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Localization of a continuous function by axis-parallel cube integrals

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ , the ambient convention under which the Lebesgue measure and integration interface used below is stated (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , let $U \subseteq R^{n}$ be open, and let $g : U \to R$ be continuous.

For a closed axis-parallel cube $Q = \prod_{i < n} [a_{i}, a_{i} + ℓ] \subseteq U$ with $ℓ > 0$ , the restriction $g ∣_{Q}$ is continuous on the compact nondegenerate rectangle $Q$ , hence bounded and Borel and Riemann integrable; denote by $\int_{Q} g d λ_{n}$ the Lebesgue integral of $g$ over $Q$ with respect to the restriction of Lebesgue measure, which by the comparison of the Riemann and Lebesgue integrals equals the Riemann integral of $g$ over $Q$ . Suppose that

$\int_{Q} g d λ_{n} = 0$

for every closed axis-parallel cube $Q \subseteq U$ with side $ℓ > 0$ . Then $g = 0$ on $U$ .

In particular, for $n = 2$ , a continuous function on an open subset of the plane whose integral over every closed axis-parallel square contained in its domain vanishes is identically zero.

The hypothesis requires every closed axis-parallel cube, with every centre and every positive side length. A single cube does not suffice, and neither do the cubes centred at one fixed point: on $U = R^{n}$ the nonzero continuous function $g (x) = x_{0} - p_{0}$ has $\int_{Q} g d λ_{n} = 0$ for every closed cube $Q$ centred at $p$ , by cancellation in its first coordinate. This family also differs from the family of closed cubes that merely contain $p$ .

Facts & Assumptions

Given: Assume ${A C}_{ω}$ . A natural number $n \geq 1$ , an open set $U \subseteq R^{n}$ , a continuous $g : U \to R$ , and the hypothesis that $\int_{Q} g d λ_{n} = 0$ for every closed axis-parallel cube $Q \subseteq U$ of positive side length, the integral being interpreted as in the Statement. Throughout, $R^{n}$ carries the Euclidean metric $d_{2}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ) and subsets carry the subspace metric (Isometry, isometric embedding, and the subspace metric on a subset).

[F1]

$U$ is open: for every $x \in U$ there is a real $r > 0$ with $B (x, r) \subseteq U$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space). The map $g$ is continuous, that is, continuous at every point of $U$ : for every $a \in U$ and every $ε > 0$ there is a real $δ > 0$ such that $∣ g (x) - g (a) ∣ < ε$ for every $x \in U$ with $d_{2} (x, a) < δ$ (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form).

[F2]

For $n \geq 1$ the Euclidean metric is $d_{2} (x, y) = \sqrt{\sum_{i < n} (x_{i} - y_{i})^{2}}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F3]

For reals $a_{i} \leq b_{i}$ the box ${x \in R^{n} : a_{i} \leq x_{i} \leq b_{i} for every i < n}$ is a compact subset of $(R^{n}, d_{2})$ (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).

[F4]

A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F5]

Every set $R$ with $R^{\circ} \subseteq R \subseteq \overline{R}$ determined by reals $a_{i} \leq b_{i}$ is Lebesgue measurable with $λ_{n} (R) = \prod_{i < n} (b_{i} - a_{i})$ ; in particular a closed cube of side $ℓ > 0$ has measure $ℓ^{n} > 0$ (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).

[F6]

A continuous map between topological spaces has Borel preimages of Borel sets (A continuous map has Borel preimages of Borel sets).

[F7]

Every continuous real function on a closed nondegenerate rectangle $Q \subseteq R^{m}$ , $m \geq 1$ , is Riemann integrable (Every continuous function on a closed nondegenerate rectangle in $R^{m}$ is Riemann integrable).

[F8]

Assume ${A C}_{ω}$ . For $m \geq 1$ and reals $a_{i} < b_{i}$ , if a bounded Borel real function $f$ on the box $Q = \prod_{i < m} [a_{i}, b_{i}]$ is Riemann integrable, then $f$ is Lebesgue integrable on $Q$ and its Lebesgue and Riemann integrals agree (Riemann–Lebesgue comparison for distribution test integrands).

[F9]

For integrable $f \leq h$ on a nondegenerate rectangle the Riemann integrals satisfy $\int_{Q} f \leq \int_{Q} h$ , and Riemann integrals are linear in the integrand (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ ).

