Translates of null sets

Question

Does there exist a null set of reals $N$ such that every null set is covered by countably many translations of $N$?

Answer 1

Certainly not. WLOG the set $E$ is $G_\delta$. Now consider open sets $E_j$ such that $E=\bigcap_j E_j$ and $\sum_j|E_j|<+\infty$. Now look at the constituting intervals of all $E_j$. Then we shall get a fixed sequence of intervals $I_k$ of total length $1$ such that every set $F$ of Lebesgue measure $0$ can be covered by the shifts of those intervals with $k\ge n$ with any $n$. Then, choosing a gauge function $h$ so that $h(x)/x\to \infty$ as $x\to 0$ but $\sum_k h(|I_k|)<+\infty$, we see that we must have each set of Lebesgue measure $0$ also of $h$-Hausdorff measure $0$, which is absurd (Cantor type construction, or something else like that).

Cool problem! I'll use it for the very next measure theory qualifier :-).

Answer 2

Here is an answer for the Cantor space $C$, the set of functions from $\omega$ to $2$. The plan is to show that for each null set $X \subseteq C$ there is a measure $0$ set $C_{a} \subseteq C$ such that $C_{a}$ is homeomorphic with $C$, and for any translate $X'$ of $X$, $X' \cap C_{a}$ is a null set in the measure induced by this homeomorphism. I haven't thought about whether this example can be converted into one for the real reals.

Basic open sets in the Cantor space are represented by functions $\sigma \colon n \to 2$, for some $n \in \omega$, where $[\sigma]$ denotes the set of $f \in C$ such that $\sigma \subseteq f$. For each such $\sigma$, the measure of $[\sigma]$, $\mu([\sigma])$, is $2^{-n}$. Given a set $a \subseteq \omega$, let $C_{a}$ be the set of $f \in C$ such that $f(n) = 0$ for each $n \in a$. If $a$ is infinite, $\mu(C_{a}) = 0$. Each $C_{a}$ is naturally homeomorphic to $C$, via deleting the coordinates in $a$. This homeomorphism induces the measure $\mu_{a}$ on $C_{a}$, where, for $\sigma$ as above, $\mu_{a}([\sigma] \cap C_{a})$ is $2^{|n \cap a|}\mu([\sigma])$ (which is $2^{-|n \setminus a|}$) if $\sigma(m) = 0$ for all $m \in a \cap n$, and $0$ otherwise.

Given a null set $X$, we may fix for each rational $r \in (0,1)$ a sequence $\langle \sigma^{r}_{i} : i \in \omega \rangle$ such that (1) each $\sigma^{r}_{i}$ is a function from some $s^{r}_{i} \in \omega$ to $2$ (2) $X \subseteq \bigcup_{i \in \omega} [\sigma^{r}_{i}]$ and (3) $\sum\{ 2^{-s^{r}_{i}} : i \in \omega\} < r$.

For any $a \subseteq \omega$, and any translate $X'$ of $X$, $\mu_{a}(X' \cap C_{a})$ is at most $$\sum\{ 2^{-|s^{r}_{i} \setminus a|} : i \in \omega \}.$$ It suffices then to find an infinite $a \subseteq \omega$ and a sequence $\langle r_{k} : k \in \omega \rangle$ of rationals from $(0,1)$ such that the sequence of values $$\sum\{ 2^{-|s^{r_{k}}_{i} \setminus a|} : i \in \omega \}$$ goes to $0$.

Given a sequence $\bar{r} = \langle r_{k} : k \in \omega \rangle$ of rationals in $(0,1)$, let $a_{\bar{r}} \subseteq \omega$ (enumerated in increasing order as $\langle a_{j} : j \in \omega \rangle$) be such that for each $j\in \omega$, and each $k \leq j$, $$\sum\{ 2^{-s^{r_{k}}_{i}} : i \in \omega \setminus a_{j}\} < r_{k}/2^{2j}.$$ Then for each $k \in \omega$, $\sum\{ 2^{-|s^{r_{k}}_{i} \setminus a|} : i \in \omega \}$ is at most $$(2^{k}r_{k}) + \sum\{ (2^{j+1}r_{k})/2^{2j} : k \leq j < \omega\}$$ which is at most $2^{k+2}r_{k}$ (if I've done the math correctly; the first term comes from considering the terms for $i < a_{k}$, the rest for $i \geq a_{k}$).

Then if we choose the $r_{k}$'s so that $2^{k+2}r_{k}$ goes to $0$, $C_{a_{\bar{r}}}$ is the desired null set.