Exercises — Measure theory — Lebesgue measure (intro)

(a) TRUE. (B3) se, ${\mu }^{\ast }([a,b])=b-a$, toh $5-2=3$. (b) TRUE. Empty set ko kisi bhi $\epsilon >0$ ke liye length $\epsilon$ ke interval se cover kiya ja sakta hai, toh ${\mu }^{\ast }(\varnothing )\le \epsilon \to 0$, aur outer measure kabhi negative nahi hota. (c) FALSE. Har $\ell (I_n)=b-a\ge 0$ hai, toh har cover-sum $\ge 0$ hai, toh infimum $\ge 0$ hai. Outer measure hamesha $\ge 0$ hota hai. (d) FALSE. (Tool A) ke mutabik, ${\mu }^{\ast }$ ka poora point (aur strength) yahi hai ki yeh har subset $A\subseteq \mathbb{R}$ ke liye defined hai.

${\mu }^{\ast }(A)={\mu }^{\ast }(A\cap E)+{\mu }^{\ast }(A\cap E^c).$ Simple words mein: $E$ har set $A$ ko cleanly ek inside-piece $A\cap E$ aur ek outside-piece $A\cap E^c$ mein cut karta hai jinke outer measures exactly add hote hain — koi leak nahi (yeh Tool C hai). Sigma-algebra dekho yeh samajhne ke liye ki aise $E$ ka collection complements aur countable unions ke under kyun closed hai.

HUM KYA karte hain: 4 points $p_k$ mein se har ek ko ek tiny interval $(p_k-\frac{\delta }{2},p_k+\frac{\delta }{2})$ se cover karo jiska length $\delta$ ho. KYUN: har point ko sirf ek interval chahiye; (B2) se outer measure zyaada se zyaada total cover length hai, aur finitely many tiny intervals ek aise total dete hain jise hum control kar sakte hain. ${\mu }^{\ast }(\{1,4,9,16\})\le 4\delta \stackrel{\delta \to 0}{\to }0.$ Outer measure $\ge 0$ hai, toh ${\mu }^{\ast }(\{1,4,9,16\})=0$. Kisi bhi finite set ka measure $0$ hota hai.

Key observation: $A$ countable hai (yeh $n$ se index ki gayi ek list hai). Countable vs uncountable dekho. HUM KYA karte hain: $n$-vein point $\frac{1}{n}$ ko $\frac{\epsilon }{2^n}$ length ke interval se cover karo. $\epsilon /2^n$ choice KYUN: lengths tab ek geometric series form karti hain jiska total finite aur controllable hota hai; sub-additivity (B2) se, ${\mu }^{\ast }(A)\le {\sum }_{n=1}^{\infty }\frac{\epsilon }{2^n}=\epsilon \cdot \frac{1/2}{1-1/2}=\epsilon \stackrel{\epsilon \to 0}{\to }0.$ Toh ${\mu }^{\ast }(A)=0$. Yeh tab bhi kaam karta hai jab points $0$ ke paas pile up (accumulate) ho rahe hoon — accumulation length nahi banata.

Figure mein kya dikha hai (neeche): real line jisme points $1,\frac{1}{2},\frac{1}{3},\dots$ (magenta dots) $0$ ki taraf crowd kar rahe hain, har ek ek violet covering interval mein wrapped hai jiska width har step mein half hota hai ($\epsilon /2^n$). Picture yeh visible banati hai ki dots $0$ ke paas pile up ho rahe hain phir bhi, total violet length geometric sum $\epsilon$ hai — arbitrarily small. Isko aise padho: "infinitely many dots, finite (shrinkable) total ink."

