Measure theory — Lebesgue measure (intro): Worked examples

4.10.25 · D3 · Maths › Advanced Topics (Elite Level) › Measure theory — Lebesgue measure (intro)

Yeh page ek drill hai. Hum Lebesgue outer measure ki machinery ko parent note Measure theory — Lebesgue measure (intro) se lete hain aur har tarah ke set pe apply karte hain jo tumhare saamne aa sakta hai. Har example se pehle tumhe ek Forecast line milti hai — answer cover karo aur guess karo. Agar tumhara intuition galat nikla, toh wahi toh poora point hai.

Yahan sirf do cheezein use hoti hain (aur named facts ki ek chhoti list), aur hum unhe dobara earn karte hain taaki koi bhi lost na ho:

Definition "Infimum" ka matlab kya hai (

$μ^{*}$ se pehle zaroori hai) Maan lo tumhare paas numbers ka ek poora collection hai — jaise, saari possible total lengths jo tum kisi set ke har legal cover se pa sakte ho. Us collection ka ek lower bound woh koi bhi number hai jo us collection ke har number ke barabar ya neeche ho. Infimum (likha jaata hai $in f$ ) un saare lower bounds mein se sabse bada hota hai — sabse ooncha floor jo tum bina collection ke kisi member ke upar pooche utha sakte ho.

Agar collection mein actually uska sabse chhota element maujood hai, toh $in f$ bas wahi minimum hai.
Agar nahi hai (jaise total lengths $2 ε$ for every $ε > 0$ : woh $0$ ki taraf shrink kar sakte hain par kabhi $0$ ke barabar nahi ho sakte), toh $in f$ woh limiting value $0$ hai jiske paas woh jaate hain par pahunchte nahi. Formally: $m = in f S$ ka matlab hai (i) $m \leq x$ for all $x \in S$ , aur (ii) $m$ se bada koi number bhi lower bound nahi hai. Yahi doosri condition ise sabse tight floor banati hai.

Drill se pehle, yahan named toolkit hai jis par har example tikta hai. Har fact plain words mein diya gaya hai taaki kuch bhi define hone se pehle use na ho.

Formula Puri jagah use hone wale Named facts (P1–P5)

(P1) Monotonicity. Agar $A \subseteq B$ toh $μ^{*} (A) \leq μ^{*} (B)$ . ( $B$ ka cover $A$ ko bhi cover karta hai, isliye $A$ chhota hi hoga.)
(P2) Countable sub-additivity. Kisi bhi sets $A_{1}, A_{2}, \dots$ ke liye, $μ^{*} (⋃_{n} A_{n}) \leq \sum_{n} μ^{*} (A_{n})$ . (Unke covers ko jodo; total sirf over-count kar sakta hai.)
(P3) Interval length. $μ^{*} ([a, b]) = b - a$ (open/half-open ke liye bhi same). (School wali "length" se match karta hai.)
(P4) Measurability & measure $μ$ . Ek set $E$ measurable hota hai jab woh har test set $A$ ko saaf split karta ho: $μ^{*} (A) = μ^{*} (A \cap E) + μ^{*} (A \cap E^{c})$ (Carathéodory). Aise $E$ ke liye hum likhte hain $μ (E) := μ^{*} (E)$ — yahi Lebesgue measure hai. Measurable sets par $μ^{*}$ equals $μ$ ; extra fact jo humein milta hai woh hai countable additivity: disjoint measurable $E_{1}, E_{2}, \dots$ ke liye, $μ (⨆_{n} E_{n}) = \sum_{n} μ (E_{n})$ (P2 mein " $\leq$ " hai; measurability ise " $=$ " tak upgrade karta hai). Saare intervals, aur isliye saare Borel sets, measurable hain. Ek useful consequence: measurable $A, B$ ke liye, inclusion–exclusion hold karta hai, $μ (A \cup B) = μ (A) + μ (B) - μ (A \cap B)$ ( $A \cup B$ ko disjoint measurable pieces $A ∖ B$ , $A \cap B$ , $B ∖ A$ mein split karo aur add karo).
(P5) Translation invariance. Kisi set ko slide karne se uska measure nahi badalta: $μ (A + t) = μ (A)$ . (Interval ko shift karne se uski width same rehti hai.)

