# A curve whose image has positive measure

It is well-known that there are continuous curves $f:I \to \mathbb R^2$ (where $I \subset \mathbb R$ is an interval) whose image have positive measure (e.g Peano curve). I have read somewhere that if we require the curve to be differentiable evrywhere then this cannot happen; but if we require it to be almost everywhere differentiable, then it can happen! How could one proceed to prove the first statement and give a counterexample for the second?

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There is no discrepancy because there measure on $I$ is not related to the measure on $\mathbb{R}^2$ for the purposes of this problem. –  Trevor Wilson Sep 12 '12 at 2:58
My first thought is that the standard Peano and Hilbert space-filling curves appear to be differentiable outside a countable set, being the limits of curves with only finitely many points of non-differentiability. –  Kevin Carlson Sep 12 '12 at 3:00
Yes I did not say there was a disreptancy, I just found the second statement counterintuitive. –  Bernard Sep 12 '12 at 3:01
@KevinCarlson: The (graphs of the) standard nowhere differentiable functions like the blancmange or Weierstrass functions (restricted to a bounded interval) are also uniform limits of curves with only finitely many points of non-differentiability. –  t.b. Sep 12 '12 at 3:03
@t.b. indeed. So much for the easy route. –  Kevin Carlson Sep 12 '12 at 3:15

Let $\phi : [0, 1] \to [0, 1]$ be the Cantor-Lebesgue function, and $\alpha : [0, 1] \to \Bbb{R}^n$ be a space-filling curve.

Since $\phi$ is stationary outside the Cantor set, it is locally constant almost everywhere. That is, $\beta := \alpha \circ \phi$ is also locally constant everywhere, allowing it to be differentiable a.e.. (In fact, $\beta' = 0$ a.e.!) On the other hand, $\phi$ is continuous and surjective. Thus $\beta$ is also a continuous path and the image of $\beta$ coincides with $\alpha$. That is, $\beta$ is also a space-filling curve. Therefore this serves as a counter-example.

Let $f : [0, 1] \to \Bbb{R}^n$ be a differentiable curve in $\Bbb{R}^n$ ($n \geq 2$). Let $\Gamma = f([0, 1])$ be the image of $f$ in $\Bbb{R}^n$.

Assuming that $|f'|$ is Lebesgue integrable, we can prove the first statement.

Theorem. [7.21, Rudin] If $f : [a, b] \to \Bbb{R}$ is everywhere differentiable and its derivative $f'$ is Lebesgue integrable, then $$f(b) - f(a) = \int_{a}^{b} f'(t) \; dt.$$

This theorem immediately implies the following corollary:

Corollary. Let $f : [0, 1] \to \Bbb{R}^n$ be everywhere differentiable and its derivative $f'$ is Lebesgue integrable. Then for any $\epsilon > 0$, then there exists a partition $\Pi = \{ 0 = x_0 < \cdots < x_N = 1 \}$ such that for $$\epsilon_k = \sup \{|f(x) - f(y)| : x, y \in [x_{k-1}, x_k] \}, \quad (1 \leq k \leq N)$$ we have $$\epsilon_k \leq \epsilon \quad \text{and} \quad \sum_{k=1}^{N} \epsilon_k \leq \| f' \|_1.$$

Proof. Since $f'$ is Lebesgue integrable, it is absolutely continuous. Thus there exists $\delta > 0$ such that whenever a measurable subset $E \subset [0, 1]$ satisfies $|E| < \delta$, we have $\int_E |f'| < \epsilon$. Now let $\Pi = \{x_k\}$ be a partition of $[0, 1]$ into subintervals with length less than $\delta$. Then for each $x_{k-1} \leq x < y \leq x_k$, $$|f(y) - f(x)| = \left| \int_{x}^{y} f' \right| \leq \int_{x}^{y} |f'| \leq \int_{x_{k-1}}^{x_k} |f'| < \epsilon.$$ Thus $$\epsilon_k \leq \int_{x_{k-1}}^{x_k} |f'| < \epsilon$$ and the conclusion follows. ////

Remark. If $f'$ is bounded, then it is Lebesgue integrable. Also, in this case, the conclusion of the Corollary follows directly by mean value theorem.

Now let $\epsilon > 0$ and $\Pi$ be a corresponding partition of $[0, 1]$ by Corollary. Then we can cover $\Gamma$ by balls $B_{2\epsilon_k}(f(x_k))$ for $k = 1, \cdots, N$. Thus

$$|\Gamma| \leq \sum_{k=1}^{N} \left|B_{2\epsilon_k}(f(x_k))\right| \leq \sum_{k=1}^{N} C_n \epsilon_k^n \leq C_n \epsilon^{n-1} \sum_{k=1}^{N} \epsilon_k \leq C_n \epsilon^{n-1} \| f' \|_1,$$

where $C_n$ is a constant depending only on the dimension $n$. (In fact, we can choose $C_n = |B_2|$.) Thus taking $\epsilon \to 0$, we have the desired result.

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WOW! That was not as difficult as I thought, but still cool. Thanks! So how do we proceed to prove the first statement (which intuitively seems obvious)? –  Bernard Sep 12 '12 at 2:59

The other direction is called (mini)-Sard's Theorem. Page 205 in Appendix 1 of Guilleman and Pollack, Differential Topology. The mini version is just this: Let $U$ be an open set of $\mathbb R^n,$ and let $f:U \rightarrow \mathbb R^m$ be a smooth map. Then, if $m > n,$ we can conclude that $f(U)$ has measure zero in $\mathbb R^m.$

The full Sard's Theorem is about critical points, while we make no demands about the relative dimensions. Let $f:X \rightarrow Y$ be a smooth map of manifolds, and let $C$ be the set of critical points of $f$ in $X.$ Then $f(C)$ has measure zero in $Y.$

They say their proof is almost verbatim from John W. Milnor, Topology from the Differentiable Viewpoint. It appears Guillemin and Pollack, also Milnor, assume $C^\infty.$ However, Milnor refers back to Pontryagin(1955), English translation (1959). Evidently Pontryagin worked with weaker conditions, but we are not told what. http://en.wikipedia.org/wiki/Sard%27s_theorem

Alright, I've been looking stuff up elsewhere. The full Sard's theorem does depend on relative dimension. I currently think that mini-Sard make be true both for $C^1$ and for differentiable everywhere, which is a pretty weak hypothesis.

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Provided $f$ is at least $C^1$ or some such regularity hypothesis. –  t.b. Sep 12 '12 at 3:17
@t.b., yes, thanks. It would appear the optimal result is Pontryagin (1955). I guess I can order a photocopy, but it takes a few days. –  Will Jagy Sep 12 '12 at 3:25
The regularity assumption $C^1$ for mini-Sard is enough as proved already in Sard's original paper, Theorem 4.1. I don't know off-hand how much this can be weakened. Whitney's example in A function not constant on a connected set of critical points is definitely worth having a look at. –  t.b. Sep 12 '12 at 3:33
@t.b. thanks. I've printed out Sard 1942, Thm. 4.1 says critical points. Critical point is when the rank of the matrix of first partials is smaller than both dimensions... –  Will Jagy Sep 12 '12 at 3:45
Oops... Guess it's bed time for me. Sorry about that. –  t.b. Sep 12 '12 at 3:50