Measure theory — Lebesgue measure (intro): Worked examples

This page is a drill. We take the Lebesgue outer measure machinery from the parent note Measure theory — Lebesgue measure (intro) and hit every kind of set it can throw at you. Before each example you get a Forecast line — cover the answer and guess. If your gut is wrong, that's the whole point.

Everything here uses just two objects (plus a short list of named facts), and we re-earn all of them so nobody is lost:

Definition What "infimum" means (needed before

$μ^{*}$ ) Suppose you have a whole collection of numbers — say, all the possible total lengths you can get from every legal cover of a set. A lower bound of that collection is any number that sits at or below every number in it. The infimum (written $in f$ ) is the greatest of all those lower bounds — the highest floor you can slide up to without poking above any member of the collection.

If the collection actually contains its smallest element, $in f$ is just that minimum.
If it doesn't (e.g. total lengths $2 ε$ for every $ε > 0$ : they can shrink toward $0$ but never equal $0$ ), the $in f$ is the limiting value $0$ that they approach but never reach. Formally: $m = in f S$ means (i) $m \leq x$ for all $x \in S$ , and (ii) no number bigger than $m$ is also a lower bound. This second clause is what makes it the tightest floor.

Before the drill, here is the named toolkit every example leans on. Each fact is stated in plain words so nothing is used before it is defined.

Formula Named facts (P1–P5) used throughout

(P1) Monotonicity. If $A \subseteq B$ then $μ^{*} (A) \leq μ^{*} (B)$ . (A cover of $B$ also covers $A$ , so $A$ can only be smaller.)
(P2) Countable sub-additivity. For any sets $A_{1}, A_{2}, \dots$ , $μ^{*} (⋃_{n} A_{n}) \leq \sum_{n} μ^{*} (A_{n})$ . (Glue their covers together; the total can only over-count.)
(P3) Interval length. $μ^{*} ([a, b]) = b - a$ (same for open/half-open). (Matches school "length".)
(P4) Measurability & measure $μ$ . A set $E$ is measurable when it splits every test set $A$ cleanly: $μ^{*} (A) = μ^{*} (A \cap E) + μ^{*} (A \cap E^{c})$ (Carathéodory). For such $E$ we write $μ (E) := μ^{*} (E)$ — the Lebesgue measure. On measurable sets $μ^{*}$ equals $μ$ ; the extra fact we gain is countable additivity: for disjoint measurable $E_{1}, E_{2}, \dots$ , $μ (⨆_{n} E_{n}) = \sum_{n} μ (E_{n})$ (the " $\leq$ " is P2; measurability upgrades it to " $=$ "). All intervals, hence all Borel sets, are measurable. A useful consequence: for measurable $A, B$ , inclusion–exclusion holds, $μ (A \cup B) = μ (A) + μ (B) - μ (A \cap B)$ (split $A \cup B$ into the disjoint measurable pieces $A ∖ B$ , $A \cap B$ , $B ∖ A$ and add).
(P5) Translation invariance. Sliding a set doesn't change its measure: $μ (A + t) = μ (A)$ . (Shifting an interval keeps its width.)

The scenario matrix

Every set you meet falls into one of these cells. The examples below are labelled with the cell they knock out, so by the end no scenario is unseen.

Cell	Class of set	Key question it tests	Example
A	Single point (degenerate interval)	Does a "zero-width" object have length?	Ex 1
B	Finite set of points	Do finitely many zeros stay zero?	Ex 2
C	Countably infinite set	Does the $ε / 2^{n}$ budget survive infinity?	Ex 3
D	A genuine interval $[a, b]$	Does $μ^{*}$ agree with school "length"?	Ex 4
E	Overlapping / union of intervals	Sub-additivity vs the true answer	Ex 5
F	Uncountable-yet-measure-zero (Cantor)	"Uncountable" $\neq =$ "big"	Ex 6
G	Real-world word problem	Total defect length in manufacturing	Ex 7
H	Exam twist: a "fat" (positive-measure) removed set	When density/measure intuitions clash	Ex 8
I	Limiting / pathological	The Vitali set — why we can't measure it	Ex 9

Cells cover: degenerate (A), finite (B), countable (C), the base interval (D), unions/overlap (E), uncountable-null (F), applied (G), exam twist (H), and the limiting non-measurable case (I).

Cell A — a single point

Worked example Example 1 —

$μ^{*} ({7})$ Statement: Find the outer measure of the one-point set ${7}$ .

Forecast: Is the length of a single dot $0$ , some tiny positive number, or undefined?

