1. ## 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

2. ## 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 $k \to \infty$?

3. ## Re: Real Analusis

Dear SlipEternal,

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

Kind regards

4. ## 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 $|f(x)-\chi_{c_n}(x)|>1$.

5. ## Re: Real Analysis

any one can complete the solution

6. ## Re: Real Analysis

Originally Posted by saed
any one can complete the solution
Hopefully you can complete the solution. What are you having trouble with?

7. ## Re: Real Analysis

I need the full solution

8. ## Re: Real Analysis

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?