FALSE: every subset of $\mathbb{R}$ of measure zero is nowhere dense

False claim: every subset of $\mathbb{R}$ of measure zero (Measure zero (a countable cover by intervals of total length below every $\epsilon$) and content zero (a finite such cover)) is nowhere dense (Nowhere dense, meager (first category), residual, and second category subsets of $\mathbb{R}$).

The claim confuses two different smallness conditions. Measure zero constrains the total length of a cover; nowhere density constrains the closure. A set may be covered by intervals of total length below any $\epsilon$ and still have every real as an adherent point, and $\mathbb{Q}$ does exactly that.

• ${\mathbb{Q}}_{\mathbb{R}}$ is nonetheless meager, being a union of countably many singletons, each of which is nowhere dense ($\mathbb{Q}$ is $F_{\sigma }$, meager and not $G_{\delta }$, while the irrationals are $G_{\delta }$, residual and not $F_{\sigma }$). So the failure above is not a failure of topological smallness in every sense: it is exactly the failure of the one-step condition. Meagreness is the countable-union closure of nowhere density, and it is the notion under which $\mathbb{Q}$ is small.

• The converse implication is also false, and needs an uncountable witness: the Smith-Volterra-Cantor set is nowhere dense and not null (FALSE: every nowhere dense subset of $\mathbb{R}$ has measure zero).

• The named witness is $\mathbb{Q}$ is dense in $\mathbb{R}$ and has measure zero ↗.