Definition. A function $f:\Bbb R\to\Bbb R$ will be called potentially continuous if there is a bijection $\phi:\Bbb R\to\Bbb R$ such that $f\circ \phi$ is continuous.

So one could say a potentially continuous (p.c.) function is "a continuous function with a mixed up domain". I was wondering whether there is an easy way to characterize such functions.

Some thoughts $\DeclareMathOperator{\im}{im}$

  • If the image $\im(f)$ is not connected (i.e. no interval), then $f$ is not p.c. because even mixing the domain cannot make $f$ satisfy the intermediate value theorem.
  • Bijective functions are always p.c. because we can choose $\phi=f^{-1}$. Every injective function with an open connected image is p.c. for a similar reason. However, only having a connected image is not enough, as e.g. there are bijections, but no continuous bijections $f:\Bbb R \to [0,1]$.
  • Initially I thought a function can never be p.c. if it attains every value (or at least uncountably many values) uncountably often, e.g. like Conways base 13 function. But then I discovered this: take a Peano curve like function $c$ (or any other continuous surjection $\Bbb R\to\Bbb R^2$) and only look at the $x$-component $c_x:\Bbb R\to\Bbb R$. This is a continuous function which attains every value uncountably often.
  • The question can also be asked this way. Given a family of pairs $(r_i,\kappa_i),i\in I$ of real numbers $r_i$ and cardinal numbers $\kappa_i\le\mathfrak c$ so that $\{r_i\mid i\in I\}$ is connected. Can we find a continuous function $f:\Bbb R\to\Bbb R$ with $|f^{-1}(r_i)|=\kappa_i$?
  • There is no continuous function which attains each real number exactly once except zero which is attained twice. So, e.g. the function $$f(x)=\begin{cases}x-1&\text{for $x\in\Bbb N$}\\x&\text{otherwise}\end{cases}$$ is not p.c., even though its image is all of $\Bbb R$.
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    $\begingroup$ (+1) Interesting question. My bet is that any function with a connected range is potentially continuous. $\endgroup$ Oct 20, 2017 at 21:43
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    $\begingroup$ @JackD'Aurizio I am pretty sure I found a counter-example. See the last item above. $\endgroup$
    – M. Winter
    Oct 21, 2017 at 0:16
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    $\begingroup$ If $f^{-1}(0) = \{a,b\}$, and $f$ injective on $(a,b)$, then $f$ is strictly increasing or decreasing on $(a,b)$, so $f(a) \ne f(b)$, contradiction. $\endgroup$
    – orangeskid
    Oct 21, 2017 at 0:41
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    $\begingroup$ Is every Darboux function potentially continuous ? $\endgroup$ Dec 18, 2017 at 15:50
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    $\begingroup$ @Thomas Andrews, f and g is potentially continuous doesn't result in potentially continuous f+g. Let f(x)=x and g(x)=-x when x is irrational and g(x)=-x+1 when x is rational number. Both f and g is potentially continous since they're bijection but f+g is 0 and 1 respectively when x is irrational or rational number so it is not potentially continuous. $\endgroup$
    – Zhaohui Du
    Aug 25, 2019 at 10:43

1 Answer 1


This should only be taken as a partial answer, since the results quoted are all "mod null", i.e. statements should only be understood to be true up to a set of measure zero. Probably a reader equipped with more descriptive set theory expertise would than me would know whether this can be upgraded to pointwise statements.

It is a result quoted, for example, in Brenier's "Polar Factorization and Monotone Rearrangement of Vector-Valued Functions", that if $(X,\mu)$ is a probability space (e.g a bounded domain in $\mathbb{R}^n$ with normalized Lebesgue measure), and $u\in L^p(X,\mu)$, then there exists a unique nondecreasing rearrangement $u^* \in L^p(0,1)$. Moreover, there exists a measure-preserving map $s$ from $(0,1)$ to $(X,\mu)$ such that $u\circ s = u^*$. In the case where X is understood to be a bounded subset of $\mathbb{R}^n$, this can be thought of as a rearrangement of the domain. (Brenier actually quotes the opposite result, but the paper he cites, by Ryff, gives both directions.)

By modifying $u^*$ on a null set, we can take $u^*$ to be lower-semicontinuous. In this case, it is apparent that $u^*$ is continuous iff it has no jump discontinuities.

In the case where $n>1$ one can also get some mileage out of Sobolev space theory. It follows from Theorem 0.1 of "A Co-area Formula with Applications to Monotone Rearrangement and to Regularity" by Rakotoson and Temam that if $u\in W^{1,p}(\Omega)$, then $u^* \in W^{1,p} (0,1)$. In particular, (up to choice of pointwise representative) $u^*$ is absolutely continuous, and therefore continuous. (This is not interesting in the case where $n=1$ since we would already be assuming that u is continuous.)

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    $\begingroup$ I offered a bounty but now i realize I am expected to evaluate all the answers to decide the award . I will likely need help. I will welcome any input on whether some answer deserves the bounty. I have not yet studied this A. $\endgroup$ Oct 10, 2019 at 4:50
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    $\begingroup$ This may be an answer to a better question than the one actually asked! As the question is stated, $\phi$ can be an arbitrary bijection without anything like measure-preserving properties. All $\phi$ is guaranteed to preserve is cardinality (of fibres, for example). $\endgroup$
    – HTFB
    Oct 13, 2019 at 21:28
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    $\begingroup$ This answer does not seem to shed any light on the original question. $\endgroup$ Oct 14, 2019 at 16:58

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