A set of content zero has measure zero

Given: A set $A \subseteq \mathbb{R}$ of content zero and a real $\varepsilon > 0$.

$A$ has content zero when for every real $\eta > 0$ there are $n \in \mathbb{N}$ and reals $a_0 \le b_0, \dots, a_n \le b_n$ with $A \subseteq \bigcup_{j \le n}[a_j,b_j]$ and $\sum_{j \le n}(b_j - a_j) \le \eta$; $A$ is null when for every real $\eta > 0$ there are sequences with the analogous properties and $\sum_{k<i}(b_k - a_k) \le \eta$ for every $i \in \mathbb{N}$ (Measure zero (a countable cover by intervals of total length below every $\varepsilon$) and content zero (a finite such cover)).

$[c,c] = \{c\}$ is an interval of length $0$, and $[c,d]$ has length $d - c \ge 0$ for $c \le d$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length).

Finite sums: $\sum_{k<i} t_k = \sum_{k<n+1} t_k + \sum_{k=n+1}^{i-1} t_k$ for $n + 1 \le i$, a sum of nonnegative terms is nonnegative and is monotone in the number of nonnegative terms adjoined, and $\sum_{k<i} t_k \le \sum_{k<n+1} t_k$ whenever $i \le n+1$ and the terms are nonnegative (Finite sums and finite products, by recursion, Laws of finite sums and finite products, Series, partial sums, convergence and the sum, divergence, and the tail series).

Ordered-field arithmetic: adding a nonnegative quantity does not decrease a value, and the order is transitive (Order is preserved by adding a constant and by adding inequalities, 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.