Dear All,

I tried to solve this problem many times without success, I could be grateful if any one helps me to solve it

Attachment 30624

Printable View

- Apr 7th 2014, 11:08 AMsaedReal Analysis
Dear All,

I tried to solve this problem many times without success, I could be grateful if any one helps me to solve it

Attachment 30624 - Apr 7th 2014, 11:27 AMSlipEternalRe: Real Analusis
What have you tried? It looks like (a) is fairly straightforward. In fact, by the result of part (b), it appears that you should be able to show that $|f| \le 1$, $\mu$-a.e. How do you show that something is true $\mu$-a.e.? Start with "Given $\varepsilon>0$, ..." and end with the $\mu$-measure of the set on which it is not true is less than $\varepsilon$.

Hint: let $\displaystyle A_k = \bigcup_{n \ge k} c_n$. What can you say about $\displaystyle \int \left|f-\chi_{A_k}\right|d\mu$ as $k \to \infty$? - Apr 7th 2014, 11:05 PMsaedRe: Real Analusis
Dear SlipEternal,

my ideas are disrupted, could you give me full details.

Kind regards - Apr 8th 2014, 05:05 AMSlipEternalRe: Real Analysis
Let $U = \{x \in X \mid |f(x)|>2 \}$. Then

$\displaystyle \int_X|f-\chi_{c_n}|d\mu = \int_U|f-\chi_{c_n}|d\mu + \int_{X\setminus U}|f-\chi_{c_n}|d\mu$

If $|f(x)|>2$, then $|f(x)-\chi_{c_n}(x)|>1$. - Apr 9th 2014, 09:14 AMsaedRe: Real Analysis
any one can complete the solution

- Apr 9th 2014, 11:04 AMSlipEternalRe: Real Analysis
- Apr 11th 2014, 08:58 AMsaedRe: Real Analysis
I need the full solution

- Apr 11th 2014, 12:49 PMSlipEternalRe: Real Analysis