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Is there any

  • measurable
  • continuous
  • differentiable
  • analytic

surjective function $f:\mathbb{R}\to\mathbb{C}$?

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You may want to investigate the properties of a space-filling curve. – Haskell Curry Jan 29 '13 at 23:18
Yes, yes, no, no: There are continuous space-filling curves (usually called Peano curves). But no space-filling curve can be differentiable. – Andrés E. Caicedo Jan 29 '13 at 23:20
I think you will find relevant this question on MO, and the nice links provided there:… – Andrés E. Caicedo Apr 6 '13 at 15:26
Added the (set-theory) tag in light of the connections mentioned in my answer. – Andrés E. Caicedo Jun 29 '13 at 5:13
up vote 2 down vote accepted

Just collecting comments. We have the answers to your questions are yes, yes, no, no. See the wiki article on space-filling curves to answer the first two questions. A $C^1$ curve has to have measure zero by Sard's theorem for instance, for a more elementary argument see here. As a final note we can come very close to analytic. In particular if we compose the Lebesgue-Cantor function with a space filling curve, then we have a space-filling curve which is constant on the complement on the Cantor set and in particular analytic almost everywhere.

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nice‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌. – user59671 Feb 21 '13 at 16:13

I recently learned the following related result, which I think deserves some publicity: Say that $f:\mathbb R\to\mathbb R^2$ is a Peano function iff it is surjective. As pointed out in the other answer, and the links in the comments, no Peano function is everywhere differentiable.

In 1984, Morayne proved that the continuum hypothesis is equivalent to the existence of a Peano function $f(x) = (f_1(x),f_2(x))$ such that, for every $x\in\mathbb R$, at least one of $f_1'(x)$ and $f_2'(x)$ exists. The reference is

Michał Morayne. On differentiability of Peano type functions, Colloq. Math. 48 (2) (1984), 261–264. MR0758535 (86i:26008).

In the same reference, he proves a related negative result, showing that such functions are necessarily "pathological": Suppose that $f_1,f_2:\mathbb R\to \mathbb R$, let $f=(f_1,f_2)$, and assume that $f_1$ is a Lebesgue measurable function, and $f(\mathbb R)$ is a Lebesgue measurable subset of the plane. If for each $x\in\mathbb R$ at least one of the derivatives $f_1'(x), f_2'(x)$ exists, then $f(\mathbb R)$ has measure zero, so in particular, it cannot be the whole plane.

For an additional set theoretic connection, in recent work, Asger Törnquist and William Weiss have shown that the existence of a Peano function $f(x) = (f_1(x),f_2(x))$ such that, for every $x\in\mathbb R$, at least one of $f_1'(x)$ and $f_2'(x)$ exists, and the functions $f_1,f_2$ are $\Sigma^1_2$, is equivalent to the assertion that $\mathbb R$ is contained in $L$, Gödel's constructible universe.

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