Measure theory — Lebesgue measure (intro): Question bank

Before we start, one reminder so every prompt below is readable from line one:

$μ^{*}$ = outer measure = tightest total length of open-interval covers, defined for every set.
$μ$ = Lebesgue measure = $μ^{*}$ restricted to measurable sets, where it becomes countably additive.
"measure zero" means: for any $ε > 0$ you can cover the set with intervals whose lengths sum to less than $ε$ .

True or false — justify

A set is countable if you can list its elements $a_{1}, a_{2}, a_{3}, \dots$ (a "countable budget" of labels). Keep that picture for the density/countability traps.

T/F: A dense set must have positive measure.

False —

Q

is dense in

R

yet has measure

0

; density is a topological "everywhere-ness", measure is length, and they are independent notions.

T/F: An uncountable set must have positive measure.

False — the Cantor set is uncountable but has measure

0

(covered by

2^{n}

intervals of length

3^{- n}

, total

(2/3)^{n} \to 0

T/F: A set of measure zero must be countable.

False — same Cantor-set example: measure zero but uncountable, so "measure zero" does not force countability.

T/F: Every countable set has measure zero.

True — enumerate

a_{1}, a_{2}, \dots

, cover

a_{n}

by an interval of length

ε / 2^{n}

; the total is

\sum ε / 2^{n} = ε \to 0

T/F:

μ^{*}

is defined only for measurable sets.

False —

μ^{*}

is defined for every subset of

R

; that generality is its whole point. It is

μ

(the restriction) that lives only on measurable sets.

T/F: Outer measure is countably additive on all subsets of

R

False — on all sets

μ^{*}

is only sub-additive (

\leq

); full additivity holds only after restricting to the

σ

-algebra of measurable sets.

T/F: If

A \subseteq B

then

μ^{*} (A) \leq μ^{*} (B)

True (monotonicity) — every cover of

B

also covers

A

, so

A

's set of cover-sums contains

B

's, and an infimum over a larger collection is smaller-or-equal.

T/F:

μ^{*} (A \cup B) = μ^{*} (A) + μ^{*} (B)

always holds for disjoint

A, B

False — this can fail for non-measurable pieces; disjointness alone is not enough, you need at least one of them measurable (the Vitali set is the counterexample).

T/F: Every subset of

R

is Lebesgue measurable.

False — the Vitali set, built via the Axiom of Choice, violates the Carathéodory split, so it cannot be measured consistently.

T/F: A single point

{p}

has measure zero.

True — cover it by

(p - ε, p + ε)

of length

2 ε

; since this works for all

ε > 0

, the measure is

0

T/F: An interval

[a, b]

with

a < b

can have measure zero.

False — by the compactness argument

μ ([a, b]) = b - a > 0

; a genuine interval always carries its length.

T/F: If

μ^{*} (A) = 0

then

A

is measurable.

True — measure-zero sets are automatically measurable (they split every test set with no leak, since

μ^{*} (A \cap E) \leq μ^{*} (A) = 0

Spot the error

Each item states a "proof" or claim. Say what breaks.

Q = {q_{1}, q_{2}, \dots}

is a union of points each of measure zero, so a finite additivity argument gives

μ^{*} (Q) = 0

Error: finite additivity handles only finitely many pieces; you need countable sub-additivity to sum infinitely many point-covers. The

ε / 2^{n}

geometric budget is exactly the countable ingredient.

"Since

[0, 1]

contains the measure-zero set

Q \cap [0, 1]

, and the rest is 'just irrationals which are like a bigger version of

Q

[0, 1]

should also be measure zero."

Error: the irrationals in

[0, 1]

are uncountable, so the

ε / 2^{n}

trick does not apply to them; indeed

μ ([0, 1]) = 1

and the irrationals carry all of that length.

"To show

μ^{*} ([a, b]) \geq b - a

, just note the cover

(a - ε, b + ε)

has length

\to b - a

Error: that cover proves the "

\leq

" direction, not "

\geq

". The "

\geq

" is the subtle part and needs compactness (finite subcover) — you cannot leave a gap in

[a, b]

uncovered.

E

is measurable because

μ^{*} (A) \leq μ^{*} (A \cap E) + μ^{*} (A \cap E^{c})

for every

A

Error: the "

\leq

" holds for every set for free by sub-additivity, so it proves nothing. Measurability is the reverse "

\geq

" (equivalently equality) that must be earned.

