1. ## about the riemann function

proof the reimann function (if x is rational then f(x)=1/q, if x is irrational
then f(X)=0)is not a step function,and for any essilope, construct a step
function k:[0,1] ~>R s.t. ||f-k||=sup{|f(t)-k(t)|:t is
in[0,1]}<essilope,then show f is regulated

2. ## Re: about the riemann function

Have you shown that it's not a step function yet? Start with the fact that a step function is constant on an interval.

For the second part, you just make the intervals small enough, and then if I remember the definition of regulated correctly, the third one follows easily.

By essilope, I assume you mean epsilon. I've never heard it called that before.

- Hollywood

3. ## Re: about the riemann function

About the second part, if i make the intervals small enough . how should i choose the value of the step function to satisfy the conditon|f-k|<epsilon?,cause every irrational member is between 2 rational number, and so as rational number, there are infinitely many intervals here, could u give me some details, many thanks.

4. ## Re: about the riemann function

Hmmmm... I think I read the question wrong earlier.

If the intervals are required to have positive length, I think there's a problem. Any interval containing, say, 1/2 will also contain an irrational number, so whatever you choose for k(x) in that interval, |f-x| is going to be greater than or equal to 1/4.

For any given epsilon, the function is only greater than epsilon for finitely many x. So maybe you can have a zero length interval at each of those values and set f=0 for the intervals between them.

Does that work?

- Hollywood

5. ## Re: about the riemann function

i don't think so ,if f=o then sup|f-k|=1>1/2 not for every esillope, the hints for this part is suggestion q<1/esillope, but i dont know how to use it

6. ## Re: about the riemann function

What I mean is, for example if $\epsilon=0.3$, set:

$k(x)= \left\{\begin{array}{ll}1 & : x=0\\0 & : x \in (0,\frac{1}{3})\\\frac{1}{3} & : x=\frac{1}{3}\\0 & : x \in (\frac{1}{3},\frac{1}{2})\\\frac{1}{2} & : x=\frac{1}{2}\\0 & : x \in (\frac{1}{2},\frac{2}{3})\\\frac{1}{3} & : x=\frac{2}{3}\\0 & : x \in (\frac{2}{3},1)\\1 & : x=1\end{array}\right.$

In the five places where $f(x)\ge{\epsilon}$, I have set $k(x)=f(x)$ so $|f(x)-k(x)|=0<\epsilon$. Everywhere else, $k(x)=0$, so $|f(x)-k(x)|=|f(x)|<\epsilon$.

- Hollywood

7. ## Re: about the riemann function

i think thats right , but i faid to build up the general case,in terms of esillope,could u help me? i mean for a random any esillope.

8. ## Re: about the riemann function

The function is just

$k(x)= \left\{\begin{array}{ll}f(x) & : f(x)\ge\epsilon\\0 & : f(x)<\epsilon\end{array}\right.$

The trick is in showing that it's a step function - you need to show that $f(x)\ge\epsilon$ for only finitely many x, and that this means that k(x) fits the definition of a step function. The inequality $|f(x)-k(x)|<\epsilon$ is of course obvious from the construction.

By the way, it seems that "essilope" is not common usage in the UK either. As far as I can tell, you are the only person in the world that uses the term.

- Hollywood

9. ## Re: about the riemann function

i get wrong with the symbolic here, but many thanks 4 ur help!!!!