorem for Borel sets which says that every uncountable Borel set $$B$$ contains a subset $$B'$$ homeomorphic to the standard Cantor set $$C$$. The final item above shows that $$B'$$ can then be partitioned into uncountably many homeomorphic copies of $$C_1$$. We write $$B\supseteq B'=\bigsqcup_{i\in I}C^i$$ where each $$C^i\cong C_1$$ is homeomorphic. By this homeomorphism, each $$C^i$$ is also uncountable compact, and hence $$\mu(C^i)>0$$. It follows that $$\mu(B)\ge\sum_{i\in I}\mu(C^i)=\infty$$ since the last sum is an uncountable sum of positive summands.

Addendum, per our previous Discord conversation:

From now on a Borel measure on $$\mathbb{R}$$ is, as you defined, any measure space $$(\mathbb{R},\mathcal{F},\mu)$$ where the measurable sets $$\mathcal{F}\supseteq\mathbf{B}(\mathbb{R})$$ contain all Borel sets and possibly more.

A weird measure is such a measure where the $$\mu$$-null sets $$\textsf{null}_\mu\subseteq\mathcal{F}$$ coincide with $$[\mathbb{R}]^{\le\aleph_0}\subseteq\mathbf{B}(\mathbb{R})$$ the set of all countable subsets of $$\mathbb{R}$$, and moreover there exists $$X\in\mathcal{F}$$ measurable with $$0<\mu(X)<\infty$$.

Claim. The existence of a weird measure is independent of $$\textsf{ZFC}$$:

1. Under $$\textsf{ZFC}+\textsf{CH}$$, there exists a weird measure $$(\mathbb{R},\mathcal{S},\mu)$$ with uncountable $$X\in\mathcal{S}$$ such that $$\mu(X)=1$$;
2. Under $$\textsf{ZFC}+\textsf{MA}_{\aleph_1}+\lnot(\textsf{CH})$$, there exists no weird measure, and so for every measure space $$(\mathbb{R},\mathcal{F},\mu)$$ with $$\mathbf{B}(\mathbb{R})\subseteq\mathcal{F}$$ and $$\textsf{null}_\mu=[\mathbb{R}]^{\le\aleph_0}$$, and for every $$X\in\mathcal{F}$$ uncountable, necessarily $$\mu(X)=\infty$$.

Sketches.

(1) We prove a stronger statement where $$\textsf{CH}$$ is replaced with $$\mathop{\textsf{add}}(\textsf{null})=\mathfrak{c}$$ and in the definition of $$\mu$$ we replace with $$\textsf{null}_\mu=[\mathbb{R}]^{<\mathfrak{c}}$$, rewording everything else accordingly (eg. for every instance of "countable" we say "less than continuum" instead). Clearly this implies part (1) of the claim.

Using a usual $$\mathop{\textsf{add}}(\textsf{null})=\mathfrak{c}$$ construction, take an increasing cofinal sequence $$\langle A_i:i<\mathfrak{c}\rangle$$ in the poset $$(\textsf{null},{\subseteq})$$, and let $$\langle K_i:i<\mathfrak{c}\rangle$$ enumerate all the Lebesgue non-null compact sets. Set $$X=\{x_i:i<\mathfrak{c}\}$$ with $$x_i\in K_i\smallsetminus A_i$$ so that $$X\cap A\ne\varnothing$$ for all Lebesgue non-null $$A$$ and $$\left|X\cap N\right|<\mathfrak{c}$$ for all Lebesgue null $$N$$.

Let $$\nu$$ be the Gaussian probability measure on $$\mathbb{R}$$, and restrict $$\nu$$ to $$X$$ by setting $$\nu^*(X\cap B)=\nu(B)$$ for every Borel $$B$$. Note $$\nu^*$$ is well-defined with $$\nu^*(X)=1$$ and $$\textsf{null}_{\nu^*}=[X]^{<\mathfrak{c}}$$. We let $$\mathcal{S}=\mathbf{B}(\mathbb{R})\oplus X$$ be the Borel $$\sigma$$-algebra enriched with $$X$$, and set $$\mu(A)=\nu^*(A\cap X)+\begin{cases}0,&\left|A\smallsetminus X\right|<\mathfrak{c}\\\infty,&\left|A\smallsetminus X\right|=\mathfrak{c}\end{cases}$$ Then $$\mu$$ satisfies $$\textsf{null}_\mu=[\mathbb{R}]^{<\mathfrak{c}}$$, with $$\mu(X)=1$$ witnessing weirdness. $$\square$$

(2) It suffices to show every uncountable $$X\in\mathcal{F}$$ contains a subpartition $$X\supseteq\bigsqcup_{i<\omega_1}X_i$$ with each $$X_i\in\mathcal{F}$$ uncountable. Then $$\mu(X)\ge\sum_{i<\omega_1}\mu(X_i)=\infty$$ works, since $$\mu(X_i)>0$$ for all $$i$$.

It's known that $$\textsf{MA}_{\aleph_1}$$ implies for every $$A\sqcup B\subseteq\mathbb{R}$$ with $$\left|A\right|=\left|B\right|=\aleph_1$$, there exists a $$G_\delta$$ set $$A\subseteq C\subseteq\mathbb{R}\smallsetminus B$$. My earlier MSE post uses a variation of a proof of this fact.

Take any subpartition $$X\supseteq \bigsqcup_{i<\omega_1} A_i$$ where each $$\left|A_i\right|=\aleph_1$$. Take $$G_\delta$$ sets $$\langle C_i:i<\omega_1\rangle$$ with $$A_i\subseteq C_i\subseteq\mathbb{R}\smallsetminus\bigsqcup_{i. Then $$X_i=X\cap\left(C_i\smallsetminus\bigcup_{j works, as $$A_i\subseteq X_i$$ and $$\bigsqcup_{i<\omega_1}X_i\subseteq X$$. $$\square$$

Remark.

Of course you might notice now that asking the same question but with every instance of "countable" replaced with "less than $$\mathfrak{c}$$" compels one to work in $$\textsf{ZFC}+\lnot(\textsf{MA})$$ in case (2) (in fact $$\mathop{\textsf{add}}(\textsf{null})<\mathfrak{c}$$). I have not given thought to what happens then, but a quick gut feeling says large cardinals are possibly relevant.

• This is an interesting construction. Is there a neat way to see that the subset $B'$ which exists by the perfect set theorem is Borel measurable? I ask because it's presented as an uncountable union of sets, whose measurability I also wonder about. May 20 at 21:56
• @AlexOrtiz Thanks for the comment. Any continuous image of a compact set is compact, and that in particular includes every homeomorphic image of a Cantor set. May 20 at 21:58