Upper bound ("$\le$"): $(-\epsilon ,1+\epsilon )$ aur $(3-\epsilon ,4+\epsilon )$ se cover karo; (B2) se ${\mu }^{\ast }\le (1+2\epsilon )+(1+2\epsilon )=2+4\epsilon \to 2$. Toh ${\mu }^{\ast }\le 2$. Lower bound ("$\ge$") — ${\mu }^{\ast }$ ke terms mein aur split test ke saath sahi se stated: $[0,1]$ aur $[3,4]$ dono intervals hain, isliye measurable (Tool C) hain, aur disjoint hain. Test set $A=[0,1]\cup [3,4]$ par $E=[0,1]$ ka Carathéodory split test apply karo: ${\mu }^{\ast }(A)={\mu }^{\ast }(A\cap E)+{\mu }^{\ast }(A\cap E^c)={\mu }^{\ast }([0,1])+{\mu }^{\ast }([3,4])=1+1=2,$ jahaan dono pieces ko (B3) se evaluate kiya gaya hai. Dhyan do yeh ${\mu }^{\ast }$ ke baare mein statement hai throughout — equality (sirf "$\le$" nahi) exactly wahi hai jo $[0,1]$ ki measurability deti hai; sub-additivity (B2) akele sirf "$\le$" deta. Dono bounds combine karke: ${\mu }^{\ast }([0,1]\cup [3,4])=2$. $■$

Subtlety: cover ka dense hona (uske intervals $[0,1]$ ke har hisse mein reach karte hain) ka matlab yeh nahi ki woh $[0,1]$ ko fill karta hai. Covering intervals overlap karte hain aur gaps chhodte hain, aur unki total length — jo (B2) se ${\mu }^{\ast }$ ko hamesha care karta hai — tiny ho sakti hai chahe unka union topologically spread out ho. Concrete, quantified picture: rationals $q_1,q_2,\dots$ enumerate karo aur $q_n$ ke around $\epsilon /2^n$ length ka interval dalo. Union $U={\bigcup }_n{I}_n$ ek dense open set hai, phir bhi (B2) se ${\mu }^{\ast }(U)\le {\sum }_n\frac{\epsilon }{2^n}=\epsilon .$ $\epsilon =0.01$ lo: $U$ mein $[0,1]$ ka har rational hai, toh uske intervals har sub-interval mein poke karte hain, phir bhi ${\mu }^{\ast }(U)\le 0.01$. $[0,1]$ mein uska complement ek nowhere-dense set (Tool D) hai measure $\ge 0.99$ ke saath — Cantor set ka ek "fat" cousin. Density topology hai; measure size hai; yeh genuinely independent hain.

Figure mein kya dikha hai (neeche): segment $[0,1]$ jisme rationals scatter hue hain (violet), har ek ek magenta interval se capped hai jiska length $\epsilon /2^n$ hai. Chahe magenta line ke har part ke paas appear kare (dense), magenta ka total width $\le \epsilon$ hai; untouched peach gaps positive-measure leftover hain. Isko aise padho: "har jagah reach karta hai, par almost kuch nahi occupy karta."

HUM KYA karte hain: construction ke stage $n$ par, $C$ mein $2^n$ closed intervals hain jinmein se har ek ki length $3^{-n}$ hai. KYUN yeh $C$ ko cover karta hai: middle-third removal sirf points delete karta hai, toh $C$ har stage par surviving intervals ke andar rehta hai. Stage $n$ par total length ((B2) se): ${\mu }^{\ast }(C)\le 2^n\cdot 3^{-n}={\left(\frac{2}{3}\right)}^n\stackrel{n\to \infty }{\to }0.$ Toh ${\mu }^{\ast }(C)=0$. Shocking kyun hai: $C$ uncountable hai (yeh base-3 digits ke zariye infinite $\{0,2\}$-strings se biject karta hai), phir bhi measure $0$ hai. Parent ka Example 3 compare karo: $[0,1]$ uncountable hai aur measure $1$ bhi. Conclusion: "uncountable" aur "positive measure" independent hain — tum ek ko doosre ke bina rakh sakte ho. Cantor set aur Countable vs uncountable dekho.

