## On strong measure zero sets

I meant to write a longer blog entry on strong measure zero sets (on the real line $\mathbb R$), but it is getting too long, so it may take me more than I expected. For now, let me record here an argument showing the following:

Theorem. If $X$ is a strong measure zero set and $F$ is a closed measure zero set, then $X+F$ has measure zero.

The argument is similar to the one in

Janusz Pawlikowski. A characterization of strong measure zero sets, Israel J. Math., 93 (1), (1996), 171-183. MR1380640 (97f:28003),

where the result is shown for strong measure zero subsets of $\{0,1\}^{\mathbb N}$. This is actually the easy direction of Pawlikowski’s result, showing that this condition actually characterizes strong measure zero sets, that is, if $X+F$ is measure zero for all closed measure zero sets $F$, then $X$ is strong measure zero. (Since this was intended for my analysis course, and I do not see how to prove Pawlikowski’s argument without some appeal to results in measure theory, I am only including here the easy direction.) Pawlikowski’s argument actually generalizes an earlier key result of Galvin, Mycielski, and Solovay, who proved that a set $X$ has strong measure zero iff it can be made disjoint from any given meager set by translation, that is, iff for any $G$ meager there is a real $r$ with $X+r$ disjoint from $G$.

I proceed with the (short) proof after the fold.

Recall that $X\subseteq \mathbb R$ is (or has) measure zero iff for any $\epsilon>0$ then is an open covering $\mathcal C$ of $X$ such that $\sum_{I\in\mathcal C}\mathrm{lh}(I)<\epsilon$.

Similarly, $X$ has (or is) strong measure zero iff for all sequences $(\epsilon_n)_n$ of positive reals there is an open covering $\mathcal C=\{I_n\mid n\in\mathbb N\}$ of $X$ such that $\mathrm{lh}(I_n)<\epsilon_n$.

Here, $\mathrm{lh}((a,b))=b-a$, and that $\mathcal C$ is an open cover of $X$ means that each $I$ in $\mathcal C$ is an open interval, and $\bigcup C=\bigcup\{ I\mid I\in\mathcal C\}\supseteq X$.

The following apparently stronger version of the definition of strong measure zero is useful in the proof.

Lemma. A set $A$ is strong measure zero iff for all sequences $(\epsilon_n)_{n\in\mathbb N}$ of positive numbers there are open intervals $I_n$ with $\mathrm{lh}(I_n)<\epsilon_n$ for all $n$ and such that $A\subseteq\limsup_n I_n$, that is, $A\subseteq\bigcap_n\bigcup_{m>n}I_m$.

Proof. Split $\mathbb N$ into countably many infinite sets $B_n$. Since $A$ is measure zero, there are intervals $I_n$ for $n\in B_n$, with $\mathrm{lh}(I_n)<\epsilon_n$ and $A\subseteq \bigcup_{n\in B_n} I_n$. The sequence $(I_n)_{n\in\mathbb N}$ is as required. $\mathsf{QED}$

We now proceed with the proof of the theorem. Assume that $X$ is strong measure zero and that $F$ is closed and measure zero. We may assume that $F$ is compact, since (easily) the countable union of (strong) measure zero sets has (strong) measure zero. It follows that for every $n$ there is a set $F_n$ made up of a finite number $k_n$ of open intervals $I_{n,m}$, $m, such that $F\subseteq F_n$ and $\sum_m\mathrm{lh}(I_{n,m})<\frac1{2^n}$. (To see this, simply cover $F$ with open intervals whose lengths add up to less that $1/2^n$, and use compactness to extract a finite subcover.)

Let $\epsilon_n$ be small enough that $k_n\epsilon_n+\frac1{2^n}<(0.9)^n$. Let $(J_n)_n$ be a sequence of open intervals, with $\mathrm{lh}(J_n)<\epsilon_n$ and $X\subseteq\bigcap_n\bigcup_{m>n} J_m$. The point is that $X+F\subseteq\bigcap_n\bigcup_{m>n}(F_m+J_m)$, and this representation shows that $X+F$ indeed has measure zero, since $\sum_n(0.9)^n$ converges, so its tails approach zero. $\mathsf{QED}$

The result fails in general without the assumption that $F$ is closed. For example, it is consistent that there are Luzin sets $X$ such that $X+X=\mathbb R$. But any Luzin set is strong measure zero. On the other hand, it is consistent that every strong measure zero set $X$ is countable, in which case of course $X+F$ has measure zero whenever $F$ has measure zero.