"The Vitali set has some outer measure

m > 0

; translating it gives infinitely many disjoint copies inside

[0, 1]

, so their measures sum to something finite — no contradiction."

Error: countably many disjoint translates of the same positive measure sum to

\infty

, yet they fit in a bounded set, so the sum should be finite — that contradiction (also

m = 0

fails) is precisely why the Vitali set is non-measurable.

μ^{*}

and

μ

are just two names for the same function, so the distinction is pedantic."

Error: they are the same only on measurable sets;

μ^{*}

extends to all sets but loses additivity there, while

μ = μ^{*} ∣_{M}

is the additive object. The distinction is where all the pathology hides.

Why questions

Why do we take an infimum (not a supremum) in the definition of

μ^{*} (A)

Because a cover overestimates size — the true length is at most any cover's total — so we squeeze downward to the tightest cover; a supremum would run off to infinity.

Why must the covering intervals be countably many, not finitely many?

Finite covers can't tightly wrap infinite structures like

Q

; countably many intervals with a geometric

ε / 2^{n}

budget can, which is exactly what upgrades the theory beyond Riemann integration.

Why do we restrict from "all sets" to measurable sets at all?

Because you provably cannot have all four wishlist properties (interval-length, additivity, translation invariance, defined-everywhere) on every subset; restricting to a

σ

-algebra rescues countable additivity by throwing out the leaky sets.

Why is the Carathéodory split required for every test set

A

, not just for

A = R

Because measurability must be robust —

E

has to cut cleanly no matter what you probe it with; testing only against

R

is too weak to guarantee closure under countable unions.

Why does chopping the

y

-axis (Lebesgue) succeed where chopping the

x

-axis (Riemann) fails?

Chopping the range asks "for which

x

f

near height

y

?", producing sets whose measure we can evaluate even when

f

oscillates wildly; the domain-chopping of Riemann needs the function tame on each strip.

Why does the Axiom of Choice appear in this topic at all?

It is used to select one representative from each of uncountably many cosets to build the Vitali set — a purely non-constructive act — which is why non-measurable sets exist yet can never be written down explicitly.

Why can

Q

be "dense yet negligible"?

Density means points sit arbitrarily close everywhere, but you can still bag each of the countably many points in an interval of your choosing; the total length can be made tiny, so it occupies no length.

Edge cases

Handle these boundary/degenerate inputs directly.

What is

μ^{*} (\emptyset)

, and why?

0

— the empty set is covered by the empty family (or an interval of length

ε \to 0

), so its infimum of cover-lengths is

0

What is

μ^{*} (R)

+ \infty

— no countable collection of intervals of finite total length can cover the whole line; measure is allowed to take the value

\infty

Is a "degenerate interval"

[a, a] = {a}

measure zero?

Yes — it is a single point, covered by

(a - ε, a + ε)

, so its measure is

0

, consistent with

b - a = 0

Can a set have

μ^{*} = 0

yet be dense in

R

Yes —

Q

itself is both dense and measure zero; density and measure impose no constraint on each other.

If a set has measure zero, is every subset of it also measurable with measure zero?

Yes — by monotonicity a subset has outer measure

\leq 0

, hence

= 0

, and measure-zero sets are automatically measurable (this is completeness of Lebesgue measure).

What happens to countable additivity for a finite disjoint union — is it a special case?

Yes — countable additivity contains finite additivity by padding the union with empty sets (

μ (\emptyset) = 0

), so nothing is lost; countable is the strictly stronger statement.

Does translation invariance hold for non-measurable sets under

μ^{*}

Yes —

μ^{*}

is translation invariant on all sets (covers translate to covers of equal length); the trouble is not invariance but that additivity fails there.

Recall Fast self-check

Give one set that is uncountable but measure zero. ::: The Cantor set.
Give one set that is dense but measure zero. ::: The rationals $Q$ .
Which direction of the Carathéodory equation is the "free" one? ::: The " $\leq$ ", from sub-additivity.
What tool builds a non-measurable set? ::: The Axiom of Choice (Vitali set).
What value can measure legitimately take besides finite numbers? ::: $+ \infty$ (e.g. $μ (R)$ ).