# Real Analysis

• Apr 7th 2014, 10:08 AM
saed
Real 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, 10:27 AM
SlipEternal
Re: 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, 10:05 PM
saed
Re: Real Analusis
Dear SlipEternal,

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

Kind regards
• Apr 8th 2014, 04:05 AM
SlipEternal
Re: 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, 08:14 AM
saed
Re: Real Analysis
any one can complete the solution
• Apr 9th 2014, 10:04 AM
SlipEternal
Re: Real Analysis
Quote:

Originally Posted by saed
any one can complete the solution

Hopefully you can complete the solution. What are you having trouble with?
• Apr 11th 2014, 07:58 AM
saed
Re: Real Analysis
I need the full solution
• Apr 11th 2014, 11:49 AM
SlipEternal
Re: Real Analysis
Quote:

Originally Posted by saed
I need the full solution

As I asked above, what are you having trouble with? The full solution to a problem like that takes time and effort. If you are not willing to put in the time and effort to describe what you are having trouble with, why would anyone on this forum help you at all?