FALSE: every set of measure zero has content zero

Given: The set $E:={\mathbb{Q}}_{\mathbb{R}}\cap [0,1]$, where ${\mathbb{Q}}_{\mathbb{R}}$ is the image of $\mathbb{Q}$ in $\mathbb{R}$ (The rationals embed densely in the reals).

The false claim: every subset of $\mathbb{R}$ of measure zero has content zero.

$\mathbb{Q}\approx \mathbb{N}$ and a subset of an at most countable set is at most countable, so $E$ is at most countable, hence null ($\mathbb{Q}$ is countably infinite, Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable, Every at most countable subset of $\mathbb{R}$ has measure zero, The rationals embed densely in the reals).

${\mathbb{Q}}_{\mathbb{R}}$ is dense in $\mathbb{R}$: strictly between any two reals lies a rational (Both $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable, The rationals embed densely in the reals).

$[c,d]$ is a closed set, a finite union of closed sets is closed, and $\bar{A}$ is the set of points every neighbourhood of which meets $A$, so a closed set containing $A$ contains $\bar{A}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets, The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$).

If $[a,b]\subseteq {\bigcup }_{j\le n}[c_j,d_j]$ with $a\le b$ and $c_j\le d_j$, then ${\sum }_{j\le n}(d_j-c_j)\ge b-a$ (If finitely many intervals cover a closed bounded interval $[a,b]$, the sum of their lengths is at least $b-a$).

$A$ has content zero when for every real $\epsilon >0$ it has a finite cover by closed intervals of total length at most $\epsilon$ (Measure zero (a countable cover by intervals of total length below every $\epsilon$) and content zero (a finite such cover)).

Every nonempty finite set of reals has a maximum and a minimum (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

Ordered-field arithmetic: $0<1$, so $2>0$ and $2^{-1}>0$ and $2^{-1}<1$; adding a constant and multiplying by a positive preserve an inequality; the order is total and transitive (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.