[F10]

Lower and upper Darboux sums of a bounded $f$ over a grid $P$ are $L (f, P) = \sum_{i} m_{i} vol (Q_{i})$ and $U (f, P) = \sum_{i} M_{i} vol (Q_{i})$ , with $m_{i} = \inf f [Q_{i}]$ and $M_{i} = \sup f [Q_{i}]$ ; the cells $Q_{i}$ of a grid partition have $\sum_{i} vol (Q_{i}) = vol (Q)$ ; and the Darboux integrals are the suprema and infima of these sums, the function being Riemann integrable when they agree (Lower and upper Darboux sums over a grid partition in $R^{m}$ , Grid partitions of a rectangle in $R^{m}$ , their cells, refinements and mesh, The lower and upper Darboux integrals over a nondegenerate rectangle in $R^{m}$ ).

[F11]

If a grid $P^{'}$ refines a grid $P$ , then $L (f, P) \leq L (f, P^{'}) \leq U (f, P^{'}) \leq U (f, P)$ (Refinement raises multidimensional lower sums and lowers upper sums, with a quantitative boundary-slab estimate).

[F12]

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

Proof

1.1givenF8F9algebra

Suppose that $g (p) \neq 0$ for some $p \in U$ . Replacing $g$ by $- g$ if necessary, assume $g (p) > 0$ : the replacement preserves the hypothesis, because for every admissible cube the two Riemann integrals satisfy $\int_{Q} (- g) = - \int_{Q} g$ and the Lebesgue integrals equal them by [F8], so $\int_{Q} (- g) d λ_{n} = - \int_{Q} g d λ_{n} = 0$ , and it preserves the conclusion $g = 0$ ; both $g$ and $- g$ are continuous.

1.2F1F2choosealgebra

By continuity of $g$ at $p$ with $ε = g (p) / 2 > 0$ there is $δ_{1} > 0$ with $∣ g (x) - g (p) ∣ < g (p) / 2$ for every $x \in U$ with $d_{2} (x, p) < δ_{1}$ ; by openness of $U$ there is $δ_{2} > 0$ with $B (p, δ_{2}) \subseteq U$ . Put $δ : = \min {δ_{1}, δ_{2}} > 0$ and $ℓ : = δ / \sqrt{n} > 0$ , and consider the closed axis-parallel cube $Q : = \prod_{i < n} [p_{i} - ℓ / 2, p_{i} + ℓ / 2]$ . For $x \in Q$ we have $d_{2} (x, p)^{2} = \sum_{i < n} (x_{i} - p_{i})^{2} \leq n (ℓ / 2)^{2} = δ^{2} / 4$ , so $d_{2} (x, p) \leq δ / 2 < δ \leq δ_{1}$ and $x \in B (p, δ_{2}) \subseteq U$ ; thus $Q \subseteq U$ and $g (x) > g (p) - g (p) / 2 = g (p) / 2$ for every $x \in Q$ .

1.3F2F10constructalgebra

It remains to verify the sharpness clause. Fix $p \in R^{n}$ , put $a = ℓ / 2 > 0$ and $Q = \prod_{i < n} [p_{i} - a, p_{i} + a]$ , and let $h (x) = x_{0} - p_{0}$ on $R^{n}$ . Let $Φ (x) : = (2 p_{0} - x_{0}, x_{1}, \dots, x_{n - 1})$ ; then $Φ$ maps $Q$ onto $Q$ , $Φ \circ Φ$ is the identity, and for each cell $A$ of a grid whose first-coordinate partition is symmetric about $p_{0}$ the mirror cell $Φ (A)$ has $vol (Φ (A)) = vol (A)$ , $\inf h [Φ (A)] = - \sup h [A]$ and $\sup h [Φ (A)] = - \inf h [A]$ .

2.1F3F4F5F6F7F8step 1.2

The set $Q$ is a nonempty compact subset of $R^{n}$ and the restriction $g ∣_{Q}$ is continuous for the subspace metric, so it is bounded and attains its extrema; its Borel measurability is [F6] applied to the continuous map $g ∣_{Q}$ , and continuity on the closed nondegenerate rectangle gives Riemann integrability. By [F8] the Lebesgue integral $\int_{Q} g d λ_{n}$ exists and equals the Riemann integral of $g ∣_{Q}$ , and by [F5] $λ_{n} (Q) = ℓ^{n} > 0$ .

