LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Localization: a continuous function whose integrals vanish over a neighbourhood-basis family is identically zero

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ , which supplies the measure properties of Lebesgue measure used below (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Let $W \subseteq R^{3}$ be open and let $H : W \to R$ be continuous in the Euclidean metric: for every $y \in W$ and $ε > 0$ there is $δ > 0$ such that $z \in W$ , $∥ z - y ∥_{2} < δ$ imply $∣ H (z) - H (y) ∣ < ε$ . Let $F$ be a family of nonempty open subsets of $W$ such that

$\begin{matrix} for every y_{0} \in W and every neighbourhood V of y_{0} there is A \in F with y_{0} \in A \subseteq V . & (L) \end{matrix}$

Suppose the signed Lebesgue integral is defined and $\int_{A} H d x = 0$ for every $A \in F$ (Integral over a measurable subset, Integrable real and complex functions, and their integrals, Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ ). Then $H = 0$ on $W$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, the open set $W$ , the continuous $H$ , the family $F$ satisfying (L), and the defined signed integrals $\int_{A} H = 0$ .

[F1]

If $y_{0} \in W$ and $A \subset W$ is open with $y_{0} \in A$ , then some $σ > 0$ has $B_{σ} (y_{0}) \subseteq A$ ; if $σ$ is chosen also so that $\overline{B} (y_{0}, ρ) \subseteq W$ for every $ρ < σ$ , then for every $0 < r < σ$ the half-open box

$Q_{r} : = \prod_{j = 0}^{2} (y_{0, j} - \frac{r}{2}, y_{0, j} + \frac{r}{2}]$

has $\overline{Q_{r}} \subseteq B_{σ} (y_{0}) \subseteq A$ because every point of $\overline{Q_{r}}$ is within distance $\frac{\sqrt{3}}{2} r < σ$ of $y_{0}$ ; its Lebesgue measure is $λ_{3} (Q_{r}) = r^{3} > 0$ (Half-open boxes in $R^{n}$ and their volume, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F2]

Order and linearity of the Lebesgue integral: if $0 \leq g \leq h$ pointwise and both are integrable on a measurable set, then $\int g \leq \int h$ ; and $\int - H = - \int H$ whenever $H$ is integrable (Monotonicity and nonnegative homogeneity of the nonnegative integral, The Lebesgue integral is linear on $L^{1} (μ)$ ).

[F3]

If $H$ is continuous and positive throughout a nonempty compact set $K \subseteq W$ , then it attains its minimum on $K$ and that minimum is positive (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point). Every closed box $\overline{Q_{r}}$ is compact by Heine--Borel (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

Proof

technique · direct; a positive value would force a positive integral over a small box

1.1givenF1F3choose

Suppose for contradiction that $H (y_{0}) \neq 0$ for some $y_{0} \in W$ . If $H (y_{0}) > 0$ , continuity and openness give $σ > 0$ with $\overline{B} (y_{0}, σ) \subseteq W$ and $H > 0$ on this closed ball; if $H (y_{0}) < 0$ , apply the same argument to $- H$ . Put $V : = B_{σ} (y_{0})$ .

2.1step 1.1givenF1F3

Choose $A \in F$ with $y_{0} \in A \subseteq V$ by (L). Since $A$ is open, [F1] gives a box $Q_{r}$ with $\overline{Q_{r}} \subseteq A$ . In the positive case, $H > 0$ on $V$ , so [F3] applied to the compact set $\overline{Q_{r}} \subset V$ gives $ε : = \min_{\overline{Q_{r}}} H > 0$ ; in the negative case apply [F3] to $- H$ .

3.1step 2.1F1F2algebra∎

In the case $H > 0$ on $V$ : since $0 \leq ε 1_{Q_{r}} \leq H$ on $A$ , [F2] and [F1] give $\int_{A} H d x \geq \int_{Q_{r}} H d x \geq ε λ_{3} (Q_{r}) > 0,$ contradicting $\int_{A} H = 0$ . In the case $H < 0$ on $V$ , applying the same argument to $- H$ , whose integral over $A$ also vanishes by the linearity clause of [F2], gives $\int_{A} (- H) > 0$ , again a contradiction. Hence $H (y_{0}) = 0$ for every $y_{0} \in W$ .

Remarks

The family $F$ is exactly the family for which vanishing region integrals are available in the applications: images of small reference boxes under a diffeomorphism, small Euclidean balls, or bounded $C^{1}$ domains shrinking to a point. The hypothesis (L) is what makes "choose $A$ inside any neighbourhood" legitimate; no countable choice is used, since $A$ is chosen for a single specified neighbourhood.
For an $R^{n}$ -valued continuous $H$ the statement and proof apply to each component separately; the applications in this collection use the scalar or the componentwise form.

Depends on

Dependency tree · 2 levels

Sources

R. Abeyaratne, Continuum Mechanics, Volume II of Lecture Notes on the Mechanics of Solids (MIT, 1988, revised 2020), Sections 4.1–4.2 (the localization step from a global balance law to a field equation; conservation of mass)