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(1) Let (, A, i) be a measure space. {AnE: Ae A} is a o-algebra of E, contained in (a) Fix E E A. Prove that Ap = A. (b) Let

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IF YOU HAVE ANY DOUBTS COMMENT BELOW I WILL BE TTHERE TO HELP YOU..ALL THE BEST..

(A)  Fix E E A. Prove that Ap = {AnE: A E A} is a a-algebra of E, contained in A.

SOL::-

Recall the definition of a \sigma-algebra, \mathcal A is said to be an \sigma-algebra of \Omega, if

i) EA

ii) \mathcal A is closed under complement.

iii) \mathcal A is closed under countable union.

Note that i) E E A=> E E A

ii) Let ВЕАВ Е, then to show BC E A . Note that by def there exists A E A, such that B=E\cap A. Now since \mathcal A is an sigma algebra, implies E, AE A C , now Bc E B E-(An E)= En (ANE) = En (Ac U E°) E A. as (Ac UE) E A .

iii) Take any countable family с Ар { В Б , then for all i, there exists A, E A, be such that Bi AnEUB En (JA) e AB iEN iEN

Hence \mathcal A_E is a \sigma-algebra of E.

B)  Let be the restriction of u to AE. Prove that E is a measure on AE

SOL::-

Recall \mu is said to be a measure on a \sigma-algebra \mathcal A, if

i) \mu(A)\ge 0, for all A E A, ii) \mu(\phi)=0, iii) \mu(\bigcup_{i\in \mathbb Z} A_i)= \sum_{i \in \mathbb Z}}\mu(A_i) , for pairwise disjoint A_i's

Note that A\in \mathcal A_E=> A\in \mathcal A=> \mu_E(A)=\mu(A)\ge 0

Also note that = 0 LE

and \mu_E(\bigcup_i A_i)=\mu(\bigcup_i A_i)=\sum_i\mu(A_i)=\sum_i \mu_E(A_i) , hence a measure.  

C) Suppose that f -> R* is measurable (with respect to A). Let g = f|e be the restriction of f to E. Prove that g: E -> R is measurable (with respect to AE

SOL::-

Note that it is enough to show that inverse image of open set U\subset \mathbb R , is in the sigma algebra.

Note that f is measurable implies f^{-1}(U)\in \mathcal A , then ΤU)- U uEEAΕ -1 (U) Λ ΕΕ A. g , hence measurable.

D) Suppose that f is integrable on E. Prove that f du g duE AnE AnE AnE for any A CA. (You must show that f and g are integrable on A intersection E

d) Note that since f is integrable there exists simple functions \{f_i\} such that \int_Ef_id\mu\to \int_Efd\mu uniformely.

Now for simple function \int_{E\cap A}f_id\mu\le \int_{E\cap A}f_id\mu , gives us fiduo lim io0JEnA thus \int _{E\cap A} f d\mu =\lim_{i\to \infty }\int_{E\cap A}f_id\mu<\infty exists and hence f is integrable on E\cap A. And now f|_{E\cap A}=g , gives us g is integrable as so is the function f|_{E\cap A}, and integration both side we get gdHE EnA fEnadu EnA

I HOPE YOU UNDERSTAND..

PLS RATE THUMBS UP..ITS HELPS ME ALOT..

THANK YOU...!!

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