The scenario matrix

Har set jo tum miloge woh inn cells mein se kisi ek mein aata hai. Neeche ke examples us cell ke saath labelled hain jo woh knock out karta hai, taaki end mein koi scenario unseen na rahe.

Cell	Set ki class	Key question jo test hota hai	Example
A	Single point (degenerate interval)	Kya "zero-width" object ki length hoti hai?	Ex 1
B	Points ka finite set	Kya finitely many zeros zero hi rehte hain?	Ex 2
C	Countably infinite set	Kya $ε / 2^{n}$ budget infinity tak survive karta hai?	Ex 3
D	Ek genuine interval $[a, b]$	Kya $μ^{*}$ school wali "length" se agree karta hai?	Ex 4
E	Overlapping / union of intervals	Sub-additivity vs sahi answer	Ex 5
F	Uncountable-yet-measure-zero (Cantor)	"Uncountable" $\neq =$ "bada"	Ex 6
G	Real-world word problem	Manufacturing mein total defect length	Ex 7
H	Exam twist: ek "fat" (positive-measure) removed set	Jab density/measure intuitions clash karein	Ex 8
I	Limiting / pathological	Vitali set — hum ise measure kyun nahi kar sakte	Ex 9

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

Cell A — ek single point

Worked example Example 1 —

$μ^{*} ({7})$ Statement: Ek point wale set ${7}$ ka outer measure nikalo.

Forecast: Ek single dot ki length $0$ hai, koi tiny positive number hai, ya undefined hai?

Step 1. ${7}$ ko ek single open interval $I = (7 - ε, 7 + ε)$ se cover karo. Yeh step kyun? Ek point ki koi width nahi hoti, isliye ek chhota interval usse swallow kar leta hai. $ε$ ek free knob hai jise hum shrink kar sakte hain.

Step 2. Uski length hai $ℓ (I) = (7 + ε) - (7 - ε) = 2 ε$ . Toh $μ^{*} ({7}) \leq 2 ε$ . Yeh step kyun? $μ^{*}$ saari covers par infimum hai, isliye woh $\leq$ is particular cover ke total se hoga.

Step 3. Yeh har $ε > 0$ ke liye hold karta hai. Ek non-negative number jo har positive number se neeche ho woh $0$ hi hoga (uska infimum-floor $0$ hai). Isliye $μ^{*} ({7}) = 0$ . Yeh step kyun? $μ^{*} \geq 0$ hamesha (lengths non-negative hoti hain), aur yeh kisi bhi positive se exceed nahi kar sakta — $0$ par squeeze ho jaata hai.

Verify: Sanity check — ek point interval $[7, 7]$ hai, aur (P3) deta hai $μ^{*} = 7 - 7 = 0$ . ✔

Figure s01 — ise proof ki tarah padho, decoration ki tarah nahi. Amber dot point ${7}$ hai. Teeno stacked bars cover $(7 - ε, 7 + ε)$ hain teen shrinking $ε$ ke liye, har ek pe uski length $2 ε$ likhi hai (Step 2). Stack ko upar padhte waqt tum dekh rahe ho $ε \to 0$ : har bar ke saath printed bar-length exactly woh " $\leq 2 ε$ " bound hai, aur bars visibly dot par collapse ho rahi hain — woh collapse hi Step 3 ka "har positive number se neeche $\Rightarrow 0$ " hai.

Figure — Measure theory — Lebesgue measure (intro)

Cell B — finitely many points

Cell C — ek countably infinite set

Worked example Example 3 —

$μ^{*} (N)$ aur $μ^{*} (Q \cap [0, 1])$ Statement: (a) saare natural numbers ${1, 2, 3, \dots}$ ; (b) $[0, 1]$ ke andar rationals.