Step 1. Cover ${7}$ by the single open interval $I = (7 - ε, 7 + ε)$ . Why this step? A point has no width, so one small interval already swallows it. We keep $ε$ as a free knob we can shrink.

Step 2. Its length is $ℓ (I) = (7 + ε) - (7 - ε) = 2 ε$ . So $μ^{*} ({7}) \leq 2 ε$ . Why this step? $μ^{*}$ is the infimum over all covers, hence $\leq$ this one particular cover's total.

Step 3. This holds for every $ε > 0$ . A non-negative number that is below every positive number must be $0$ (its infimum-floor is $0$ ). Hence $μ^{*} ({7}) = 0$ . Why this step? $μ^{*} \geq 0$ always (lengths are non-negative), and it can't exceed anything positive — squeezed to $0$ .

Verify: Sanity — a point is the interval $[7, 7]$ , and (P3) gives $μ^{*} = 7 - 7 = 0$ . ✔

Figure s01 — read it as the proof, not decoration. The amber dot is the point ${7}$ . The three stacked bars are the cover $(7 - ε, 7 + ε)$ for three shrinking $ε$ , each labelled with its length $2 ε$ (Step 2). Reading up the stack you are watching $ε \to 0$ : the bar-length printed beside each bar is exactly the " $\leq 2 ε$ " bound, and the bars visibly collapse onto the dot — that collapse is Step 3's "below every positive number $\Rightarrow 0$ ".

Figure — Measure theory — Lebesgue measure (intro)

Cell B — finitely many points

Cell C — a countably infinite set

Worked example Example 3 —

$μ^{*} (N)$ and $μ^{*} (Q \cap [0, 1])$ Statement: (a) all natural numbers ${1, 2, 3, \dots}$ ; (b) the rationals inside $[0, 1]$ .

Forecast: Now there are infinitely many points. Does the size finally become positive? Does it matter that (b) is dense — sitting arbitrarily close to every real in $[0, 1]$ ?

Step 1. In each part the set is countable: list its members $q_{1}, q_{2}, q_{3}, \dots$ . Why this step? A countable list is exactly what lets us hand out a shrinking budget indexed by $n$ .

Step 2. Cover $q_{n}$ by an interval of length $\frac{ε}{2 ^{n}}$ . Why this step? The lengths $\frac{ε}{2}, \frac{ε}{4}, \frac{ε}{8}, \dots$ form a geometric series — a known finite sum, so the total won't blow up.

Step 3. Total $= n = 1 \sum \infty \frac{ε}{2 ^{n}} = ε \cdot \frac{1/2}{1 - 1/2} = ε$ . So $μ^{*} \leq ε$ (infimum $\leq$ this cover), and letting $ε \to 0$ gives $μ^{*} = 0$ for both (a) and (b). Why this step? The geometric sum $\sum 2^{- n} = 1$ is the engine: infinitely many points, still a finite (shrinkable) total.

Verify: The partial sum after $N$ terms is $ε (1 - 2^{- N}) < ε$ , confirming the series is bounded by $ε$ . ✔

Punchline for (b): $Q \cap [0, 1]$ is dense — no gap in $[0, 1]$ avoids it — yet its length is $0$ . Density (a topological idea) and measure (a size idea) are unrelated. See Countable vs uncountable.

Figure s02 — the picture is the geometric-series argument. The amber dots are the points $x = 1/ n$ crowding toward $0$ ; each is wrapped in a cyan box of width $ε / 2^{n}$ , so the boxes shrink by a factor of $2$ each step (Step 2). Reading the widths left-to-right you are literally reading the terms $ε /2, ε /4, \dots$ of the series in Step 3; the amber caption sums them to $ε$ and then sends that to $0$ .

Cell D — a genuine interval

Worked example Example 4 —

$μ^{*} ([2, 6])$ , the " $\geq$ " direction done carefully Statement: Confirm $μ^{*} ([2, 6]) = 4$ , and see why the lower bound is the hard part.

Forecast: Upper bound is easy. But could a clever pile of tiny intervals cover $[2, 6]$ with total length less than $4$ ?

