Let $H^p(A)=\lim_{\delta\to 0}H^p_{\delta}(A)$ be the $p$-dimensional Hausdorff measure, where $H^p_{\delta}(A)=\inf\{\sum_{i}diam(E_i)^p:A\subset\bigcup_{i}E_i,diam(E_i)<\delta\}$


Suppose $f:A\to\mathbb{R}^k$ satisfies Lipschitz condition: $\|f(x)-f(y)\|\leq M\|x-y\|$ for constant $M$. Show that for p-dimensional Hausdorff measure, $H^p((f(A))\leq M^p \cdot H^{p}(A)$.

Here is my proof:

Let $A\subset\bigcup_{j}E_j$, and $diam(E_j)<\delta$, satisfying $H^p(A)+\epsilon\ge\sum_{j}diam^p(E_j)$.

By Lipschitz condition, $diam((f(E_j))\leq M\cdot diam(E_j)$ and $H^p_{\delta}(A)+\epsilon\ge\frac{1}{M^p}\sum_{j}diam^{p}(f(E_j))$.

Since $f(A)\subset\bigcup_{j}f(E_j)$, and $diam(f(E_j))\leq M\cdot\delta$,

$M^p(H^p_{\delta}(A)+\epsilon)\ge H_{M\delta}^{p}(f(A))$. Let $\epsilon, \delta\to 0$, $H^p((f(A))\leq M^p\cdot H^p(A)$.

Can someone help me check whether the proof is correct?


Suppose $f:[a,b]\to\mathbb{R}$ has continuous derivatives. Let $A=\{(x,f(x):x\in[a,b]\}\subset\mathbb{R}^2$. Show that $H^1(A)=\int_{a}^{b}\big((1+f')^2\big)^{1/2}dx$.

By the first question, $b-a=H^{1}([a,b])\leq H^{1}(A)\leq M\cdot H^{1}([a,b])=M(b-a)$. How to use this to show the equation?

By the way, I am not so sure about how to solve question 2. Please leave your solutions if possible.



1 Answer 1


Your proof for (1) is alright except you wrote $M$ instead of $M^p$ at the first character of the last line.

For (2), first note that $A$ is compact and since we can suppose the $E_i$ to be open sets, given a cover of $A$ by $E_i$'s, we can find a finite subcover.
But then, we can also find a uniform neighbourhood of $f$ such that the graphs of all the functions near $f$ are contained in the finite subcover we found above.

Consider a partition $P=\{x_0, x_1, ..., x_n \}$ of $[a, b]$.

Let $m_i = \inf_{x_{i - 1} \le x \le x_i} f(x)$ and $M_i = \sup_{x_{i - 1} \le x \le x_i} f(x)$.
Analogously, let $m'_i = \inf_{x_{i - 1} \le x \le x_i} f'(x)$ and $M'_i = \sup_{x_{i - 1} \le x \le x_i} f'(x)$.

As a consequence of the initial remarks, and the fact that $f$ is $C^1$, we can take $E_i = [x_{i-1}, x_i] \times [m_i, M_i]$.
Observe that $diam E_i = ((x_{i-1} - x_i)^2 + (M_i - m_i)^2)^\frac{1}{2}$, since it's a rectangle.

From mean value theorem, we have:

$((x_{i-1} - x_i)^2 + (m'_i)^2(x_{i-1} - x_i)^2)^\frac{1}{2} \le diam E_i \le ((x_{i-1} - x_i)^2 + (M'_i)^2(x_{i-1} - x_i)^2)^\frac{1}{2}$.

Summing over $i$ and letting $|P| \to 0$, we get to $H_{\delta}^1(A)=\int_a^b(1 + f'(x)^2)^\frac{1}{2}dx$(again, we can do that because $f'$ is continuous on a compact set, hence integrable). It follows immediately that $H^1(A)=\int_a^b(1 + f'(x)^2)^\frac{1}{2}dx$.

  • $\begingroup$ Thanks! But how do you show that $A$ is compact? $\endgroup$
    – user758472
    Commented May 9, 2020 at 4:10
  • $\begingroup$ $A$ is the image of the compact set $[a, b]$ by the continuous function $x \mapsto (x, f(x))$. $\endgroup$ Commented May 9, 2020 at 4:12
  • $\begingroup$ Should it be $m_i^2$ and $M_i^2$ in the mean value theorem step? $\endgroup$
    – user758472
    Commented May 10, 2020 at 10:54
  • $\begingroup$ Yep. I'll edit it. $\endgroup$ Commented May 10, 2020 at 17:02

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