CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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A nondecreasing function that is not right-continuous can fail countable additivity

Statement refuted

That the interval prescription $μ ((a, b]) = F (b) - F (a)$ still defines a countably additive Borel measure when $F$ is merely nondecreasing and not right-continuous.

Facts & Assumptions

Given: The nondecreasing function

$F (x) : = {\begin{cases} 0, & x \leq 0, \\ 1, & x > 0. \end{cases}$

[L1]

Measures are continuous from above on decreasing sets when the first set has finite measure. (Continuity from above when one set has finite measure)

Counterexample

technique · direct

1.1

If a Borel measure $μ$ satisfied $μ ((a, b]) = F (b) - F (a)$ , then [given] $μ ((0, 1 / n]) = 1$ for every $n$ , because $F (1 / n) = 1$ and $F (0) = 0$ .

given

2.1

The sets $(0, 1 / n]$ decrease to $\emptyset$ , and [step 1.1, L1] $μ ((0, 1]) = 1 < + \infty$ . Therefore [L1] would force $μ ((0, 1 / n]) \to μ (\emptyset) = 0$ , contradicting step 1.1. So the interval prescription fails countable additivity for this nondecreasing non-right-continuous $F$ .

step 1.1L1∎

Depends on

Continuity from above when one set has finite measure

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 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 B. Folland, Real Analysis, 2nd ed., Section 1.5 (standard reference, not scraped)