5. null comeager sets exist

problem

Let $X$ be a Polish space without isolated points and let $μ$ be an atomless measure on $X$ (atomless means that every singleton is $μ$-null). Show that $X$ has a $μ$-null comeager subset.

aside

Do we also need some “finiteness”/“boundedness” assumption—for each $x∈X$, there is an open set containing $x$ with finite measure? Otherwise, consider this: let $X=ℝ$ and define a measure $μ$ where all countable sets are null and uncountable sets are infinite-measure. Then any null set is (by definition) countable, and therefore a countable union of singletons, each of which is nowhere dense, so the set must also be meager.

proof

Let $Q⊂X$ be a countable dense set, and let $𝒰$ be a countable basis of open sets of $X$. Fix $q∈Q$, and let $𝒰_q=\{U∈𝒰:q∈U\}$.

For all $r∈X∖\{q\}$, there exists an open set containing $q$ but not $r$, so

$$⋂_{U∈𝒰_q} U = \{q\}.$$

Therefore by measure continuity and atomless-ness,

$$\inf_{U∈𝒰_q} μ(U) = μ(\{q\}) = 0.$$

Now we will construct a $μ$-null comeager subset of $X$. (To do so, we will first construct a sequence of open and dense sets $V_n$ with measure tending to zero, and then take $S=⋂_{n∈ℕ}V_n$.)

First fix $n∈ℕ$. Let $\{q_i\}_{i∈ℕ}$ be an enumeration of $Q$, and for each $i∈ℕ$ let $B_i∈𝒰_{q_i}$ such that $μ(B_i)<1/(2^{i+1} n)$. Then let

$$V_n = ⋃_{i∈ℕ} B_i \qquad⟹\qquad μ(V_n) ≤ ∑_{i∈ℕ} μ(B_i) = \frac 1 n.$$

At the same time, observe that $V_n$ is a union of open sets and is therefore open, and $Q⊆V_n$ so $V_n$ is dense.

So let $S≔⋂_{n∈ℕ}V_n$. $S$ is by construction a countable intersection of open dense sets and therefore comeager, but also $μ(S)≤μ(V_n)=1/n$ for each $n$, so $μ(S)=0$.