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

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Finite quadratic variation does not imply finite total variation

Statement refuted

The implication "a continuous path with finite quadratic variation along the dyadic meshes of a compact interval has finite total variation on that interval" is false. Brownian motion provides the witness on $[0, 1]$ : almost every Brownian path has dyadic quadratic sums converging uniformly to elapsed time there while its total variation on that interval is infinite.

Facts & Assumptions

Given: AC and a standard Brownian motion $B$ .

[F1]

Almost surely, simultaneously: $t \mapsto B_{t}$ is continuous on $[0, 1]$ , the dyadic partial quadratic-variation processes of $B$ on $[0, 1]$ converge uniformly to $t$ , and the variation sums of $B$ are unbounded above on every nondegenerate compact interval. This follows by intersecting the full-measure continuity event in the Brownian definition with the full-measure event carrying both variation conclusions. Brownian motion Brownian one- and quadratic variation

[F2]

The Brownian one- and quadratic-variation corollary in [F1] is stated under AC and uses it in its supplier and countable-event interfaces; the present counterexample carries that exact hypothesis forward. Brownian one- and quadratic variation The Axiom of Choice

Counterexample

technique · direct

1.1

The probability-one event in [F1] is nonempty; fix an outcome $ω$ in that event, so the same chosen path is continuous and has both stated variation properties.

F1given

2.1

For the path $t \mapsto B_{t} (ω)$ on $[0, 1]$ the dyadic quadratic sums are finite for each mesh and converge uniformly to elapsed time, whereas the supremum of the absolute-increment sums over partitions of $[0, 1]$ is $+ \infty$ ; both assertions refer to the same fixed continuous path.

step 1.1F1

3.1

Hence the refuted implication fails: the witness is continuous on $[0, 1]$ , has finite quadratic variation there in the stated dyadic sense, and nevertheless is not of bounded variation there. AC is the standing hypothesis required by the Brownian corollary [F1], as recorded in [F2]; after its nonempty probability-one event is supplied, fixing one witness makes no additional choice-family construction.

F1F2step 2.1∎

Source notes

Lawler, Section 2.8, records the dichotomy between divergence of the absolute-increment sums and convergence of the squared-increment sums for Brownian paths; the counterexample packages the two properties on the fixed horizon $[0, 1]$ as the failure of an implication about deterministic continuous paths.

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Sources

Gregory F. Lawler, Stochastic Calculus: An Introduction with Applications, Section 2.8 (standard reference, not scraped)