## Introduction Lebesgue measure quantifies the $n$-dimensional “volume’’ of subsets of $\mathbb R^{n}$. Whenever a set has [topological](/page/Topology) dimension strictly less than $n$, its Lebesgue measure equals $0$. For example, a circle sitting in $\mathbb R^{3}$ has $3$-dimensional Lebesgue measure $0$ because a one-dimensional curve occupies no volume. The natural question is therefore: How can we assign a meaningful $s$-dimensional size to a subset of $\mathbb R^{n}$ when $0\le s < n$? A first attempt is to parameterise the set by $s$ variables and integrate over $\mathbb R^{s}$. Two difficulties appear immediately. * **Jacobian distortion** A parameterisation changes lengths and areas, so a corrective factor is required. * **Absence of global charts** Many [sets](/page/Set) (for example fractals) possess no single or even finitely many smooth parameterisations. Hausdorff measure overcomes both problems by defining size in a coordinate-free manner that works for *all* subsets of a [metric space](/page/Metric%20Space). ## Formal Definition [definition:Hausdorff measure] Let $(X,d)$ be a metric space, let $E\subset X$, and let $s\ge0$.For any set $A\subset X$ define \begin{align*} \operatorname{diam}A=\sup\{d(x,y):x,y\in A\}. \end{align*} For $\delta > 0$ set test \begin{align*}\mathcal{H}^{s}_{\delta}(E)=\inf\Bigl\{\sum_{k=1}^{\infty}(\operatorname{diam}A_{k})^{s}:E\subset\bigcup_{k=1}^{\infty}A_{k}\operatorname{diam}A_{k} < \delta\Bigr\}. \end{align*} Define \begin{align*} \mathcal H^{s}(E)=\lim_{\delta\downarrow0}\mathcal H^{s}_{\delta}(E). \end{align*} The [limit](/page/Limit) exists in $[0,\infty]$ because $\mathcal H^{s}_{\delta}(E)$ decreases as $\delta$ decreases. [/definition] Intuitively, $\mathcal H^{s}(E)$ is obtained by covering $E$ with sets of arbitrarily small diameter, adding the $s$-th power of each diameter, and keeping the smallest possible total cost. A first sanity check is to compare $\mathcal H^{s}$ with Lebesgue measure. [proposition] Let $E\subset\mathbb R^{n}$ and let $\lambda^{n}$ be $n$-dimensional Lebesgue measure. 1. If $s > n$, then $\mathcal H^{s}(E)=0$. 2. If $s < n$ and $E$ has non-empty interior, then $\mathcal H^{s}(E)=\infty$. 3. There exists a dimensional constant $c_{n} > 0$ such that \begin{align*} \mathcal H^{n}(E)=c_{n}\,\lambda^{n}(E)\quad\text{for every Borel set }E\subset\mathbb R^{n}. \end{align*} [/proposition] [proof] *Step 1. Case $s > n$.* Fix $\varepsilon > 0$. Because $E$ is a subset of $\mathbb R^{n}$, there exists a countable collection of closed cubes $\{Q_{k}\}_{k=1}^{\infty}$ that covers $E$ and satisfies \begin{align*} \sum_{k=1}^{\infty}\lambda^{n}(Q_{k}) < \varepsilon. \end{align*} For each $k$, the diameter $\operatorname{diam}Q_{k}$ equals the edge length $l_{k}$ times $\sqrt n$. Therefore \begin{align*} \sum_{k=1}^{\infty}(\operatorname{diam}Q_{k})^{s} =(\sqrt n)^{s}\sum_{k=1}^{\infty}l_{k}^{s} \le(\sqrt n)^{s}\sum_{k=1}^{\infty}l_{k}^{n} =(\sqrt n)^{s-n}\sum_{k=1}^{\infty}\lambda^{n}(Q_{k}) < (\sqrt n)^{s-n}\varepsilon. \end{align*} Since $\varepsilon$ was arbitrary, $\mathcal H^{s}(E)=0$. *Step 2. Case $s < n $ with $\operatorname{int}E\neq\varnothing$.* Choose a closed ball $\overline B_{r}(x_{0})\subset E$. For any cover $\{A_{k}\}$ of $E$ with $\operatorname{diam}A_{k} < \delta$, the same family covers $\overline B_{r}(x_{0})$. Each set $A_{k}$ lies inside some ball of radius $\operatorname{diam}A_{k}$, so the volume of $A_{k}$ is bounded above by $c_{n}\operatorname{diam}(A_{k})^{\,n}$, where $c_{n}$ is the volume of the $n$-dimensional unit ball. Because the volumes of the $A_{k}$ sum to at least the volume of $\overline B_{r}(x_{0})$, \begin{align*} c_{n}r^{n}\le \sum_{k=1}^{\infty}c_{n}\operatorname{diam}(A_{k})^{\,n}. \end{align*} Hence \begin{align*} \sum_{k=1}^{\infty}\operatorname{diam}(A_{k})^{s} \ge\sum_{k=1}^{\infty}\operatorname{diam}(A_{k})^{\,n} \ge r^{n}. \end{align*} The right side is independent of $\delta$, so $\mathcal H^{s}_{\delta}(E)\ge r^{n}$ for every $\delta$. Consequently $\mathcal H^{s}(E)=\infty$. *Step 3. Identification with Lebesgue measure when $s=n$.