$A \subseteq \mathbb R$ is strong measure zero if given any sequence $( \epsilon_n )_{n \in \mathbb N}$ of positive reals there is a sequence $( I_n )_{n \in \mathbb N}$ of open intervals such that $I_n$ has length at most $\epsilon_n$ for each $n$, and $A \subseteq \bigcup_{n \in \mathbb N} I_n$.

I have read that the Cantor set is an example of a null set which is not strong measure zero.

Can anyone explain why or direct me to a reference which contains a proof?


Lemma. If $X \subseteq [0,1]$ is a strong measure zero set, and $f : [0,1] \to [0,1]$ is continuous, then $f[X]$ is also strong measure zero.

proof. Note that $f$ is uniformly continuous, so for each $\varepsilon > 0$ there is a $\delta ( \varepsilon ) > 0$ such that if $I \subseteq [0,1]$ is an interval of length $\leq \delta ( \varepsilon )$, then $f[I]$ has length $\leq \varepsilon$. Given a sequencce $\{ \varepsilon_i \}_{i \in \mathbb{N}}$ of positive reals, we know we can cover $X$ by a sequence $\{ I_i \}_{i \in \mathbb{N}}$ of intervals such that $I_i$ has length $\leq \delta(\varepsilon_i)$. Then $f[X] \subseteq \bigcup_{i \in \mathbb{N}} f[I_i]$.

Recall now that the Cantor function is continuous, and the image of the Cantor set under this function is the unit interval $[0,1]$ which is clearly not strong measure zero (or even Lebesgue measure zero).

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  • $\begingroup$ Perfectly clear. Thank you! $\endgroup$ – topsi Aug 20 '14 at 13:33

For a more direct approach: Suppose that $I_1 , I_2 , \ldots$ are open intervals in $\mathbb{R}$ such that $I_n$ has length $3^{-n}$.

Note that $I_1$ must be disjoint from either $[ 0 , \frac 1 3 ]$ or $[ \frac 2 3 , 1 ]$ (or both, I guess, if $I_1$ was chosen particularly badly). Let us label one of these intervals disjoint from $I_1$ by $A_1$.

Now $I_2$ must be disjoint from at least one of the two closed intervals obtained from deleting the open middle third of $A_1$: let us label one of these $A_2$.

And we continue like this. For each $n$ the open interval $I_{n+1}$ must be disjoint from one of the two closed intervals obtained from deleting the middle third of $A_n$, and we label one of these $A_{n+1}$.

So we have a decreasing sequence $A_1 \supseteq A_2 \supseteq \cdots$ of nonempty compact subsets of $\mathbb{R}$, and so their intersetion must be nonempty. It is clear that any point of this intersection (actually, there can only be one) must belong to the Cantor set, and is not in $\bigcup_{i=1}^\infty I_n$. Thus the sequence of open intervals we began with cannot cover the Cantor set.

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  • $\begingroup$ Thank you for this detailed direct proof! $\endgroup$ – topsi Aug 21 '14 at 11:13

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