(a) $n$-vein interval ki length $2\cdot \frac{\epsilon }{2^{n+1}}=\frac{\epsilon }{2^n}$ hai. Countable sub-additivity (B2) se, ${\mu }^{\ast }(U_{\epsilon })\le {\sum }_{n=1}^{\infty }\frac{\epsilon }{2^n}=\epsilon .$ (b) $\epsilon =\frac{1}{2}$ ke saath: ${\mu }^{\ast }(U_{1/2})\le \frac{1}{2}$. Ab $[0,1]=(U_{1/2}\cap [0,1])\cup K$ jahaan $K=[0,1]\setminus U_{1/2}$. (B3) se ${\mu }^{\ast }([0,1])=1$ use karke aur sub-additivity (B2) se, $1={\mu }^{\ast }([0,1])\le {\mu }^{\ast }(U_{1/2}\cap [0,1])+{\mu }^{\ast }(K)\le \frac{1}{2}+{\mu }^{\ast }(K),$ toh ${\mu }^{\ast }(K)\ge \frac{1}{2}$. Yeh nonempty hai (uska measure positive hai). ✔ (c) $U_{1/2}$ open aur dense hai (isme $[0,1]$ ka har rational hai) phir bhi measure $\le \frac{1}{2}$ hai; leftover $K$ nowhere dense hai (Tool D: uske closure mein koi interval nahi, kyunki har interval ek rational se milta hai aur isliye $U_{1/2}$ se milta hai) phir bhi measure $\ge \frac{1}{2}$ hai. Toh "topologically big" (dense open) aur "measure-big" opposite directions mein point kar sakte hain. Yeh Problem 3.1 ke peeche ka rigorous engine hai.

HUM KYA check karein (Tool C): har test set $T$ ke liye, ${\mu }^{\ast }(T)={\mu }^{\ast }(T\cap A)+{\mu }^{\ast }(T\cap A^c).$ Step 1 — "$\le$" free hai sub-additivity (B2) se: ${\mu }^{\ast }(T)={\mu }^{\ast }((T\cap A)\cup (T\cap A^c))\le {\mu }^{\ast }(T\cap A)+{\mu }^{\ast }(T\cap A^c)$. Step 2 — "$\ge$": Monotonicity (B1) se, kyunki $T\cap A\subseteq A$ hai, ${\mu }^{\ast }(T\cap A)\le {\mu }^{\ast }(A)=0$ milta hai, toh ${\mu }^{\ast }(T\cap A)=0$. Aur $T\cap A^c\subseteq T$ se (B1) se ${\mu }^{\ast }(T\cap A^c)\le {\mu }^{\ast }(T)$ milta hai. Add karke: ${\mu }^{\ast }(T\cap A)+{\mu }^{\ast }(T\cap A^c)\le 0+{\mu }^{\ast }(T)={\mu }^{\ast }(T).$ Combine: dono inequalities equality deti hain, toh $A$ measurable hai. Ek subset $B\subseteq A$ ke liye: (B1) se ${\mu }^{\ast }(B)\le {\mu }^{\ast }(A)=0$, toh ${\mu }^{\ast }(B)=0$, aur same argument se $B$ measurable hai. $■$ Payoff: isi liye Cantor set (measure $0$) aur uske saare subsets automatically measurable hain — jab length hi nahi hai toh koi leak possible nahi.