Step 1. Upper bound. The single open interval $J = (2 - ε, 6 + ε)$ satisfies $[2, 6] \subseteq J$ , so it is a legal cover with total length $ℓ (J) = 4 + 2 ε$ . Because $μ^{*}$ is the infimum over all covers, it is $\leq$ the total of this cover: $μ^{*} ([2, 6]) \leq 4 + 2 ε$ . (Equivalently, by monotonicity (P1) $μ^{*} ([2, 6]) \leq μ^{*} (J)$ , and by (P2)/(P3) $μ^{*} (J) \leq ℓ (J) = 4 + 2 ε$ .) Let $ε \to 0$ : $μ^{*} \leq 4$ . Why this step? A single interval containing $[2, 6]$ is one legal cover, and the infimum can never exceed any legal cover's total — that is precisely what (P1)/(P2) formalise. Shrinking $ε$ pushes the bound to $4$ .

Step 2. Lower bound (the subtle half). $[2, 6]$ is compact (closed and bounded). Any open cover has a finite subcover — finitely many intervals that already cover $[2, 6]$ . Why this step? With infinitely many intervals you could fear infinite cancellation tricks; compactness forces the problem down to a finite pile, where "no gaps allowed" is provable.

Step 3. Those finitely many intervals leave no gap in $[2, 6]$ ; walking from $2$ to $6$ you're always inside some interval, so their lengths must total at least the distance travelled, $6 - 2 = 4$ . Hence every cover totals $\geq 4$ , so $μ^{*} \geq 4$ . Why this step? If the finite lengths summed to less than $4$ , some sub-interval of $[2, 6]$ would be uncovered — contradiction.

Step 4. $\leq 4$ and $\geq 4$ give $μ^{*} ([2, 6]) = 4$ .

Verify: Matches (P3): $ℓ ([a, b]) = b - a = 6 - 2 = 4$ . ✔

Cell E — union of overlapping intervals

Worked example Example 5 —

$μ^{*} ((0, 3) \cup (2, 5))$ Statement: Two intervals that overlap on $(2, 3)$ .

Forecast: Naïvely add lengths $3 + 3 = 6$ ? Or is the true answer smaller because they share ground?

Step 1. The union of $(0, 3)$ and $(2, 5)$ is the single interval $(0, 5)$ (they touch through the shared piece $(2, 3)$ ). Why this step? Overlap means the two intervals merge into one contiguous stretch; drawing the number line makes this obvious.

Step 2. So $μ^{*} ((0, 3) \cup (2, 5)) = μ^{*} ((0, 5)) = 5 - 0 = 5$ by (P3). Why this step? The measure of a single open interval is its width.

Step 3. Compare with sub-additivity (P2): $μ^{*} \leq μ^{*} ((0, 3)) + μ^{*} ((2, 5)) = 3 + 3 = 6$ . Indeed $5 \leq 6$ — the inequality is strict here because of the overlap. Why this step? This is exactly why $μ^{*}$ is only sub-additive on general covers: double-counting the shared part inflates the bound. The true measure never double-counts.

Verify: Intervals are measurable (P4), so we may apply inclusion–exclusion (P4): $μ ((0, 3) \cup (2, 5)) = μ ((0, 3)) + μ ((2, 5)) - μ ((2, 3)) = 3 + 3 - 1 = 5$ . ✔ Matches Step 2. (Note: this equality is legal only because all three sets are measurable — inclusion–exclusion is an additivity fact, not a property of raw $μ^{*}$ .)

Figure s03 — it shows why " $3 + 3$ " over-counts. The cyan bar is $(0, 3)$ , the amber bar is $(2, 5)$ , and the shaded white band marks the overlap $(2, 3)$ of length $1$ that both bars claim. The single white bar below is the true union $(0, 5)$ of length $5$ — the shaded band is precisely the " $- 1$ " inclusion–exclusion correction from the Verify line, made visible.

Cell F — uncountable, yet measure zero

Worked example Example 6 — the

Cantor set has $μ^{*} = 0$ Statement: Build $C$ by repeatedly deleting middle-thirds of $[0, 1]$ . Show $μ^{*} (C) = 0$ even though $C$ is uncountable.

Forecast: An uncountable set (as many points as all of $[0, 1]$ !) — surely it has positive length?

Step 1. At stage $n$ , $C$ is covered by $2^{n}$ closed intervals each of length $3^{- n}$ . Why this step? Deleting the open middle third of each surviving interval doubles the count ( $2^{n}$ ) and thirds the width ( $3^{- n}$ ).

Step 2. Total covering length at stage $n$ : $2^{n} \cdot 3^{- n} = (\frac{2}{3})^{n}$ . By monotonicity (P1), since $C$ sits inside this cover, $μ^{*} (C) \leq (2/3)^{n}$ . Why this step? $C$ sits inside all stages, so it sits inside stage $n$ for every $n$ — pick whichever $n$ you like.

