advanced30 min studyLesson 183 in Mathematics

Lebesgue measure and nonmeasurable sets

Separate topology, cardinality, and measure, then identify the obstruction behind a Vitali set.

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What you’ll understand

  • Distinguish Lebesgue and Borel measurability through completion.
  • Prove that an uncountable compact set can have measure zero.
  • Explain why translation invariance and countable additivity obstruct measuring a Vitali selection.

Length is compatible with far more sets than elementary calculus suggests, yet extending it to every subset creates contradictions. Understanding both sides of that statement is useful. Null sets can be dense or uncountable, so neither sparsity in a picture nor cardinality predicts measure. At the other extreme, a nonmeasurable selection is not a mysterious interval of unknown length; its translated copies force incompatible requirements on any proposed translation-invariant countably additive extension of interval length.

The measurable domain includes null subsets

Lebesgue measure on the real line is a complete, translation-invariant measure agreeing with interval length. Every open set is a countable union of disjoint open intervals, and all Borel sets are measurable. Completing Borel measure adds every subset of every Borel null set. More precisely, each Lebesgue measurable set differs from a Borel set by a subset of a Borel null set. This description allows some functions that are not Borel measurable to be Lebesgue measurable after null-set modification.

Several notions of size remain independent. The rational numbers are dense but null because they are countable. The complement of the rationals in any interval has full measure in that interval but contains no open subinterval consisting entirely of irrational points. Countable additivity does not apply to the uncountable union of singleton points making up an interval. Each singleton having measure zero therefore does not force every set to have measure zero. The quantifier countable is doing essential work.

m(E△B)=0for some Borel Bwhenever E is Lebesgue measurablem(E\triangle B)=0\quad\text{for some Borel }B\quad\text{whenever }E\text{ is Lebesgue measurable}
Here E△B denotes symmetric difference. The null difference is itself Lebesgue measurable; it need not be a Borel set.↔ Scroll wide equations horizontally.

A compact uncountable set with no length

Construct the middle-thirds Cantor set by removing the middle third of each surviving interval repeatedly. After n stages, the remaining set Cₙ consists of 2ⁿ closed intervals of length 3⁻ⁿ. Its total length is therefore (2/3)ⁿ. The final intersection is closed and bounded, hence compact, and lies inside every Cₙ. Monotonicity gives its measure at most every one of these lengths. Taking n arbitrarily large proves that the Cantor set is null without requiring a list of its points.

To see uncountability, associate each infinite binary sequence with the ternary number whose digits are twice those binary digits. The resulting numbers lie in the Cantor set, and distinct binary sequences produce distinct numbers under this restricted digit construction. Thus uncountably many choices fit inside a set of measure zero. The endpoints' alternative ternary descriptions do not destroy this injection when one uses the all-zero-or-two representation consistently. A finite stage has positive length; the limit set has zero length.

C=⋂n=0∞Cn,0≤m(C)≤m(Cn)=2n3−n⟶0C=\bigcap_{n=0}^{\infty}C_n,\qquad 0\le m(C)\le m(C_n)=2^n3^{-n}\longrightarrow0
Every finite-stage set is measurable. Monotonicity alone supplies the displayed estimate, so the nullity proof does not depend on counting the limiting points.↔ Scroll wide equations horizontally.

Why a Vitali selection cannot be assigned a length

On [0,1], declare two points equivalent when their difference is rational. Using the axiom of choice, select one representative from each equivalence class and call the selection V. Its rational translates V+q for rational q in [−1,1] are pairwise disjoint: an overlap would identify two selected representatives of the same class, forcing them equal and then forcing the translation amounts equal. The union of these translates covers [0,1] and is contained in [−1,2]. Both containment statements matter to the contradiction.

If V were measurable with measure zero, countable additivity would give measure zero to that union, contradicting its covering [0,1]. If V had positive measure, infinitely many disjoint translates would give the bounded union infinite measure, contradicting containment in [−1,2]. Infinite measure for V is already impossible since V lies in [0,1]. The conclusion is failure of measurability, not failure of outer measure to exist. The construction uses choice and the rational equivalence relation; an arbitrary complicated-looking subset is not automatically nonmeasurable.

Measuring the finite stages rather than counting points

The Cantor construction starts with [0,1]. Compute the total lengths after stages 1, 2, and n, and use them to determine the measure of the limiting set.

  1. Stage 1 leaves two intervals of length 1/3, so its total length is 2/3. Stage 2 leaves four intervals of length 1/9, so its total length is 4/9.

  2. At stage n, each previous interval contributes two intervals one third as long. The recurrence Lₙ=(2/3)Lₙ₋₁ with L₀=1 gives Lₙ=(2/3)ⁿ.

  3. The limiting compact set lies inside every finite-stage set. Its nonnegative measure is bounded by (2/3)ⁿ for every n, so its measure is zero, even though binary digit choices show it is uncountable.

The result

The Cantor set is an uncountable compact null set. Cardinality and Lebesgue measure answer different questions.

Explore the model

Interactive lab

Interactive lab · function

Null changes cannot change an integral

Use the coefficient a and endpoint x controls for the quadratic and predict its area on [0,x]. Changing its values at countably many selected inputs would leave that Lebesgue integral unchanged. The plot shows only the continuous representative; it does not render Cantor sets, Vitali selections, or a test for measurability.

Null changes cannot change an integralThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn integral is insensitive to measurable null changes even when pointwise behavior changes dramatically.

Check your understanding

Why does the union of the singleton points in [0,1] not prove the interval null?

Try it yourself

Let C be the usual Cantor set. Explain why every subset of C is Lebesgue measurable, and why that statement does not imply that every subset of [0,1] is measurable.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use completeness for the first claim and the finite-stage length estimate to identify the required null container.

Compare with the solution

The Cantor set is a measurable null set, so completeness makes each of its subsets measurable with measure zero. The interval [0,1] has positive measure; completeness says nothing about arbitrary subsets of a positive-measure set. The Vitali construction supplies subsets of that interval that cannot be measured consistently with translation invariance and countable additivity. A null container is a substantive assumption, not an expendable part of the argument.

Take it with you

  • Density, cardinality, and measure should be checked independently.
  • Completeness includes every subset of a measurable null set.
  • The Vitali contradiction uses rational translates, countable additivity, translation invariance, and bounded containment.

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