Forecast: Ab infinitely many points hain. Kya ab size finally positive ho jaayegi? Kya yeh matter karta hai ki (b) dense hai — $[0, 1]$ ke har real ke arbitrarily close baithe hain?

Step 1. Dono parts mein set countable hai: apne members $q_{1}, q_{2}, q_{3}, \dots$ list karo. Yeh step kyun? Countable list exactly wahi hai jo humein $n$ se indexed shrinking budget dene deti hai.

Step 2. $q_{n}$ ko $\frac{ε}{2 ^{n}}$ length wale interval se cover karo. Yeh step kyun? Lengths $\frac{ε}{2}, \frac{ε}{4}, \frac{ε}{8}, \dots$ ek geometric series banate hain — ek known finite sum, isliye total blow up nahi karega.

Step 3. Total $= n = 1 \sum \infty \frac{ε}{2 ^{n}} = ε \cdot \frac{1/2}{1 - 1/2} = ε$ . Toh $μ^{*} \leq ε$ (infimum $\leq$ is cover), aur $ε \to 0$ karne par (a) aur (b) dono ke liye $μ^{*} = 0$ milta hai. Yeh step kyun? Geometric sum $\sum 2^{- n} = 1$ engine hai: infinitely many points, phir bhi ek finite (shrinkable) total.

Verify: $N$ terms ke baad partial sum $ε (1 - 2^{- N}) < ε$ hai, jo confirm karta hai ki series $ε$ se bound hai. ✔

Punchline for (b): $Q \cap [0, 1]$ dense hai — $[0, 1]$ mein koi gap usse avoid nahi kar sakta — phir bhi uski length $0$ hai. Density (ek topological idea) aur measure (ek size idea) unrelated hain. Dekho Countable vs uncountable.

Figure s02 — picture geometric-series argument hai. Amber dots points $x = 1/ n$ hain jo $0$ ki taraf crowd kar rahe hain; har ek ko ek cyan box mein wrap kiya hai width $ε / 2^{n}$ ke saath, isliye boxes har step mein $2$ ke factor se shrink hote hain (Step 2). Widths ko left-to-right padhne par tum literally series ke terms $ε /2, ε /4, \dots$ padh rahe ho jaise Step 3 mein hain; amber caption unhe $ε$ mein sum karta hai aur phir use $0$ par bhejta hai.

Cell D — ek genuine interval

Worked example Example 4 —

$μ^{*} ([2, 6])$ , " $\geq$ " direction carefully kiya gaya Statement: Confirm karo $μ^{*} ([2, 6]) = 4$ , aur dekho kyun lower bound hard part hai.

Forecast: Upper bound aasaan hai. Lekin kya tiny intervals ka koi clever dher $[2, 6]$ ko total length $4$ se kam mein cover kar sakta hai?

Step 1. Upper bound. Single open interval $J = (2 - ε, 6 + ε)$ satisfy karta hai $[2, 6] \subseteq J$ , isliye yeh total length $ℓ (J) = 4 + 2 ε$ wala ek legal cover hai. Kyunki $μ^{*}$ saari covers par infimum hai, woh $\leq$ is cover ke total se hai: $μ^{*} ([2, 6]) \leq 4 + 2 ε$ . (Equivalently, monotonicity (P1) se $μ^{*} ([2, 6]) \leq μ^{*} (J)$ , aur (P2)/(P3) se $μ^{*} (J) \leq ℓ (J) = 4 + 2 ε$ .) $ε \to 0$ karne par: $μ^{*} \leq 4$ . Yeh step kyun? $[2, 6]$ ko contain karne wala ek interval ek legal cover hai, aur infimum kisi bhi legal cover ke total se exceed nahi kar sakta — yahi (P1)/(P2) formalise karte hain. $ε$ shrink karne se bound $4$ tak aa jaata hai.

Step 2. Lower bound (subtle half). $[2, 6]$ compact hai (closed aur bounded). Kisi bhi open cover ka finite subcover hota hai — finitely many intervals jo already $[2, 6]$ cover kar lein. Yeh step kyun? Infinitely many intervals ke saath darr rehta hai ki infinite cancellation tricks ho sakti hain; compactness problem ko ek finite dher tak force kar deta hai, jahan "no gaps allowed" provable hai.

