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First well order $\mathbb{Q}=\{r_m\}_{m=0}^\infty$. Let $B_{m,n}$ be the open ball centered at $r_m$ with radius $2^{-(m+n)}$. Let $B=\bigcap_{n=0}^\infty\bigcup_{m=0}^\infty B_{m,n}$. Clearly $B$ has Lebesgue measure zero. But how to show it's not a union of countably many Jordan content zero sets?

A set $A\subseteq\mathbb{R}$ has Lebesgue measure zero iff $\forall\epsilon>0\exists$a countable family of open intervals $\{I_n\}_{n\in\omega}(A\subseteq\bigcup_{n\in\omega}I_n\wedge\sum_{n=0}^\infty|I_n|<\epsilon)$.

A set $A\subseteq\mathbb{R}$ has Jordan content zero iff $\forall\epsilon>0\exists$a finite family of open intervals $\{I_n\}_{n=0}^N(A\subseteq\bigcup_{n=0}^NI_n\wedge\sum_{n=0}^N|I_n|<\epsilon)$.

I found a similar topic in this question: link

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I changed this into a topology question and requires some knowledge of Baire spaces.

To say that a set has Jordan content zero implies there is a closed set with empty interior which contains that set (you can create a nest of closed sets which contain your set and decrease in Lebesgue measure, then take the intersection of all of these closed sets; if this set had interior, it would have nonzero Lebesgue measure). Thus a countable union of Jordan content zero sets is contained in a meager subset of $\mathbb{R}$.

Every subset of a meager set is meager, thus a countable union of Jordan content zero sets would be.

Now you need to see that your set $B$ is not meager. This is because a dense, $G_\delta$ subset of $\mathbb{R}$ (like yours) is comeager (its complement is a meager subset of the reals). And since the reals are a Baire space, we cannot have that $B$ is meager in addition, for otherwise the reals would be meager in themselves.

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  • $\begingroup$ Ah yes. My mistake. Fixed. $\endgroup$ – Robert Wolfe Jan 26 '14 at 11:31

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