ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedaudited 2026-09-22

How statement and proof provenance work

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A square-integrable kernel without a continuous representative

Example

Assume the Axiom of Choice (The Axiom of Choice). Let $μ$ and $ν$ be the Lebesgue measures of the intervals $[0, 1]$ on the two factors, so that $μ ([0, 1]) = ν ([0, 1]) = 1$ and $μ ([0, \frac{1}{2}]) = \frac{1}{2}$ (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, Axis-parallel rectangles in $R^{m}$ and their volume), and equip $[0, 1]^{2}$ with the completed product measure $\overline{μ \times ν}$ (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique, The completed product measure). Let

$k (x, y) : = 1_{[0, 1 / 2]} (x) for (x, y) \in [0, 1]^{2},$

the product kernel $k (x, y) = a (x) \overline{b (y)}$ with $a = 1_{[0, 1 / 2]}$ and $b = 1_{[0, 1]}$ . Then $k$ is a square-integrable kernel with $∥ k ∥_{2}^{2} = \frac{1}{2}$ and rank-one kernel operator, as an identity of $L^{2}$ classes, $(T_{k} f) (x) = ⟨ f, 1_{[0, 1]} ⟩ 1_{[0, 1 / 2]} (x)$ , but no continuous function $h : [0, 1]^{2} \to C$ agrees with $k$ almost everywhere: the class of $k$ in $L^{2} (\overline{μ \times ν}; C)$ has no continuous representative. This shows that square integrability does not force the continuity hypotheses used by the earlier continuous-kernel compactness examples.

Facts & Assumptions

Given: The Axiom of Choice, the factor Lebesgue measures $μ, ν$ on $[0, 1]$ , the completed product $\overline{μ \times ν}$ on $[0, 1]^{2}$ , the kernel $k = 1_{[0, 1 / 2]} \overline{1_{[0, 1]}}$ , and a continuous $h : [0, 1]^{2} \to C$ .

[F1]

On measurable rectangles the product measure is given by $(μ \times ν) (A \times B) = μ (A) ν (B)$ , and the completion $\overline{μ \times ν}$ extends it, agreeing with it on $(A \otimes B)$ -measurable sets (For sigma-finite factors, the product measure exists, has the rectangle formula, is sigma-finite, and is unique, Measurable rectangles in a product of measurable spaces, Assuming countable choice, every measure space has a unique complete extension to its completion).

[F2]

Every nondegenerate interval in $[0, 1]$ , with any combination of included or excluded endpoints, is Lebesgue measurable and has measure equal to its positive length. In particular $μ ([0, 1]) = ν ([0, 1]) = 1$ and $μ ([0, \frac{1}{2}]) = \frac{1}{2}$ (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, Axis-parallel rectangles in $R^{m}$ and their volume).

[F3]

The preceding example computes the product kernel: $k$ is square integrable with $∥ k ∥_{2}^{2} = ∥ a ∥_{2}^{2} ∥ b ∥_{2}^{2}$ , its kernel operator satisfies $(T_{k} f) (x) = a (x) ⟨ f, b ⟩$ for each $f$ and for $μ$ -almost every $x$ , and its range admits an ordered basis of length at most one (A square-integrable separable product kernel).

[F4]

Padding a finite disjoint family by empty sets in countable additivity shows that a measure is finitely additive on disjoint measurable sets and takes values in $[0, + \infty]$ , so a measurable set containing a measurable subset of positive measure has positive measure (Measures on sigma-algebras).

[F5]

A continuous map between metric spaces is sequentially continuous: from $p_{j} \to p$ it follows that $h (p_{j}) \to h (p)$ (Metric continuity characterisations, with countable choice for the sequential converse, Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ ).

[F6]

In $[0, 1]^{2}$ every relative ball $B (p, r)$ with $r > 0$ about a point $p$ contains a product $I \times J$ of two nondegenerate intervals in $[0, 1]$ (with the boundary faces included when $p$ lies on the boundary). Explicitly, for $p = (u, v)$ choose $0 < d < \min (1, r / \sqrt{2})$ and take $I = [\max (0, u - d), \min (1, u + d)]$ , $J = [\max (0, v - d), \min (1, v + d)]$ ; both lengths are positive and every point of their product has Euclidean distance at most $\sqrt{2} d < r$ from $p$ . By [F2], $μ (I) > 0$ and $ν (J) > 0$ , so this is a measurable rectangle of positive $(μ \times ν)$ -measure and, by [F1], of the same positive completed measure (Open ball, closed ball and sphere in a metric space, Measurable rectangles in a product of measurable spaces).

