A sequence of intervals covering $[a,b]$ has total length at least $b-a$, so no interval of positive length has measure zero

Given: Reals $a\le b$, sequences $(a_k)$ and $(b_k)$ with $a_k\le b_k$ for every $k$ and $[a,b]\subseteq {\bigcup }_k[a_k,b_k]$, and a real $M$ with ${\sum }_{k<n}(b_k-a_k)\le M$ for every $n\in \mathbb{N}$. Throughout, $\theta :=2^{-1}$.

Measure zero: $A$ is null when for every real $\epsilon >0$ there is a sequence of closed intervals covering $A$ all of whose partial total lengths are $\le \epsilon$; a subset of a null set is null (Measure zero (a countable cover by intervals of total length below every $\epsilon$) and content zero (a finite such cover)).

$[c,d]=\{x:c\le x\le d\}$ has length $d-c\ge 0$ when $c\le d$; $(c,d)$ is the open interval; a closed bounded interval is bounded (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, Lower bound, bounded below, bounded set).

Every open interval $(c,d)$ is an open set and every interval $[c,d]$ is a closed set (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen, Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length).

A subset of $\mathbb{R}$ is compact exactly when it is closed and bounded (A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded); from every family of open sets whose union contains a compact set, either the set is empty and the empty subfamily covers it, or one can extract $m\in \mathbb{N}$ and members $U_0,\dots ,U_m$ of the family whose union already contains it (Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset).

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

Powers and the geometric series: ${\theta }^0=1$ and ${\theta }^{k+1}={\theta }^k\theta$, all ${\theta }^k>0$ for $\theta >0$, and ${\sum }_{k=0}^{\mathrm{\infty }}{\theta }^k=1\mathrm{/}(1-\theta )=2$ for $\theta =2^{-1}$; a series of nonnegative terms has all its partial sums at most its sum (Integer powers $a^m$, For $\mathrm{\mid }r\mathrm{\mid }<1$, ${\sum }_{k\ge 0}{r}^k=1\mathrm{/}(1-r)$, and for $\mathrm{\mid }r\mathrm{\mid }\ge 1$ the series diverges, Series, partial sums, convergence and the sum, divergence, and the tail series, A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

Finite sums: additivity, scaling by a constant, splitting, and monotonicity in the terms (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

Every finite list $k_0,\dots ,k_m$ of naturals has an upper bound $K\in \mathbb{N}$: by induction on $m$, taking $K=0$ for the empty case and replacing $K$ by whichever of $K$ and $k_{m+1}$ is the larger, the order of $\mathbb{N}$ being total (The principle of mathematical induction, Trichotomy of the order on $\mathbb{N}$, Order on the natural numbers).

Ordered-field arithmetic: $0<1$, so $2>0$ and $0<t\cdot 2^{-1}<t$ for $t>0$; 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.