2
$\begingroup$

Let $\{r_n\}_{n=1}^\infty$ be enumeration of rational numbers, define $$f(x)=\sum_{\{ n \ / \ r_n<x \}} \frac{1}{2^n}$$

$(i)$ $f$ is continuous at irrational points

$(ii)$ $\lim_{x\rightarrow a^-} f(x) = f(a)$ $\qquad$ (left continuity)

$(iii)$ $\lim_{x\rightarrow a^+} f(x) = \sum_{\{ n \ / \ r_n \leq a \}} \frac{1}{2^n}$

$(iv)$ what is $\int_0^1 f$

Just managed to show that $f$ is discontinuous at rational points. I guess that some of the above points are similar to each other, just don't know how to deal with this kind of problems. Thank you.

$\endgroup$
1
$\begingroup$

Hints:

  • (ii) and (iii) together imply (i) and that $f$ is discontinuous at rational points.

  • (ii) and (iii) are similar: Think of the set $\{ r_n : r_n <a\}$ which defines $f(a)$. What is the "limit" of $\{r_n : r_n <a^-\}$ when $a^-\to a$ (from the left) and $\{r_n : r_n <a^+\}$ when $a^+\to a$ (from the right)?

  • For (iv). Try to write $f$ as a (increasing) limit of $f_n$, where $$f_n(x) = \sum_{\{k\le n| r_k < x\}} \frac{1}{2^k}.$$ Calculate $\int_0^1 f_n(x) dx$ and then take $n\to \infty$. (Remark: we have that $f_n \to f$ uniformly, so $\lim_{n\to \infty} \int_0^1 f_n = \int_0^1 f$).

Further hints: I would say a bit more on how to tackle (ii) and (iii) as they are not that straight forward.

Take (iii) as an example (similar for (ii)). We want to show

$$ \lim_{x\to a^+} f(x) = \sum _{\{n| r_n \le a\}} \frac{1}{2^n}.$$

That $\ge $ is obvious from the definition of $f$. We want to show that $>$ is not possible. For the sake of contradiction, assume $>$ holds. Then

$$f(x) >\delta + \sum _{\{n| r_n \le a\}} \frac{1}{2^n}$$

for all $x>a$ and for some $\delta >0$. But note that

$$b_n :=\sum_{ k\ge n} \frac{1}{2^k} \to 0$$

as $n\to 0$, so there is $K$ so that

$$\sum_{n=K}^\infty \frac{1}{2^n} <\delta.$$

So what can we say about $f(x)$ if $x$ is closed to $a$ so that $r_1, \cdots, r_K \notin (a, x)$?

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thank you. The first hint is completely clear. For the second one it is clear heuristically, if we approach from the right, i.e. $a^+\rightarrow a$ then at the limit we are going to include the point $a$ and get $\{r_n : r_n \leq a\}$ , when approaching from left we would get the same set $\{r_n : r_n < a\}$... but don't have any idea how to write the proof rigorously. $\endgroup$ – user16015 Sep 24 '15 at 7:01
  • $\begingroup$ Try to do the following: For (iii), try to show the equality by showing (a) $f(x) \ge RHS$ for all $x>a$ and (b) that $>$ is not true by arguing by contradiction. @user16015 $\endgroup$ – user99914 Sep 24 '15 at 7:23
  • $\begingroup$ I have to say that this is not that straight forward. I will try to write down more hints. $\endgroup$ – user99914 Sep 24 '15 at 7:27
0
$\begingroup$

Hint for (ii): For $x$ sufficiently close to $a$ (but $<a$) we can ensure that none of the first $N$ rationals appears in $[x,a)$. Hence $f(a)-f(x)$ can be bounded from above by $\sum_{n>N}2^{-n}$.

This can be adapted for (iii).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.