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Lesson 3 of 32

Construct size from countable covers

Prove an interval-cover outer measure is subadditive.

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To assign a size to a scattered subset of the line, cover it by open intervals and add their lengths. A better cover may use many small intervals rather than one large span. Taking the infimum over all countable covers defines Lebesgue outer measure. This construction assigns a value to every subset, but it does not yet make every subset measurable or guarantee additivity on disjoint sets.

Predict

Predict the size of the rational points in the unit interval. They are dense, so any single interval containing them has length at least one. But enumerate them as q1, q2,… and cover q_n by an interval of length ε/2^n. The total length is at most ε. Since ε is arbitrary, their outer measure is zero. Density and measure describe different properties; a dense set need not occupy positive length.

Experiment

Compare simple covering strategies.

SetCover strategyResult
One pointArbitrarily short intervalOuter measure 0
Countable setSummable individual lengthsOuter measure 0
Closed interval[a, b]Slightly enlarged intervalUpper bound b−a
Same intervalCompactness lower boundEquality b−a

For the countable row, explain why assigning the same small length to every point would not provide a finite total.

Understand

Define m*(E) as the infimum of Σlength (I_j) over countable open-interval covers of E. The empty cover gives m*(empty)=0. Every cover of a larger set covers a smaller one, proving monotonicity. To prove countable subadditivity, the case of infinite sum of outer measures is immediate. Otherwise, for a chosen ε>0, cover E_n with total length at most m*(E_n)+ε2^(−n). The union of these countably many countable covers covers the union of the E_n, and the total excess is at most ε. Taking the infimum and then letting ε decrease proves m*(union E_n)≤Σm*(E_n). These three properties define an outer measure abstractly.

Apply

For a compact interval[a, b], an enlarged open interval gives the upper bound b−a. Any countable open cover has a finite subcover by compactness. A finite collection of intervals covering[a, b] has total length at least b−a: order their endpoints, merge overlaps, and note that the component covering the interval spans both endpoints. Thus every cover has sufficient total length, giving the lower bound. Other endpoint conventions follow by sandwiching compact subintervals inside slightly enlarged open intervals. Unbounded intervals have infinite outer measure by containing compact intervals of arbitrarily large length.

Continue

The infimum may not be attained by one cover, so proofs use covers within an arbitrary error tolerance rather than a nonexistent best cover. Countable subadditivity is also weaker than disjoint additivity. The next lesson selects exactly those sets that split every test set without losing outer mass, producing a genuine complete measure. This distinction is essential in the later nonmeasurability example: the outer measure exists for the problematic set even though countable additivity cannot consistently extend ordinary length to all its translations and unions.

Continue to Split every test set without losing mass.

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Practice and reasoning

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What is the Lebesgue outer measure of the rational numbers in [0,1]?

Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.

Find the outer measure of (−2,3].

Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.

Prove interval-cover outer measure is countably subadditive.

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Prove every singleton has outer measure zero.

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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.

Explain why a cover of a compact interval has total interval length at least its length.

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Why does outer-measure subadditivity not already prove a measure?

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Reflect and continue

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