False statementConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28

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FALSE: assuming countable choice, Lusin's theorem says measurable functions are continuous off a null set

Statement refuted

Assume the Axiom of Countable Choice.

Lusin's theorem says measurable functions are continuous off a null set.

Facts & Assumptions

Given: The Axiom of Countable Choice and the Dirichlet function $D : = 1_{Q \cap [0, 1]} : [0, 1] \to R$ .

[L1]

Assuming countable choice, Lusin's theorem provides large closed sets on which a measurable real-valued function is continuous. (Assuming countable choice, Lusin's theorem on finite-measure subsets of R^n)

[L2]

Both the rationals and the irrationals are dense in $R$ . (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable)

Refutation

technique · direct

1.1

The function $D$ is measurable and, by [L1], for every $ε > 0$ there is a closed set $F \subseteq [0, 1]$ with $λ ([0, 1] ∖ F) < ε$ such that $D ∣_{F}$ is continuous.

1.2

Fix $x \in [0, 1]$ . By [L2], every neighbourhood of $x$ contains both a rational point $q$ and an irrational point $u$ . Then $D (q) = 1$ and $D (u) = 0$ , so $D$ is not continuous at $x$ . Thus $D$ is nowhere continuous on $[0, 1]$ .

2.1

Step 1.1 is exactly Lusin's conclusion, while step 1.2 shows that no null set deletion can make $D$ continuous at the remaining points as a function on $[0, 1]$ . So the stated reading of Lusin's theorem is false.

step 1.1step 1.2∎

Depends on

Used by

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Sources

Richard F. Bass, Real Analysis for Graduate Students, Example 5.16 (standard reference, not scraped)