Blocker: Vitali set $V$. Ise Axiom of Choice use karke banao: $[0,1]$ ko equivalence relation $x\sim y⟺x-y\in \mathbb{Q}$ se partition karo, aur har class se ek representative choose karo — yeh choice Axiom of Choice chahti hai — $V$ banane ke liye. Contradiction: $[-1,1]$ ke rationals ko $q_1,q_2,\dots$ enumerate karo aur translates $V_k=V+q_k$ banao. Teen facts:

1. $V_k$ pairwise disjoint hain, toh unka union ek genuine ${⨆}_k{V}_k$ hai (Tool C ka symbol): do representatives jo ek rational se differ karein woh same class mein hote, contradiction.
2. $[0,1]\subseteq {⨆}_k{V}_k\subseteq [-1,2]$.
3. Translation invariance — woh wishlist property ki "ek set ko slide karna uski length nahi badalta", $\mu (A+t)=\mu (A)$ — har $k$ ke liye $\mu (V_k)=\mu (V)=:c$ force karti, kyunki har $V_k$, $V$ ko $q_k$ se slide karna hai. Ab countable additivity (disjoint-union property $\mu ({⨆}_k{E}_k)={\sum }_k\mu (E_k)$) ko (B3) ke saath apply karo: $1=\mu ([0,1])\le {\sum }_{k=1}^{\infty }c\le \mu ([-1,2])=3.$

• Agar $c=0$: middle sum $0$ hai, $\ge 1$ ka contradiction.
• Agar $c>0$: middle sum $+\infty$ hai, $\le 3$ ka contradiction. $c$ ki koi bhi value kaam nahi karti. Toh $V$ ko translation invariance aur countable additivity dono ke saath ek consistent measure assign nahi kiya ja sakta. Isliye har set measurable nahi hai — humein $\sigma$-algebra $\mathcal{M}$ tak restrict karna padega.

Figure mein kya dikha hai (neeche): interval $[-1,2]$ jisme $[0,1]$ marked hai (dashed), aur kaafi coloured rows of dots — har row same "seed" set $V$ ka ek translate $V+q_k$ hai, ek rational se left/right shift kiya gaya. Rows disjoint hain (koi dot repeat nahi hota) phir bhi saath mein $[0,1]$ ko blanket karte hain. Caption line trap read off karti hai: identical-measure disjoint copies ya toh $0$ ya $\infty$ tak sum hote hain, kabhi $1$ aur $3$ ke beech fit nahi hote.

Nahi — diagnosis zyaada subtle hai. Sub-additivity (B2) kehti hai ${\mu }^{\ast }\left({\bigcup }_k{V}_k\right)\le {\sum }_k{\mu }^{\ast }(V_k)$, ek inequality jo hamesha hold karti hai aur akele koi contradiction nahi banati. Problem yeh hai ki additivity fail hoti hai: disjoint $V_k$ ke liye hum yeh nahi rakh sakte ${\mu }^{\ast }\left({⨆}_k{V}_k\right)={\sum }_k{\mu }^{\ast }(V_k)$ translation invariance ke saath. Equality (additivity) ka loss — koi single inequality "galat" hona nahi — yahi fatal hai. Equivalently: $V$ Carathéodory ka exact-split test (Tool C) fail karta hai — yeh measure "leak" karta hai. Woh leak outer measure ke liye permitted hai (sub-additive hi woh sabse zyaada tha jo humne saare sets par promise kiya, (B2) se); measurable sets par yeh forbidden hai, aur exactly yahi wajah hai ki $V\notin \mathcal{M}$.

Kisi bhi finite ya countable set ka outer measure hota hai ::: $0$ ($n$-vein point ko $\epsilon /2^n$ length se cover karo; total $\le \epsilon \to 0$ (B2) se). Ek set uncountable ho sakta hai aur phir bhi measure zero — example ::: Cantor set, ${\mu }^{\ast }=\lim (2/3{)}^n=0$. "Dense" aur "positive measure" hain ::: independent — ek open dense set ka measure $\le \epsilon$ ho sakta hai. Agar ${\mu }^{\ast }(A)=0$ toh $A$ hai ::: measurable, aur $A$ ka har subset bhi (monotonicity (B1) split force karta hai). Woh object jo prove karta hai ki har set measurable nahi hota ::: Vitali set, Axiom of Choice se banaya; yeh translation invariance + countable additivity todta hai.