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<k_n, 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.

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