* First assume $E$ is an open cube $Q$. For any $\delta > 0$ the cover consisting solely of $Q$ shows $\mathcal H^{n}_{\delta}(Q)\le(\operatorname{diam}Q)^{n}=n^{n/2}\lambda^{n}(Q)$. Conversely, if $\{A_{k}\}$ covers $Q$ with $\operatorname{diam}A_{k} < \delta$, Vitali’s covering theorem produces a finite subcollection of disjoint balls $\{B_{k}\}_{k=1}^{m}$ such that the union of the concentric balls $5B_{k}$ still covers $Q$. The diameter of $5B_{k}$ equals $5\,\operatorname{diam}B_{k}$, and the disjointness implies \begin{align*} \lambda^{n}(Q)\le\sum_{k=1}^{m}\lambda^{n}(5B_{k}) =5^{n}\,\omega_{n}\sum_{k=1}^{m}r_{k}^{\,n} \le 5^{n}\,\omega_{n}\sum_{k=1}^{m}(\operatorname{diam}B_{k})^{n}. \end{align*} Because $\operatorname{diam}B_{k}\le\operatorname{diam}A_{k}$, this yields \begin{align*} \lambda^{n}(Q)\le 5^{n}\,\omega_{n}\sum_{k=1}^{\infty}(\operatorname{diam}A_{k})^{n}. \end{align*} Taking the infimum over all admissible covers and then the limit $\delta\downarrow0$ gives \begin{align*} \lambda^{n}(Q)\le 5^{n}\,\omega_{n}\,\mathcal H^{n}(Q). \end{align*} Combining the two bounds yields \begin{align*} c_{n}^{-1}\lambda^{n}(Q)\le\mathcal H^{n}(Q)\le c_{n}\lambda^{n}(Q) \quad\text{with }c_{n}=\max\bigl\{n^{n/2},5^{n}\omega_{n}\bigr\}. \end{align*} A standard outer regularity argument extends the equality $\mathcal H^{n}=c_{n}\lambda^{n}$ from cubes to all Borel sets. [/proof] --- It is natural to ask whether the connection between Lebesgue and Hausdorff measures is limited to $s=n$. The next theorem, a weak form of the divergence theorem, demonstrates that $\mathcal H^{\,n-1}$ plays the role of surface measure in [integration by parts](/theorems/210). [theorem:Sobolev Gauss–Green] Let $U\subset\mathbb R^{n}$ be bounded with Lipschitz [boundary](/page/Boundary) $\partial U$, and let $\nu$ denote the outward unit normal. If $u\in W^{1,1}(U)$, then for each $i=1,\dots,n$ \begin{align*} \int_{U}\partial_{x_{i}}u(x)\,dx =\int_{\partial U}\operatorname{Tr}u(x)\,\nu_{i}(x)\,d\mathcal H^{\,n-1}(x), \end{align*} where $\operatorname{Tr}u\in L^{1}(\partial U,\mathcal H^{\,n-1})$ is the Sobolev trace of $u$. [/theorem] The formula shows that $\mathcal H^{\,n-1}$ is precisely the surface element required for flux [integrals](/page/Integral) in the weak divergence theorem. --- ## Area (Jacobian) formula [theorem:Area formula] Let $D\subset\mathbb R^{\,n-1}$ be open, and let $\Phi:D\to\mathbb R^{n}$ be a $C^{1}$ immersion (its differential has full rank everywhere). For every Borel [function](/page/Function) $f:\Phi(D)\to[0,\infty]$, \begin{align*} \int_{\Phi(D)}f(x)\,d\mathcal H^{\,n-1}(x) =\int_{D} f\bigl(\Phi(\xi)\bigr)\, \sqrt{\det\bigl[(D\Phi)^{\mathsf T}D\Phi\bigr]}\, d\xi. \end{align*} [/theorem] *Discussion.* The matrix $(D\Phi)^{\mathsf T}D\Phi$ is the Gram matrix of the tangent vectors, and the square-root determinant equals the classical area element \begin{align*} \sqrt{\det\!\bigl[(D\Phi)^{\mathsf T}D\Phi\bigr]} =\lVert\partial_{\xi_{1}}\Phi\wedge\cdots\wedge\partial_{\xi_{n-1}}\Phi\rVert. \end{align*} Consequently, Hausdorff measure reduces to the usual arc-length or surface-area integral whenever the set admits a smooth parameterisation. On polyhedral faces, the Jacobian equals $1$, so $d\mathcal H^{\,n-1}$ coincides with the standard Lebesgue measure on the face.