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Prove that there exists a monotone nondecreasing function $f:[0,1] \rightarrow \mathbb{R}$ discontinuous in rationals of $[0,1]$

  • I tried to test functions of the type $$ f(x)=\begin{cases} \frac{1}{q} & \text{if } x=\frac{p}{q} \mid p,q \in \mathbb{Z}^+ \,\,\, \text{and} \,\,\, \gcd(p,q)=1 \\ x & \text{if } x \not\in \mathbb{Q} \end{cases}$$ which is discontinuous in all rationals, but is not monotone.
Ángel Mario Gallegos
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sebastian azocar
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    possible duplicate http://math.stackexchange.com/questions/296195/example-of-an-increasing-integrable-function-f0-1-to-mathbbr-which-is-di?rq=1 and http://math.stackexchange.com/questions/69317/construct-a-monotone-function-which-has-countably-many-discontinuities/69321#69321 – Marco Cantarini Feb 23 '15 at 21:19

1 Answers1

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Since $\mathbb{Q}\cap[0,1]$ is countably infinite, then there's a bijection: $$\phi:\mathbb{N}\to\mathbb{Q}\cap[0,1]$$

Now define $$f:[0,1]\to\mathbb{R},\quad f(x)=\sum_{\phi(n)\leq x}2^{-n}$$

Since the geometric series is absolutely convergent, thus rearranging terms does not change the sum, so $f$ is well defined. Clearly $f$ is non decreasing, and $$\lim_{x\to \phi(n)^-}f(x)=\sum_{\phi(k)<\phi(n)}2^{-n}<f(\phi(n))$$

Closure
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  • @Mathemagician1234 I am also curious about other examples. Will you be interested in sharing yours? – Closure Feb 23 '15 at 21:31
  • Yup,working on it now,don't type quick in Jax. Still not good at it. – Mathemagician1234 Feb 23 '15 at 21:36
  • I'm considering deleting the example since for some reason,people didn't approve. IS it wrong? Someone give me a constructive critique and not just downvote it-if it's legitimately in error,I want to know and why. If not,stop downvoting out of spite. – Mathemagician1234 Feb 24 '15 at 02:24