LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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Countable covers by closed boxes, by open boxes and by closed cubes all compute Lebesgue outer measure

Statement

Let $n \geq 1$ and assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). For $u, v \in R^{n}$ with $u_{i} \leq v_{i}$ for every $i < n$ write $[u, v]$ for the closed rectangle and $V (u, v) : = {x \in R^{n} : u_{i} < x_{i} < v_{i} for every i < n}$ for the open box, both of size $\prod_{i < n} (v_{i} - u_{i})$ (Axis-parallel rectangles in $R^{m}$ and their volume); a closed cube of side $ℓ \geq 0$ is a set $\prod_{i < n} [c_{i}, c_{i} + ℓ]$ , of size $ℓ^{n}$ . For $E \subseteq R^{n}$ put

$λ^{c l} (E) : = \inf {\sum_{k = 0}^{\infty} vol [u^{k}, v^{k}] : E \subseteq ⋃_{k} [u^{k}, v^{k}]}, λ^{o p} (E) : = \inf {\sum_{k = 0}^{\infty} \prod_{i < n} (v_{i}^{k} - u_{i}^{k}) : E \subseteq ⋃_{k} V (u^{k}, v^{k})},$

$λ^{c b} (E) : = \inf {\sum_{k = 0}^{\infty} ℓ_{k}^{n} : E \subseteq ⋃_{k} \prod_{i < n} [c_{i}^{k}, c_{i}^{k} + ℓ_{k}]},$

infima over countable covers of the stated kind, which exist because $R^{n}$ is covered by the rectangles $[- k 1, k 1]$ , by the open boxes $V (- k 1, k 1)$ and by the cubes $\prod_{i} [- k, - k + 2 k]$ . Then

$λ^{c l} (E) = λ^{o p} (E) = λ^{c b} (E) = λ_{n}^{*} (E) .$

Facts & Assumptions

Given: A natural number $n \geq 1$ , the Axiom of Countable Choice, a subset $E \subseteq R^{n}$ , and the three infima displayed in the Statement.

[L1]

$λ_{n}^{*} (E) : = \inf {\sum_{k = 0}^{\infty} μ_{0} (A_{k}) : A_{k} \in E_{n}$ for every $k$ and $E \subseteq ⋃_{k} A_{k}}$ (Lebesgue outer measure on $R^{n}$ , Elementary sets: the finite unions of half-open boxes in $R^{n}$ ).

[L2]

$B (a, b) : = {x \in R^{n} : a_{i} < x_{i} \leq b_{i} for every i < n}$ ; a box is nonempty exactly when $a_{i} < b_{i}$ for every $i < n$ ; $vol (\emptyset) : = 0$ ; and for a nonempty box with real parameters $vol (B) : = \prod_{i < n} (b_{i} - a_{i})$ (Half-open boxes in $R^{n}$ and their volume).

[L3]

Assuming countable choice, $λ_{n}^{*} (A) = μ_{0} (A)$ for every elementary set $A$ , and $λ_{n}^{*}$ is an outer measure (Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume), $μ_{0}$ being the elementary volume of The sum of the volumes of a disjoint box decomposition of an elementary set does not depend on the decomposition.

[L4]

Assuming countable choice, $λ_{n}^{*} (E) = \inf {λ_{n} (U) : U$ open and $E \subseteq U}$ (Assuming countable choice, the Lebesgue outer measure of an arbitrary subset of $R^{n}$ is the infimum of the measures of the open sets containing it).

[L5]

Every open $U \subseteq R^{n}$ is the union of an at most countable family of pairwise disjoint dyadic cubes (Every open subset of $R^{n}$ is the union of a countable pairwise disjoint family of dyadic cubes), each of the form $Q_{k, m} = {x : m_{i} 2^{- k} < x_{i} \leq (m_{i} + 1) 2^{- k} (i < n)}$ (Dyadic cubes of generation $k$ in $R^{n}$ , Integer powers $a^{m}$ ).

[L6]

$vol (Q_{k, m}) = 2^{- k n}$ (For each generation, the dyadic cubes of that generation are pairwise disjoint and cover $R^{n}$ ).

[L7]

Assuming countable choice, $λ_{n}$ is a measure on the sigma-algebra $L (R^{n})$ with $λ_{n} (B) = vol (B)$ for every half-open box (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume), so it is countably additive on pairwise disjoint measurable sequences (Measures on sigma-algebras).

[F1]

$[a, b] : = {x \in R^{m} : a_{j} \leq x_{j} \leq b_{j} (j < m)}$ and $vol [a, b] : = \prod_{j < m} (b_{j} - a_{j})$ (Axis-parallel rectangles in $R^{m}$ and their volume).

[F2]

The nonnegative extended sum of a sequence in $[0, + \infty]$ is $\sum_{k = 0}^{\infty} a_{k} : = \sup_{n \in N} s_{n}$ , the supremum of its nondecreasing partial sums (Series in the nonnegative extended real line).

[F3]

Every nonempty subset $S \subseteq N$ has a least element (The well-ordering principle).

[F4]

For every real $ε > 0$ there is a natural number $k \geq 1$ with $1 / k < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[F5]

If $∣ r ∣ < 1$ then $\sum_{k = 0}^{\infty} r^{k} = 1 / (1 - r)$ ; in particular $\sum_{k = 0}^{\infty} 2^{- k} = 2$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[F6]

For sequences of reals, $\sum_{k < n} (a_{k} + b_{k}) = \sum_{k < n} a_{k} + \sum_{k < n} b_{k}$ ; $\sum_{k < n} λ a_{k} = λ \sum_{k < n} a_{k}$ ; if $a_{k} \leq b_{k}$ whenever $0 \leq k < n$ then $\sum_{k < n} a_{k} \leq \sum_{k < n} b_{k}$ ; and $\prod_{k < n} (a_{k} b_{k}) = (\prod_{k < n} a_{k}) (\prod_{k < n} b_{k})$ (Laws of finite sums and finite products, claims 1, 2, 4 and 6; Finite sums and finite products, by recursion).

