Measure theory — Lebesgue measure (intro): Question bank

4.10.25 · D5 · Maths › Advanced Topics (Elite Level) › Measure theory — Lebesgue measure (intro)

Shuru karne se pehle, ek reminder taaki neeche ke har prompt pehli line se readable ho:

$μ^{*}$ = outer measure = open-interval covers ki tightest total length, har ek set ke liye defined hai.
$μ$ = Lebesgue measure = $μ^{*}$ ko measurable sets par restrict karna, jahan ye countably additive ban jaata hai.
"measure zero" ka matlab: kisi bhi $ε > 0$ ke liye tum set ko aise intervals se cover kar sakte ho jinki lengths ka sum $ε$ se kam ho.

True or false — justify

Ek set countable hai agar tum uske elements $a_{1}, a_{2}, a_{3}, \dots$ list kar sako (labels ka ek "countable budget"). Density/countability traps ke liye ye picture yaad rakho.

T/F: Ek dense set ka measure zaroor positive hoga.

False —

Q

R

mein dense hai phir bhi uska measure

0

hai; density ek topological "everywhere-ness" hai, measure length hai, aur ye dono independent notions hain.

T/F: Ek uncountable set ka measure zaroor positive hoga.

False — Cantor set uncountable hai lekin uska measure

0

hai (

2^{n}

intervals se cover hota hai jinki length

3^{- n}

hai, total

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

T/F: Measure zero ke set ko countable hona zaroori hai.

False — same Cantor set ka example: measure zero hai lekin uncountable hai, isliye "measure zero" countability force nahi karta.

T/F: Har countable set ka measure zero hota hai.

True —

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

enumerate karo,

a_{n}

ε / 2^{n}

length ke interval se cover karo; total

\sum ε / 2^{n} = ε \to 0

hai.

T/F:

μ^{*}

sirf measurable sets ke liye defined hai.

False —

μ^{*}

R

ke har subset ke liye defined hai; yahi uski poori point hai. Sirf

μ

(restriction) measurable sets par rehta hai.

T/F: Outer measure

R

ke sabhi subsets par countably additive hai.

False — sabhi sets par

μ^{*}

sirf sub-additive hai (

\leq

); poori additivity tabhi milti hai jab measurable sets ke

σ

-algebra par restrict karo.

T/F: Agar

A \subseteq B

ho toh

μ^{*} (A) \leq μ^{*} (B)

True (monotonicity) —

B

ka har cover,

A

ko bhi cover karta hai, isliye

A

ke cover-sums ke set mein

B

ke cover-sums bhi hain, aur zyada badi collection ka infimum chhota-ya-barabar hota hai.

T/F: Disjoint

A, B

ke liye

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

hamesha hold karta hai.

False — ye non-measurable pieces ke liye fail ho sakta hai; disjointness akela kaafi nahi, tum chahte ho ki kam se kam unme se ek measurable ho (Vitali set counterexample hai).

T/F:

R

ka har subset Lebesgue measurable hota hai.

False — Vitali set, jo Axiom of Choice se bana hai, Carathéodory split violate karta hai, isliye ise consistently measure nahi kiya ja sakta.

T/F: Ek single point

{p}

ka measure zero hota hai.

True — ise

2 ε

length ke

(p - ε, p + ε)

se cover karo; kyunki ye sabhi

ε > 0

ke liye kaam karta hai, measure

0

hai.

T/F:

a < b

ke saath interval

[a, b]

ka measure zero ho sakta hai.

False — compactness argument se

μ ([a, b]) = b - a > 0

; ek genuine interval hamesha apni length carry karta hai.

T/F: Agar

μ^{*} (A) = 0

ho toh

A

measurable hai.

True — measure-zero sets automatically measurable hote hain (ye har test set ko bina kisi leak ke split karte hain, kyunki

μ^{*} (A \cap E) \leq μ^{*} (A) = 0

Spot the error

Har item ek "proof" ya claim batata hai. Bolo kya toot raha hai.

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

aise points ka union hai jinmein se har ek ka measure zero hai, isliye finite additivity argument se

μ^{*} (Q) = 0

milta hai."

Error: finite additivity sirf finite pieces handle karta hai; infinitely many point-covers sum karne ke liye tumhe countable sub-additivity chahiye.

ε / 2^{n}

geometric budget exactly wahi countable ingredient hai.

"Kyunki

[0, 1]

mein measure-zero set

Q \cap [0, 1]

hai, aur baaki 'sirf irrationals hain jo

Q

ke bade version jaisi hain',

[0, 1]

ko bhi measure zero hona chahiye."

Error:

[0, 1]

mein irrationals uncountable hain, isliye

ε / 2^{n}

trick unpar apply nahi hoti; asliyat mein

μ ([0, 1]) = 1

hai aur irrationals saari length carry karte hain.

μ^{*} ([a, b]) \geq b - a

dikhane ke liye, bas note karo ki cover

(a - ε, b + ε)

ki length

\to b - a

hai."

Error: woh cover "

\leq

" direction prove karta hai, "

\geq

" nahi. "

\geq

" subtle part hai aur compactness (finite subcover) chahiye — tum

[a, b]

mein koi gap nahi chhod sakte.

E

measurable hai kyunki

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

har

A

ke liye."

Error: "

\leq

" har set ke liye sub-additivity se free mein milta hai, isliye ye kuch prove nahi karta. Measurability reverse "

\geq

" (equivalently equality) hai jo earn karni padti hai.