[F7]

Choice implies Countable Choice (The Axiom of Choice, AC supplies the countable and dependent choices used in Banach integration), and Countable Choice selects one point from each of countably many nonempty subsets of a metric space (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Verification

technique · direct (contradiction)

Given: The objects above, and the null set $N : = {(x, y) \in [0, 1]^{2} : h (x, y) \neq k (x, y)}$ of the assumed almost-everywhere agreement, with $\overline{μ \times ν} (N) = 0$ .

1.1

The functions $a = 1_{[0, 1 / 2]}$ and $b = 1_{[0, 1]}$ have $∥ a ∥_{2}^{2} = μ ([0, \frac{1}{2}]) = \frac{1}{2}$ and $∥ b ∥_{2}^{2} = ν ([0, 1]) = 1$ by [F2], so [F3] gives that $k$ is square integrable with $∥ k ∥_{2}^{2} = \frac{1}{2}$ , that $(T_{k} f) (x) = ⟨ f, b ⟩ 1_{[0, 1 / 2]} (x)$ , as an identity of $L^{2}$ classes. Since $T_{k} b = a \neq 0$ and the range is contained in $C a$ , its range is exactly this one-dimensional subspace, proving rank one.

F2F3

1.2

If $N$ contained a ball $B (p, r)$ with $r > 0$ , then [F1] and [F6] would produce a measurable rectangle $R \subseteq B (p, r) \subseteq N$ whose completed measure is $μ (A) ν (B) > 0$ , and the disjoint decomposition $N = R \cup (N ∖ R)$ , together with the additivity and nonnegativity of [F4], would give $\overline{μ \times ν} (N) \geq \overline{μ \times ν} (R) > 0$ , contradicting $\overline{μ \times ν} (N) = 0$ ; hence no ball with positive radius is contained in $N$ .

F1F4F6

2.1

Values on the left half. Let $p = (x, y)$ with $0 < x < \frac{1}{2}$ and $0 < y < 1$ . For each $j \geq 0$ the ball $B (p, 1 / (j + 1))$ is not contained in $N$ by [step 1.2], so it contains a point $q_{j}$ of its complement, and by [F7] the countably many points $q_{j}$ may be chosen simultaneously; then $q_{j} \to p$ by construction. For $j$ large enough $q_{j}$ lies in the rectangle $(0, \frac{1}{2}) \times (0, 1)$ on which $k = 1$ , and $q_{j} \notin N$ gives $h (q_{j}) = k (q_{j}) = 1$ ; sequential continuity [F5] therefore forces $h (p) = 1$ .

step 1.2F5F7

2.2

Values on the right half. The same argument with $(0, \frac{1}{2})$ replaced by $(\frac{1}{2}, 1)$ , where $k = 0$ , and with the same null set $N$ , gives $h (p) = 0$ for every $p = (x, y)$ with $\frac{1}{2} < x < 1$ and $0 < y < 1$ .

step 1.2F5F7

3.1

Contradiction at the interface. Let $p_{0} = (\frac{1}{2}, \frac{1}{2})$ and let $p_{j} = (\frac{1}{2} - 1 / (j + 3), \frac{1}{2})$ and $p_{j}^{'} = (\frac{1}{2} + 1 / (j + 3), \frac{1}{2})$ ; both sequences converge to $p_{0}$ in $[0, 1]^{2}$ , [step 2.1] gives $h (p_{j}) = 1$ for every $j$ , and [step 2.2] gives $h (p_{j}^{'}) = 0$ for every $j$ . Sequential continuity [F5] applied to the first sequence gives $h (p_{0}) = 1$ and applied to the second gives $h (p_{0}) = 0$ , a contradiction; therefore no continuous $h$ agrees with $k$ almost everywhere.

step 2.1step 2.2F5

4.1

Steps 1.1 and 3.1 establish all the asserted properties: square integrability with $∥ k ∥_{2}^{2} = \frac{1}{2}$ , the rank-one form of $T_{k}$ on the one hand, and the impossibility of a continuous representative on the other.

step 1.1step 3.1∎

Depends on

Used by

A Hilbert–Schmidt kernel operator is compact on L two Example

Dependency tree · two levels

79 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

Gerald Teschl, Topics in Real and Functional Analysis, version November 17, 2017 — §3.6, comparison of continuous and L2 kernels, printed pp. 93–96 (standard reference, not scraped)
Sheldon Axler, Measure, Integration & Real Analysis — null sets and continuous representatives, Chapter 7 (standard reference, not scraped)