Step 3. Woh finitely many intervals $[2, 6]$ mein koi gap nahi chhodte; $2$ se $6$ tak chalte waqt tum hamesha kisi na kisi interval ke andar ho, isliye unki lengths ka total kam se kam $4$ hona chahiye, yani $6 - 2 = 4$ ka distance. Isliye har cover ka total $\geq 4$ hai, toh $μ^{*} \geq 4$ . Yeh step kyun? Agar finite lengths $4$ se kam mein sum hote, toh $[2, 6]$ ka koi sub-interval uncovered reh jaata — contradiction.

Step 4. $\leq 4$ aur $\geq 4$ milkar dete hain $μ^{*} ([2, 6]) = 4$ .

Verify: (P3) se match karta hai: $ℓ ([a, b]) = b - a = 6 - 2 = 4$ . ✔

Cell E — overlapping intervals ka union

Worked example Example 5 —

$μ^{*} ((0, 3) \cup (2, 5))$ Statement: Do intervals jo $(2, 3)$ par overlap karte hain.

Forecast: Naively lengths add karo $3 + 3 = 6$ ? Ya sahi answer chhota hai kyunki dono shared ground rakhte hain?

Step 1. $(0, 3)$ aur $(2, 5)$ ka union single interval $(0, 5)$ hai (woh shared piece $(2, 3)$ ke through touch karte hain). Yeh step kyun? Overlap ka matlab hai dono intervals ek contiguous stretch mein merge ho jaate hain; number line draw karne par yeh obvious ho jaata hai.

Step 2. Toh $μ^{*} ((0, 3) \cup (2, 5)) = μ^{*} ((0, 5)) = 5 - 0 = 5$ by (P3). Yeh step kyun? Ek single open interval ka measure uski width hai.

Step 3. Compare karo sub-additivity (P2) se: $μ^{*} \leq μ^{*} ((0, 3)) + μ^{*} ((2, 5)) = 3 + 3 = 6$ . Indeed $5 \leq 6$ — yahan inequality strict hai overlap ki wajah se. Yeh step kyun? Exactly isliye $μ^{*}$ general covers par sirf sub-additive hai: shared part ko double-count karna bound inflate karta hai. Sahi measure kabhi double-count nahi karta.

Verify: Intervals measurable hain (P4), isliye hum inclusion–exclusion (P4) apply kar sakte hain: $μ ((0, 3) \cup (2, 5)) = μ ((0, 3)) + μ ((2, 5)) - μ ((2, 3)) = 3 + 3 - 1 = 5$ . ✔ Step 2 se match karta hai. (Note: yeh equality tabhi legal hai jab teeno sets measurable hon — inclusion–exclusion ek additivity fact hai, raw $μ^{*}$ ki property nahi.)

Figure s03 — yeh dikhata hai kyun " $3 + 3$ " over-count karta hai. Cyan bar $(0, 3)$ hai, amber bar $(2, 5)$ hai, aur shaded white band overlap $(2, 3)$ ko mark karta hai length $1$ ke saath jo dono bars claim karte hain. Neeche wala single white bar true union $(0, 5)$ hai length $5$ ke saath — shaded band exactly woh " $- 1$ " inclusion–exclusion correction hai Verify line se, visible bana diya gaya.

Cell F — uncountable, phir bhi measure zero

Worked example Example 6 —

Cantor set ka $μ^{*} = 0$ Statement: $C$ banao $[0, 1]$ ke middle-thirds repeatedly delete karke. Dikhao $μ^{*} (C) = 0$ hai jabki $C$ uncountable hai.

Forecast: Ek uncountable set (jitne points $[0, 1]$ mein hain utne hi!) — surely uski positive length hogi?

Step 1. Stage $n$ par, $C$ ko $2^{n}$ closed intervals se cover kiya jaata hai har ek length $3^{- n}$ ke saath. Yeh step kyun? Har surviving interval ka open middle third delete karne se count double ho jaata hai ( $2^{n}$ ) aur width third ho jaati hai ( $3^{- n}$ ).