2.2F5F8F10step 1.2algebra

The constant $c : = g (p) / 2 > 0$ has Riemann integral $c ℓ^{n}$ over $Q$ : for every grid $P$ of $Q$ every cell satisfies $m_{i} = M_{i} = c$ , so $L (c, P) = U (c, P) = c \sum_{i} vol (Q_{i}) = c vol (Q) = c ℓ^{n}$ by [F10]; hence the Darboux integrals of $c$ equal $c ℓ^{n}$ , the function is Riemann integrable, and by [F8] its Lebesgue integral over $Q$ is $c ℓ^{n} > 0$ .

2.3F10F11step 1.3algebra

Let $P$ be any grid of $Q$ and let $P^{''}$ be the grid with first-coordinate partition the union of the first-coordinate breakpoints of $P$ with their reflections $t \mapsto 2 p_{0} - t$ , and with the remaining coordinate partitions those of $P$ ; then $P^{''}$ refines $P$ and its first-coordinate partition is symmetric about $p_{0}$ , so its cells fall into mirror pairs and self-mirrored cells, and summing the identities of step 1.3 over those classes gives $L (h, P^{''}) = - U (h, P^{''})$ , whence $L (h, P^{''}) \leq 0 \leq U (h, P^{''})$ ; by [F11], $L (h, P) \leq L (h, P^{''}) \leq 0$ and $U (h, P) \geq U (h, P^{''}) \geq 0$ for every grid $P$ .

3.1step 2.2step 1.2step 2.1F9givencontradiction

Since $g (x) > c$ for every $x \in Q$ by step 1.2 and both $g ∣_{Q}$ and the constant $c$ are Riemann integrable on $Q$ , monotonicity [F9] gives $\int_{Q} g d λ_{n} \geq c ℓ^{n} > 0$ , where step 2.1 identifies the Lebesgue integral of $g$ with its Riemann integral and step 2.2 evaluates the constant. This contradicts the hypothesis $\int_{Q} g d λ_{n} = 0$ for the admissible cube $Q \subseteq U$ constructed in step 1.2. Hence $g (p) = 0$ for every $p \in U$ , and the planar clause is the case $n = 2$ .

3.2F7F8F12step 2.3algebra

For each integer $m \geq 1$ let $P_{m}$ be the grid with first-coordinate breakpoints $p_{0} + k a / m$ for $- m \leq k \leq m$ and single intervals of length $2 a$ in the remaining coordinates. Its first-coordinate partition is symmetric about $p_{0}$ , so $L (h, P_{m}) = - U (h, P_{m})$ by step 2.3, and direct summation over the $2 m$ cells of first-coordinate length $a / m$ gives $L (h, P_{m}) = - U (h, P_{m}) = - a^{2} (2 a)^{n - 1} / m$ , a quantity tending to $0$ ; hence $\sup_{P} L (h, P) \geq L (h, P_{m})$ for every $m$ forces $\sup_{P} L (h, P) \geq 0$ , and together with step 2.3 this gives $\sup_{P} L (h, P) = 0$ , while $\inf_{P} U (h, P) = 0$ likewise. Therefore $h$ is Riemann integrable over $Q$ with $\int_{Q} h = 0$ , and by [F8], using that $h ∣_{Q}$ is continuous hence bounded and Borel, the Lebesgue integral is $\int_{Q} h d λ_{n} = 0$ .

4.1givenstep 3.2step 3.1∎

The function $g (x) = x_{0} - p_{0}$ is continuous and nonzero on $R^{n}$ , and step 3.2 shows $\int_{Q} g d λ_{n} = 0$ for the cube $Q$ of step 1.3 centred at $p$ ; every closed cube centred at $p$ is of that form, so the family of cubes with the single centre $p$ does not satisfy the localizing hypothesis although $g \neq 0$ , and a single cube plainly cannot either. Thus the all-cubes hypothesis of the Statement is exactly what the argument of steps 1.1-3.2 uses.

Source notes

The localization statement is standard: a continuous function that is orthogonal to every axis-parallel cube integral vanishes. The proof above is written in the library's own measure and Darboux interfaces, using the multidimensional Darboux integral of The lower and upper Darboux integrals over a nondegenerate rectangle in $R^{m}$ and the comparison with the Lebesgue integral of Riemann–Lebesgue comparison for distribution test integrands; the sharpness witness is the coordinate-cancellation computation, proved directly by mirror-symmetric grid refinements rather than by a separate symmetry or change-of-variables theorem.

Localization of a continuous function by axis-parallel cube integrals

Statement

Facts & Assumptions

Proof

Source notes

Depends on

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Sources