Step 3. As $n \to \infty$ , $(2/3)^{n} \to 0$ (a ratio below $1$ raised to growing powers). So $μ^{*} (C) \leq (2/3)^{n} \to 0$ , giving $μ^{*} (C) = 0$ . Why this step? The same squeeze: $μ^{*}$ is below a quantity that collapses to $0$ .

Verify: $(2/3)^{10} \approx 0.01734$ , already tiny and shrinking. ✔ Yet $C$ is uncountable — proving "uncountable" and "positive measure" are independent (Countable vs uncountable).

Figure s04 — each row is one line of the induction in Step 1. Row $0$ is the full $[0, 1]$ ; each lower row deletes the middle third of every surviving segment, so segment-count doubles and width thirds — exactly the " $2^{n}$ intervals of width $3^{- n}$ " of Step 1. The right-hand label on each row prints its running total $(2/3)^{n}$ , marching $1 \to 0.667 \to 0.444 \to \dots$ : that column is the sequence in Step 3 collapsing to $0$ .

Cell G — a real-world word problem

Worked example Example 7 — total defect length on a wire

Statement: A $1 m$ wire has microscopic flaws at positions $x_{n} = \frac{1}{n} m$ for $n = 1, 2, 3, \dots$ . Flaw $n$ spans an interval of width $\frac{1}{100 \cdot 2 ^{n}} m$ centred at $x_{n}$ . What total length of wire is flawed, and could you polish it all away with a " $1 cm$ polishing budget"?

Forecast: Infinitely many flaws — is the total defect length infinite, or does it stay under $1 cm = 0.01 m$ ?

Step 1. The flawed set is covered by the flaw-intervals; by sub-additivity (P2) its outer measure is at most $n = 1 \sum \infty \frac{1}{100 \cdot 2 ^{n}}$ . Why this step? Outer measure of the flawed region is at most the summed widths of the covering flaw-intervals.

Step 2. Factor out $\frac{1}{100}$ : $\frac{1}{100} n = 1 \sum \infty 2^{- n} = \frac{1}{100} \cdot 1 = \frac{1}{100} m = 1 cm$ . Why this step? Same geometric series $\sum 2^{- n} = 1$ as Ex 3 — the budget engine again.

Step 3. Total flawed length $\leq 0.01 m = 1 cm$ . So yes — a $1 cm$ polishing budget suffices. Why this step? $μ^{*}$ gives a rigorous upper bound on how much material must be treated.

Verify: $\frac{1}{100} \sum_{n = 1}^{\infty} 2^{- n} = \frac{1}{100} \cdot 1 = 0.01 m$ . Units: metres throughout. ✔

Cell H — the exam twist (positive-measure removed set)

Worked example Example 8 — a "fat" removed set:

$μ ([0, 1] ∖ C)$ Statement: From $[0, 1]$ remove the Cantor set $C$ . What is the measure of what remains? Exam trap: "you removed uncountably many points, so the leftover must be small."

Forecast: After deleting an uncountable set, is the remaining measure near $0$ , or near $1$ ?

Step 1. First note $C$ is measurable: it is closed (an intersection of closed stages), hence a Borel set, and by (P4) all Borel sets are measurable — so $μ (C) = μ^{*} (C)$ is defined, and its complement $[0, 1] ∖ C$ is measurable too (measurable sets form a Sigma-algebra, closed under complement). Why this step? Additivity (P4) is only allowed for measurable pieces, so we must certify measurability before splitting.

Step 2. $[0, 1] = C ⊔ ([0, 1] ∖ C)$ , a disjoint split of two measurable sets. Countable additivity (P4) gives $μ ([0, 1]) = μ (C) + μ ([0, 1] ∖ C)$ . Why this step? Disjoint measurable pieces have measures that add exactly — the upgrade from sub-additive to additive.

Step 3. $μ ([0, 1]) = 1$ by (P3) and $μ (C) = 0$ (Ex 6, since $μ = μ^{*}$ on the measurable set $C$ ). So $μ ([0, 1] ∖ C) = 1 - 0 = 1$ . Why this step? Substitute the two known measures into the additivity equation.

Step 4. So the leftover — the union of all the deleted open middle-thirds — has full measure $1$ , yet it misses the uncountable set $C$ . Why this step? This defeats the trap: you removed uncountably many points but zero length, so nothing measurable was lost.

