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

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- April 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 - April 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 $k \to \infty$? - April 7th 2014, 11:05 PMsaedRe: Real Analusis
Dear SlipEternal,

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

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

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

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