False statementConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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FALSE: the supremum of an arbitrary family of measurable functions is always measurable

Statement

False claim. The pointwise supremum of an arbitrary family of measurable functions is measurable. The correct theorem on this page is only the sequential version.

Facts & Assumptions

Given: The Axiom of Choice, the Lebesgue measurable-space structure on $[0, 1]$ , a Vitali set $V \subseteq [0, 1]$ , and the family ${1_{{t}} : t \in V}$ of indicator functions on $[0, 1]$ .

[L1]

Assuming choice, Vitali sets exist and are not Lebesgue measurable. (Assuming choice on the cosets of $Q$ in $R$ , a Vitali set in $[0, 1]$ exists, Assuming the Axiom of Choice, a Vitali set is not Lebesgue measurable)

[L2]

The indicator of a measurable set is measurable. (An indicator function is measurable exactly when its set is measurable)

Refutation

technique · direct

1.1

Each singleton ${t}$ is measurable, so [L2] makes every [given, L2] $1_{{t}}$ measurable.

givenL2

2.1

The pointwise supremum of the family is

step 1.1L1L2

$\sup_{t \in V} 1_{{t}} = 1_{V} .$

Since $V$ is not measurable by [L1], the function $1_{V}$ is not measurable. So an uncountable supremum of measurable functions can fail to be measurable. [step 1.1, L1, L2] ∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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

Sheldon Axler, Measure, Integration and Real Analysis, Exercise 29 (standard reference, not scraped)