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Define $f:[0,1]\to \mathbb{R}$, $f(x):=0$ if $x\notin \mathbb{Q}$, $f(p/q):=1/q$, $q>0$, $p, q$ coprime integers.

Prove that $f$ is regulated.

A function $f:[a,b]\to\Bbb R$ is a regulated function if $\forall \varepsilon>0$ there is a step function $\varphi:[a,b]\to\Bbb R$ such that $sup_{x\in[a,b]}|f(x)-\varphi(x)| <\varepsilon$.

So far I've worked out that $f$ is continuous at all irrational values of $x$ and discontinuous at all rational values of $x$ but I'm unsure what to do next.

Edit: I've also worked out that because $f=0$ or $f=1/q$ and $0<1/q\leqslant 1$ then $0\leqslant f \leqslant 1$. I'm not sure if this helps?

  • If we allow the indicator function of singleton as step function, proving this problem is easy. But if we does not allow it as step function, then I think $f$ is not regulated. – Hanul Jeon Oct 25 '14 at 10:18
  • If I'm being stupid please excuse me, but what is the indicator function of singleton? –  Oct 25 '14 at 10:25
  • I intend a characteristic function of set only has one element. Wikipedia article may helpful for you. – Hanul Jeon Oct 25 '14 at 10:27
  • The indicator function $\psi_{{a}}$ of the singleton (i.e. set with a single element) ${a}$ is given by $$\psi_{{a}}(x) =\cases{1&if $x=a$ (actually if $x\in {a}$)\0&otherwise} $$ – Arthur Oct 25 '14 at 10:32
  • What set would I use for $a$ though? The obvious choice is for it to be either the rationals or irrationals but in either case $\psi_{a}(x)$ wouldn't be a step function so it wouldn't work. –  Oct 25 '14 at 10:40

1 Answers1

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A function $f: [a,b]\to \mathbb{R}$ is regulated if and only if it has RHS limit at $x\in[a,b)$ and LHS limit at $(a,b]$.

Hint: ($\impliedby$) $\forall \epsilon>0$, Let $A=\{x\in [a,b]\,|\exists \, \text{step function}\, s: [a,b]\to \mathbb{R}\,\text{such that}\, |f(z)-s(z)| <\epsilon, \forall z\in[a,x]\}$.

Try to show $A=[a,b]$.

Thomae's function has limit $0$ at all real number ( which is mentioned here), so it's regulated.

But here I also allow step function of degenerate interval.

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  • How would I prove that it has limit $0$ at all real numbers though? Is it because there is infinitely many irrationals on the interval $[0,1]$ so there's infinitely many points where $f(x)=0$? –  Oct 25 '14 at 10:51
  • @john.smith The proof is quite similar to show it's continuous at irrational number. I have updated a link in my answer. – John Oct 25 '14 at 10:56