Verify: The deleted middles have total length $\frac{1}{3} + \frac{2}{9} + \frac{4}{27} + \dots = \frac{1}{3} \sum_{k = 0}^{\infty} (\frac{2}{3})^{k} = \frac{1}{3} \cdot \frac{1}{1 - 2/3} = 1$ . ✔ Matches Step 3.

Cell I — the limiting / non-measurable case

Worked example Example 9 — why the Vitali set has NO consistent measure

Statement: On $[0, 1]$ , group reals by "differ by a rational". Using the Axiom of Choice, pick one representative from each group; call this set $V$ . Show no value of $μ (V)$ can satisfy the toolkit — so $V$ is not measurable (P4).

Forecast: Can we assign $V$ length $0$ ? Some positive number? Anything at all?

Step 1. List the rationals in $[- 1, 1]$ as $r_{1}, r_{2}, \dots$ (they are countable) and form the shifted copies $V_{k} = V + r_{k}$ . These are pairwise disjoint and all live inside $[- 1, 2]$ . Why this step? Two representatives differ by an irrational (by construction), so shifting by different rationals can never make them collide — the copies don't overlap.

Step 2. Their union covers $[0, 1]$ : every real in $[0, 1]$ differs from its representative by some rational in $[- 1, 1]$ . So $[0, 1] \subseteq ⨆_{k} V_{k} \subseteq [- 1, 2]$ . By monotonicity (P1) and (P3), taking measures gives $1 \leq μ (⨆_{k} V_{k}) \leq 3$ . Why this step? A set squeezed between two known-measure sets has measure between their measures.

Step 3. Suppose $V$ were measurable. Then each $V_{k}$ is measurable too, and countable additivity (P4) turns the middle into a sum: $μ (⨆_{k} V_{k}) = \sum_{k} μ (V_{k})$ . Why this step? Additivity is exactly what lets us break the union's measure into the copies' measures — but only if they're measurable, which is the assumption we're testing.

Step 4. Translation invariance (P5) forces $μ (V_{k}) = μ (V + r_{k}) = μ (V)$ for every $k$ — sliding by a rational can't change length. So $k \sum μ (V_{k}) = k = 1 \sum \infty μ (V)$ : infinitely many identical copies of one number $μ (V)$ . Why this step? This pins the whole infinite sum to a single unknown value $μ (V)$ .

Step 5. The contradiction — both cases fail the trap $1 \leq sum \leq 3$ from Step 2.

If $μ (V) = 0$ : the sum is $0 + 0 + 0 + \dots = 0$ , but Step 2 demands it be $\geq 1$ . Impossible.
If $μ (V) > 0$ : the sum is $\infty c + c + c + \dots = + \infty$ , but Step 2 demands it be $\leq 3$ . Impossible.

Step 6. No value of $μ (V)$ survives, so the assumption " $V$ is measurable" was false. Hence $V$ is not measurable — exactly why we restrict $μ$ to the Sigma-algebra $M$ instead of demanding it on all sets. And recall (from the box above) the whole set $V$ existed only via the Axiom of Choice — the pathology is inseparable from that non-constructive choice. Why this step? The pathology is the concrete reason the wishlist can't hold for every subset of $R$ .

Verify: Logic check — the required interval for the sum is $[1, 3]$ ; the value $0$ (from $μ (V) = 0$ ) lies below it and $+ \infty$ (from $μ (V) > 0$ ) lies above it, so neither case lands inside $[1, 3]$ . ✔ No consistent measure exists.

Active recall

Recall One line per cell — can you produce the answer?

Cell A/B: measure of any finite point-set? ::: $0$
Cell C: measure of $Q \cap [0, 1]$ , and why density doesn't matter? ::: $0$ ; density is topological, measure is size
Cell D: hard direction of $μ^{*} ([a, b]) = b - a$ relies on which property? ::: compactness → finite subcover, no gaps
Cell E: why can $μ^{*} (A \cup B) < μ^{*} (A) + μ^{*} (B)$ , and which fact lets you compute the exact value? ::: overlap double-counts in the sub-additive bound (P2); inclusion–exclusion (needs measurability, P4)
Cell F: Cantor set measure and cardinality? ::: measure $0$ , yet uncountable
Cell H: $μ ([0, 1] ∖ C)$ , and which property lets you split it? ::: $1$ (full measure); countable additivity (P4)
Cell I: why is the Vitali set unmeasurable, which two properties clash, and what built it? ::: additivity (P4) + translation invariance (P5) force the sum to be $0$ or $\infty$ , never in $[1, 3]$ ; the Axiom of Choice built it non-constructively