I meant to write a longer blog entry on strong measure zero sets (on the real line ), 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 is a strong measure zero set and is aclosedmeasure zero set, then 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 . This is actually the easy direction of Pawlikowski’s result, showing that this condition actually characterizes strong measure zero sets, that is, if is measure zero for all closed measure zero sets , then 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 has strong measure zero iff it can be made disjoint from any given meager set by translation, that is, iff for any meager there is a real with disjoint from .

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

Recall that *is* (or *has*) *measure zero *iff for any then is an open covering of such that .

Similarly, *has* (or* is*) *strong measure zero* iff for all sequences of positive reals there is an open covering of such that .

Here, , and that is an open cover of means that each in is an open interval, and .

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

Lemma.A set is strong measure zero iff for all sequences of positive numbers there are open intervals with for all and such that , that is, .

** Proof.** Split into countably many infinite sets . Since is measure zero, there are intervals for , with and . The sequence is as required.

We now proceed with the proof of the theorem. Assume that is strong measure zero and that is closed and measure zero. We may assume that is compact, since (easily) the countable union of (strong) measure zero sets has (strong) measure zero. It follows that for every there is a set made up of a finite number of open intervals , , such that and . (To see this, simply cover with open intervals whose lengths add up to less that , and use compactness to extract a finite subcover.)

Let be small enough that . Let be a sequence of open intervals, with and . The point is that , and this representation shows that indeed has measure zero, since converges, so its tails approach zero.

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