Step 2. Stage $n$ par total covering length: $2^{n} \cdot 3^{- n} = (\frac{2}{3})^{n}$ . Monotonicity (P1) se, kyunki $C$ is cover ke andar hai, $μ^{*} (C) \leq (2/3)^{n}$ . Yeh step kyun? $C$ saare stages ke andar hai, isliye woh stage $n$ ke andar hai har $n$ ke liye — jo $n$ chahiye pick karo.

Step 3. Jab $n \to \infty$ , $(2/3)^{n} \to 0$ ( $1$ se kam ratio growing powers tak raise hota hai). Toh $μ^{*} (C) \leq (2/3)^{n} \to 0$ , jo deta hai $μ^{*} (C) = 0$ . Yeh step kyun? Same squeeze: $μ^{*}$ ek aisi quantity se neeche hai jo $0$ par collapse ho rahi hai.

Verify: $(2/3)^{10} \approx 0.01734$ , already tiny aur shrink ho raha hai. ✔ Phir bhi $C$ uncountable hai — prove karta hai ki "uncountable" aur "positive measure" independent hain (Countable vs uncountable).

Figure s04 — har row Step 1 ke induction ki ek line hai. Row $0$ poora $[0, 1]$ hai; har neeche wali row har surviving segment ka middle third delete karti hai, isliye segment-count double aur width third ho jaati hai — exactly Step 1 ka " $2^{n}$ intervals of width $3^{- n}$ ". Har row par right-hand label uska running total $(2/3)^{n}$ print karta hai, $1 \to 0.667 \to 0.444 \to \dots$ chalte hue: woh column hi Step 3 ki sequence hai jo $0$ par collapse ho rahi hai.

Cell G — ek real-world word problem

Worked example Example 7 — ek wire par total defect length

Statement: Ek $1 m$ wire mein microscopic flaws hain positions $x_{n} = \frac{1}{n} m$ par $n = 1, 2, 3, \dots$ ke liye. Flaw $n$ ek interval span karta hai width $\frac{1}{100 \cdot 2 ^{n}} m$ ke saath $x_{n}$ par centred. Wire ka kul kitna hissa flawed hai, aur kya tum ise ek " $1 cm$ polishing budget" se polish kar sakte ho?

Forecast: Infinitely many flaws hain — kya total defect length infinite hai, ya $1 cm = 0.01 m$ se neeche rehti hai?

Step 1. Flawed set flaw-intervals se cover hota hai; sub-additivity (P2) se uska outer measure at most hai $n = 1 \sum \infty \frac{1}{100 \cdot 2 ^{n}}$ . Yeh step kyun? Flawed region ka outer measure covering flaw-intervals ki summed widths se at most hai.

Step 2. $\frac{1}{100}$ factor out karo: $\frac{1}{100} n = 1 \sum \infty 2^{- n} = \frac{1}{100} \cdot 1 = \frac{1}{100} m = 1 cm$ . Yeh step kyun? Same geometric series $\sum 2^{- n} = 1$ jaisi Ex 3 mein hai — wahi budget engine phir se.

Step 3. Total flawed length $\leq 0.01 m = 1 cm$ . Toh haan — ek $1 cm$ polishing budget kaafi hai. Yeh step kyun? $μ^{*}$ ek rigorous upper bound deta hai ki kitna material treat karna padega.

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

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

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

$μ ([0, 1] ∖ C)$ Statement: $[0, 1]$ se Cantor set $C$ remove karo. Jo bachta hai uska measure kya hai? Exam trap: "tumne uncountably many points remove kiye, isliye bacha hua chhota hona chahiye."

Forecast: Ek uncountable set delete karne ke baad, remaining measure $0$ ke paas hai ya $1$ ke paas?

