$\mathbb{Q}\cap [0,1]$ has measure zero and not content zero, although it is bounded

Refuted claim: every set of measure zero has content zero (FALSE: every set of measure zero has content zero).

The witness is $E:={\mathbb{Q}}_{\mathbb{R}}\cap [0,1]$, the rationals of the unit interval (The rationals embed densely in the reals, Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length). It is at most countable, hence null; it is bounded; and every finite family of intervals covering it has total length at least $1$, because the union of finitely many closed intervals is closed and contains the closure of $E$, which is all of $[0,1]$. The refutation is carried out in full in FALSE: every set of measure zero has content zero; this item records the witness and says what makes it work.

• Boundedness is not what is missing. $E$ is bounded, and its failure is as large as it can be: no finite cover does better than the trivial cover of $[0,1]$ by itself. What $E$ lacks is closedness, and hence compactness; adding it repairs the implication completely (For a compact subset of $\mathbb{R}$, measure zero and content zero coincide).

• The closure is the whole obstruction. Content zero is insensitive to passing to the closure, since a finite union of closed intervals is closed, whereas measure zero is not: $E$ is null and $\bar{E}=[0,1]$ is not (A sequence of intervals covering $[a,b]$ has total length at least $b-a$, so no interval of positive length has measure zero). That single asymmetry is the entire difference between the two notions.

• Compare the compact case. The Cantor set is uncountable and null, and being compact it also has content zero (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points). So the failure here is not about cardinality: a much larger null set can have content zero, and a countable one need not.