"Vitali set ka kuch outer measure

m > 0

hai; use translate karne se

[0, 1]

ke andar infinitely many disjoint copies milte hain, isliye unke measures kisi finite cheez mein sum hote hain — koi contradiction nahi."

Error: same positive measure ke countably many disjoint translates

\infty

mein sum hote hain, phir bhi wo ek bounded set mein fit hote hain, isliye sum finite hona chahiye — wahi contradiction (aur

m = 0

bhi fail hota hai) exactly kyun Vitali set non-measurable hai.

μ^{*}

aur

μ

ek hi function ke do naam hain, isliye distinction pedantic hai."

Error: ye measurable sets par hi same hain;

μ^{*}

sabhi sets tak extend hota hai lekin wahan additivity khota hai, jabki

μ = μ^{*} ∣_{M}

additive object hai. Distinction wohi jagah hai jahan saari pathology chhupi hai.

Why questions

Hum

μ^{*} (A)

ki definition mein supremum nahi, infimum kyun lete hain?

Kyunki ek cover size ko overestimate karta hai — true length kisi bhi cover ke total se zyada se zyada itni hai — isliye hum tightest cover ki taraf neechay squeeze karte hain; supremum infinity tak bhaag jaata.

Covering intervals countably many kyun hone chahiye, finitely many nahi?

Finite covers infinite structures jaise

Q

ko tightly wrap nahi kar sakte; countably many intervals ek geometric

ε / 2^{n}

budget ke saath kar sakte hain, jo exactly theory ko Riemann integration se aage upgrade karta hai.

Hum "sabhi sets" se measurable sets par restrict kyun karte hain?

Kyunki provably tum har subset par charon wishlist properties (interval-length, additivity, translation invariance, defined-everywhere) ek saath nahi rakh sakte;

σ

-algebra par restrict karna leaky sets ko hataakar countable additivity bachaa leta hai.

Carathéodory split har ek test set

A

ke liye required kyun hai, sirf

A = R

ke liye nahi?

Kyunki measurability robust honi chahiye —

E

ko cleanly cut karna chahiye chahe tum use kisi bhi cheez se probe karo; sirf

R

ke against test karna countable unions ke under closure guarantee karne ke liye bahut weak hai.

y

-axis chop karna (Lebesgue) kyun succeed karta hai jahan

x

-axis chop karna (Riemann) fail hota hai?

Range chop karna poochhta hai "kin

x

ke liye

f

height

y

ke paas hai?", aisi sets produce karta hai jinka measure hum evaluate kar sakte hain chahe

f

wildly oscillate kare; Riemann ka domain-chopping function ko har strip par tame chahiye.

Is topic mein Axiom of Choice bilkul aata kyun hai?

Iska use Vitali set banane ke liye uncountably many cosets mein se ek-ek representative select karne mein hota hai — ek purely non-constructive act — isliye non-measurable sets exist karte hain phir bhi explicitly likhe nahi ja sakte.

Q

"dense phir bhi negligible" kyun ho sakta hai?

Density ka matlab hai points har jagah arbitrarily close baithte hain, lekin tum phir bhi apni choice ke interval mein countably many points mein se har ek ko bag kar sakte ho; total length choti banaayi ja sakti hai, isliye ye koi length occupy nahi karta.

Edge cases

In boundary/degenerate inputs ko directly handle karo.

μ^{*} (\emptyset)

kya hai, aur kyun?

0

— empty set empty family se cover hota hai (ya

ε \to 0

length ke interval se), isliye uske cover-lengths ka infimum

0

hai.

μ^{*} (R)

kya hai?

+ \infty

— koi bhi countable collection finite total length ke intervals ki poori line cover nahi kar sakti; measure

\infty

value le sakta hai.

Kya ek "degenerate interval"

[a, a] = {a}

measure zero hai?

Haan — ye ek single point hai,

(a - ε, a + ε)

se cover hota hai, isliye uska measure

0

hai, jo

b - a = 0

ke consistent hai.

Kya kisi set ka

μ^{*} = 0

ho sakta hai phir bhi

R

mein dense ho?

Haan —

Q

khud dense bhi hai aur measure zero bhi; density aur measure ek doosre par koi constraint impose nahi karte.

Agar kisi set ka measure zero hai, toh kya uska har subset bhi measurable hai measure zero ke saath?

Haan — monotonicity se subset ka outer measure

\leq 0

hoga, isliye

= 0

hoga, aur measure-zero sets automatically measurable hote hain (ye Lebesgue measure ki completeness hai).

Finite disjoint union ke liye countable additivity kya hota hai — kya ye ek special case hai?

Haan — countable additivity mein finite additivity bhi aa jaata hai kyunki union mein empty sets padding kar do (

μ (\emptyset) = 0

), isliye kuch nahi khota; countable zyada strong statement hai.

Kya

μ^{*}

ke liye non-measurable sets par translation invariance hold karta hai?

Haan —

μ^{*}

sabhi sets par translation invariant hai (covers same length ke covers mein translate hote hain); takleef invariance mein nahi balki wahan additivity fail hone mein hai.

Recall Fast self-check

Ek aisa set batao jo uncountable ho lekin measure zero ho. ::: Cantor set.
Ek aisa set batao jo dense ho lekin measure zero ho. ::: Rationals $Q$ .
Carathéodory equation ka kaun sa direction "free" wala hai? ::: " $\leq$ ", sub-additivity se.
Non-measurable set banane mein kaun sa tool use hota hai? ::: Axiom of Choice (Vitali set).
Measure finite numbers ke alaawa legitimately kaun si value le sakta hai? ::: $+ \infty$ (jaise $μ (R)$ ).