Step 1. Pehle note karo $C$ measurable hai: yeh closed hai (closed stages ka intersection), isliye ek Borel set hai, aur (P4) se saare Borel sets measurable hote hain — toh $μ (C) = μ^{*} (C)$ defined hai, aur uska complement $[0, 1] ∖ C$ bhi measurable hai (measurable sets ek Sigma-algebra banate hain, complement ke under closed). Yeh step kyun? Additivity (P4) sirf measurable pieces ke liye allowed hai, isliye hume split karne se pehle measurability certify karni hogi.

Step 2. $[0, 1] = C ⊔ ([0, 1] ∖ C)$ , do measurable sets ka disjoint split. Countable additivity (P4) deta hai $μ ([0, 1]) = μ (C) + μ ([0, 1] ∖ C)$ . Yeh step kyun? Disjoint measurable pieces ke measures exactly add hote hain — sub-additive se additive tak ka upgrade.

Step 3. $μ ([0, 1]) = 1$ by (P3) aur $μ (C) = 0$ (Ex 6, kyunki measurable set $C$ par $μ = μ^{*}$ ). Toh $μ ([0, 1] ∖ C) = 1 - 0 = 1$ . Yeh step kyun? Do known measures additivity equation mein substitute karo.

Step 4. Toh jo bachta hai — saare deleted open middle-thirds ka union — ka full measure $1$ hai, phir bhi woh uncountable set $C$ ko miss karta hai. Yeh step kyun? Yeh trap defeat karta hai: tumne uncountably many points remove kiye lekin zero length, toh koi measurable cheez nahi gayi.

Verify: Deleted middles ki total length $\frac{1}{3} + \frac{2}{9} + \frac{4}{27} + \dots = \frac{1}{3} \sum_{k = 0}^{\infty} (\frac{2}{3})^{k} = \frac{1}{3} \cdot \frac{1}{1 - 2/3} = 1$ hai. ✔ Step 3 se match karta hai.

Cell I — limiting / non-measurable case

Intuition Pehle:

Axiom of Choice kya hai, aur yahan kyun zaroori hai Ab tak ki har construction explicit thi — hum covering intervals ka naam le sakte the. Vitali set alag hai. Ise banane ke liye humein ek saath infinitely many arbitrary choices karni padti hain (uncountably many groups mein se har ek se ek representative), koi rule nahi batata ki kaun sa choose karo. Woh principle jo aise simultaneous, ruleless selection ki ijazat deta hai woh hai Axiom of Choice: yeh assert karta hai ki ek choice function exist karta hai, kabhi exhibit kiye bina. Isliye Vitali set non-constructive hai — koi formula nahi, koi picture nahi. Wahi non-constructive origin exactly iska reason hai ki yeh har "nice" property ko dodge kar sakta hai aur unmeasurable ho jaata hai. (Is example ke saath koi figure nahi hai precisely isliye kyunki set draw nahi kiya ja sakta.)

Worked example Example 9 — Vitali set ka koi consistent measure kyun NAHI hai

Statement: $[0, 1]$ par, reals ko "rational se differ karte hain" ke basis par group karo. Axiom of Choice use karke, har group se ek representative choose karo; is set ko $V$ kaho. Dikhao ki $μ (V)$ ki koi bhi value toolkit ko satisfy nahi kar sakti — isliye $V$ not measurable (P4) hai.

Forecast: Kya hum $V$ ko length $0$ assign kar sakte hain? Koi positive number? Kuch bhi?

Step 1. $[- 1, 1]$ mein rationals ko $r_{1}, r_{2}, \dots$ list karo (woh countable hain) aur shifted copies $V_{k} = V + r_{k}$ banao. Yeh pairwise disjoint hain aur saari $[- 1, 2]$ ke andar rehti hain. Yeh step kyun? Do representatives ek irrational se differ karte hain (construction se), isliye alag rationals se shift karne par woh kabhi collide nahi ho sakte — copies overlap nahi karti.