[F7]

An at most countable family may always be presented as a sequence (Finite, countably infinite, countable, uncountable).

Proof

technique · direct

1.1

For a natural number $r$ , reals $0 \leq p_{i} \leq q_{i}$ $(i < r)$ and a real $V \geq 1$ with $q_{i} \leq V$ for every $i < r$ , one has $\prod_{i < r} q_{i} - \prod_{i < r} p_{i} \leq V^{r} \sum_{i < r} (q_{i} - p_{i})$ : at $r = 0$ both products are $1$ and both sides are $0$ , and the passage from $r$ to $r + 1$ uses $\prod_{i < r + 1} q_{i} - \prod_{i < r + 1} p_{i} = q_{r} (\prod_{i < r} q_{i} - \prod_{i < r} p_{i}) + (q_{r} - p_{r}) \prod_{i < r} p_{i}$ with $\prod_{i < r} p_{i} \leq V^{r}$ , so the estimate follows by induction on $r$ .

F1F6

1.2

For real $u_{i} \leq v_{i}$ one has $V (u, v) \subseteq B (u, v) \subseteq [u, v]$ and $vol B (u, v) = vol [u, v] = \prod_{i < n} (v_{i} - u_{i})$ , the box being empty and the product zero together when some $u_{i} = v_{i}$ ; moreover $[u, v] \subseteq V (u - θ 1, v + θ 1)$ for every real $θ > 0$ , whose size is $\prod_{i < n} (v_{i} - u_{i} + 2 θ)$ , and a closed cube of side $ℓ$ is the closed rectangle $[c, c + ℓ 1]$ of size $ℓ^{n}$ .

L2F1F6

2.1

$λ^{c l} (E) \leq λ^{c b} (E)$ , because every closed-cube cover is a closed-rectangle cover with the same terms.

step 1.2F1

2.2

$λ_{n}^{*} (E) \leq λ^{o p} (E)$ , because an open-box cover $E \subseteq ⋃_{k} V (u^{k}, v^{k})$ gives the elementary cover $E \subseteq ⋃_{k} B (u^{k}, v^{k})$ whose covering cost $\sum_{k} μ_{0} (B (u^{k}, v^{k}))$ has exactly the same terms.

step 1.2L1L2L3

2.3

$λ^{o p} (E) \leq λ^{c l} (E)$ : given a closed-rectangle cover and a real $ε > 0$ , let $m_{k}$ be the least natural number with $\prod_{i < n} (v_{i}^{k} - u_{i}^{k} + 2 / (m_{k} + 1)) \leq vol [u^{k}, v^{k}] + ε 2^{- k}$ , which exists by step 1.1 with $V$ a real at least $1$ bounding all $v_{i}^{k} - u_{i}^{k} + 2$ and by the Archimedean property; the open boxes $V (u^{k} - θ_{k} 1, v^{k} + θ_{k} 1)$ with $θ_{k} : = 1 / (m_{k} + 1)$ cover $E$ , and each partial sum of their sizes is at most $\sum_{k < N} vol [u^{k}, v^{k}] + ε \sum_{k < N} 2^{- k} \leq \sum_{k = 0}^{\infty} vol [u^{k}, v^{k}] + 2 ε$ , so $λ^{o p} (E)$ is at most that closed cover's total plus $2 ε$ , for every positive real $ε$ .

step 1.1step 1.2F2F3F4F5F6

2.4

$λ^{c b} (E) \leq λ_{n}^{*} (E)$ : the inequality is trivial when $λ_{n}^{*} (E) = + \infty$ , and otherwise, given a real $ε > 0$ , outer regularity supplies an open $U \supseteq E$ with $λ_{n} (U) \leq λ_{n}^{*} (E) + ε$ , the dyadic decomposition writes $U$ as a disjoint union of an at most countable family of dyadic cubes, presented as a sequence $(Q_{k_{j}, m^{j}})_{j}$ and padded with copies of $\emptyset$ if it is finite, countable additivity gives $\sum_{j} λ_{n} (Q_{k_{j}, m^{j}}) = λ_{n} (U)$ , and each $Q_{k_{j}, m^{j}}$ is contained in the closed cube $\prod_{i < n} [m_{i}^{j} 2^{- k_{j}}, m_{i}^{j} 2^{- k_{j}} + 2^{- k_{j}}]$ of side $2^{- k_{j}}$ and size $2^{- k_{j} n} = λ_{n} (Q_{k_{j}, m^{j}})$ , a padding term contributing the degenerate cube of side $0$ .

step 1.2L4L5L6L7F1F7

3.1

The four quantities therefore satisfy $λ^{c l} (E) \leq λ^{c b} (E) \leq λ_{n}^{*} (E) \leq λ^{o p} (E) \leq λ^{c l} (E)$ , so all four are equal.

step 2.1step 2.2step 2.3step 2.4∎

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97 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John K. Hunter, Measure Theory (UC Davis lecture notes), Chapter 2 (standard reference, not scraped)
T. Tao, An Introduction to Measure Theory (GSM 126), Section 1.2 (standard reference, not scraped)