Question

(1) Let (, A, /i) be a measure space = {AnE: A E A} is a o-algebra of E, contained in (a) Fix E E A. Prove that AE A. (b) Let be the restriction of u to AE. Prove that uE is a measure on Ag (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) (d) Suppose that f is integrable on E. Prove that g diE, ANE ANE for any A C A. (You must show that f and g are integrable on AnE.)

Answer 1

(a) 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 $B^c=E-B=E-(A\cap E)=E\cap (A\cap E)^c=E\cap (A^c\cup E^c)\in \mathcal A_E,$ as $(A^c\cup E^c)\in \mathcal A$ .

iii) Take any countable family $\{B_i\}\subset \mathcal A_E$ , then for all i, there exists
A, E A
, be such that $B_i=A_i\cap E=> \bigcup_{i\in \mathbb N} B_i=E\cap ( \bigcup_{i\in \mathbb N} A_i)\in \mathcal A_E$

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

b) 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 $\mu_E(\phi)=\mu(\phi)=0$

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

Note that is measurable implies $f^{-1}(U)\in \mathcal A$ , then $g^{-1}(U)=f^{-1}(U)\cap E\in \mathcal A_E$ , hence measurable.

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 $\lim_{i\to \infty }\int_{E\cap A}f_id\mu<\infty$ 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 $\int_{E\cap A}f|_{E\cap A}d\mu=\int _{E\cap A}gd\mu_E$ , hence done.

Feel free to comment if you have any doubts. Cheers!