Step 2. Unka union $[0, 1]$ cover karta hai: $[0, 1]$ ka har real apne representative se kisi rational $[- 1, 1]$ mein differ karta hai. Toh $[0, 1] \subseteq ⨆_{k} V_{k} \subseteq [- 1, 2]$ . Monotonicity (P1) aur (P3) se, measures lene par $1 \leq μ (⨆_{k} V_{k}) \leq 3$ milta hai. Yeh step kyun? Do known-measure sets ke beech squeezed set ka measure unke measures ke beech hota hai.

Step 3. Maan lo $V$ measurable hai. Toh har $V_{k}$ bhi measurable hai, aur countable additivity (P4) middle ko ek sum mein badal deta hai: $μ (⨆_{k} V_{k}) = \sum_{k} μ (V_{k})$ . Yeh step kyun? Additivity exactly wahi hai jo union ke measure ko copies ke measures mein break karne deti hai — lekin sirf tab jab woh measurable hon, jo woh assumption hai jise hum test kar rahe hain.

Step 4. Translation invariance (P5) force karta hai $μ (V_{k}) = μ (V + r_{k}) = μ (V)$ har $k$ ke liye — rational se slide karna length change nahi kar sakta. Toh $k \sum μ (V_{k}) = k = 1 \sum \infty μ (V)$ : ek number $μ (V)$ ki infinitely many identical copies. Yeh step kyun? Yeh poore infinite sum ko ek single unknown value $μ (V)$ par pin kar deta hai.

Step 5. Contradiction — dono cases Step 2 ke trap $1 \leq sum \leq 3$ fail karte hain.

Agar $μ (V) = 0$ : sum hai $0 + 0 + 0 + \dots = 0$ , lekin Step 2 maangta hai $\geq 1$ . Impossible.
Agar $μ (V) > 0$ : sum hai $\infty c + c + c + \dots = + \infty$ , lekin Step 2 maangta hai $\leq 3$ . Impossible.

Step 6. $μ (V)$ ki koi bhi value survive nahi karti, isliye assumption " $V$ measurable hai" galat thi. Isliye $V$ not measurable hai — exactly isliye hum $μ$ ko Sigma-algebra $M$ tak restrict karte hain aur saare sets par demand nahi karte. Aur yaad raho (upar wale box se) poora set $V$ sirf Axiom of Choice ke zariye exist karta tha — pathology us non-constructive choice se alag nahi ki ja sakti. Yeh step kyun? Yeh pathology concrete reason hai ki wishlist $R$ ke har subset ke liye hold nahi kar sakti.

Verify: Logic check — sum ke liye required interval $[1, 3]$ hai; value $0$ (from $μ (V) = 0$ ) usse neeche hai aur $+ \infty$ (from $μ (V) > 0$ ) usse upar hai, isliye koi bhi case $[1, 3]$ ke andar nahi aata. ✔ Koi consistent measure exist nahi karta.

Active recall

Recall Har cell ke liye ek line — kya tum answer produce kar sakte ho?

Cell A/B: kisi bhi finite point-set ka measure? ::: $0$
Cell C: $Q \cap [0, 1]$ ka measure, aur density kyun matter nahi karta? ::: $0$ ; density topological hai, measure size hai
Cell D: $μ^{*} ([a, b]) = b - a$ ki hard direction kis property par rely karti hai? ::: compactness → finite subcover, no gaps
Cell E: $μ^{*} (A \cup B) < μ^{*} (A) + μ^{*} (B)$ kyun ho sakta hai, aur exact value compute karne ke liye kaun sa fact kaam aata hai? ::: overlap sub-additive bound mein double-count karta hai (P2); inclusion–exclusion (measurability chahiye, P4)
Cell F: Cantor set ka measure aur cardinality? ::: measure $0$ , phir bhi uncountable
Cell H: $μ ([0, 1] ∖ C)$ , aur use split karne ke liye kaun si property kaam aati hai? ::: $1$ (full measure); countable additivity (P4)
Cell I: Vitali set unmeasurable kyun hai, kaun si do properties clash karti hain, aur ise kisne banaya? ::: additivity (P4) + translation invariance (P5) sum ko $0$ ya $\infty$ force karte hain, kabhi $[1, 3]$ mein nahi; Axiom of Choice ne ise non-constructively banaya