Hausdorff Measure And Lebesgue Measure

Two Measures on Euclidean Space

Lebesgue measure describes volume in $\mathbb{R}^n$, while $n$-dimensional Hausdorff measure is constructed using covers whose cost depends on their diameters. These constructions look different, but in the dimension of the ambient space they describe the same volume up to a fixed constant. The constant depends on the convention used in defining Hausdorff measure; keeping track of it is essential when comparing formulas.

Throughout, $n\geq 1$. Write $\operatorname{diam}(U)=\sup\{|x-y|:x,y\in U\}$ for the diameter of a set $U\subseteq\mathbb{R}^n$, where $|\cdot|$ is Euclidean distance. For $E\subseteq\mathbb{R}^n$, $\delta>0$, and the unnormalized diameter convention, define

\mathcal{H}^n_\delta(E) =\inf\left\{\sum_{i=1}^{\infty}\bigl(\operatorname{diam}(U_i)\bigr)^n: E\subseteq\bigcup_{i=1}^{\infty}U_i,\ \operatorname{diam}(U_i)\leq\delta\right\}, \qquad \mathcal{H}^n(E)=\lim_{\delta\downarrow0}\mathcal{H}^n_\delta(E).

The infimum is over countable covers by subsets of $\mathbb{R}^n$; empty sets may be omitted. The limit exists in $[0,\infty]$, since decreasing $\delta$ restricts the allowed covers and thus makes $\mathcal{H}^n_\delta(E)$ nondecreasing. Here $\mathcal{H}^n$ denotes the unnormalized Hausdorff outer measure. Its restriction to Borel sets is a measure, and its completion includes subsets of Borel null sets.

Let $\mathcal{L}$ be the Lebesgue sigma-algebra on $\mathbb{R}^n$, and let $m$ be Lebesgue measure. Define $\omega_n=m(B(0,1))$, the volume of the Euclidean unit ball. The normalization constant below is specific to the diameter-power convention just given.

Standard Euclidean proportionality theorem: For every Lebesgue measurable set $E\in\mathcal{L}$, $$ \mathcal{H}^n(E)=\frac{\omega_n}{2^n}\,m(E), $$ with the equality understood in $[0,\infty]$. In particular, the normalized Hausdorff measure $$ \mathcal{H}^n_{\mathrm{norm}}(E)=\frac{2^n}{\omega_n}\mathcal{H}^n(E) $$ equals $m(E)$ for every $E\in\mathcal{L}$.

This proportionality is a standard result of geometric measure theory. Its proof relies on geometric covering results, including the isodiametric inequality, together with the identification of translation-invariant volume on Euclidean space. Those geometric ingredients are not proved here; the theorem is being used as a named standard result, not as a claim that the covering argument is immediate. The formula also specifies exactly which normalization gives equality: multiplying the unnormalized $\mathcal{H}^n$ by $2^n/\omega_n$ removes the factor.

What the Proportionality Says

The theorem concerns $n$-dimensional Hausdorff measure in $\mathbb{R}^n$. It does not say that $\mathcal{H}^s$ agrees with $m$ when $s\neq n$, nor that Hausdorff measure in every dimension is ordinary volume. For example, in $\mathbb{R}^2$, the factor for unnormalized $\mathcal{H}^2$ is $\omega_2/2^2=\pi/4$; in $\mathbb{R}^3$, it is $\omega_3/2^3=(4\pi/3)/8=\pi/6$.

The theorem is stated for Lebesgue measurable sets, not just Borel sets. One way to see why this extension is legitimate is to use the standard description of Lebesgue measure as the completion of its restriction to Borel sets. If a Lebesgue measurable $E$ differs from a Borel set $B$ only inside a Borel null set $Z$, then the Borel-set equality gives $\mathcal{H}^n(Z)=0$. The outer-measure inequalities then imply $\mathcal{H}^n(E)=\mathcal{H}^n(B)$, just as $m(E)=m(B)$. Thus completing the measures does not change the formula.

In particular, for Lebesgue measurable $E$, $m(E)=0$ if and only if $\mathcal{H}^n(E)=0$. If $0<m(E)<\infty$, then $0<\mathcal{H}^n(E)<\infty$ as well. Infinite measure is also preserved: $m(E)=\infty$ implies $\mathcal{H}^n(E)=\infty$, because the proportionality constant is positive and finite.

Worked Example: An Annulus in the Plane

Let $E=\{x\in\mathbb{R}^2:1\leq |x|\leq2\}$. This is a Borel set. Its Lebesgue measure is the area between circles of radii $2$ and $1$, so

m(E)=\pi(2^2)-\pi(1^2)=4\pi-\pi=3\pi.

In dimension $2$, $\omega_2=\pi$ and $2^2=4$. Therefore the unnormalized Hausdorff measure is

\mathcal{H}^2(E)=\frac{\pi}{4}\,m(E) =\frac{\pi}{4}\cdot 3\pi =\frac{3\pi^2}{4}.

The normalized value is $\mathcal{H}^2_{\mathrm{norm}}(E)=(4/\pi)(3\pi^2/4)=3\pi$, which agrees with $m(E)$. Check: multiplying the unnormalized answer by the factor $\pi/4$ gives $(\pi/4)(3\pi)=3\pi^2/4$, as required by the theorem.

Translations and Dilations

The cover definition also gives an exact transformation rule. This rule is useful independently of the proportionality theorem: it says how Hausdorff measure changes under a similarity of Euclidean space.

Proposition: For $E\subseteq\mathbb{R}^n$, $b\in\mathbb{R}^n$, and $a>0$, $$ \mathcal{H}^n(aE+b)=a^n\mathcal{H}^n(E), $$ where $aE+b=\{ax+b:x\in E\}$. The equality includes the value $+\infty$. Translation alone therefore leaves $\mathcal{H}^n$ unchanged.

Proof. Fix $\delta>0$. The map $x\mapsto ax+b$ is a bijection of $\mathbb{R}^n$, and for every set $U$, $\operatorname{diam}(aU+b)=a\operatorname{diam}(U)$. It sends each countable cover of $E$ with diameters at most $\delta$ to a cover of $aE+b$ with diameters at most $a\delta$; the sum of the $n$th powers of the diameters is multiplied by $a^n$. Conversely, applying $y\mapsto (y-b)/a$ to any such cover of $aE+b$ gives a cover of $E$ with diameters at most $\delta$, and divides its cost by $a^n$. Taking infima in both directions yields $\mathcal{H}^n_{a\delta}(aE+b)=a^n\mathcal{H}^n_\delta(E)$. Letting $\delta\downarrow0$, the left-hand side tends to $\mathcal{H}^n(aE+b)$, since $a\delta\downarrow0$, and the right-hand side tends to $a^n\mathcal{H}^n(E)$. This also holds if the limit is infinite. $\square$

When $a=0$, the image $aE+b$ is either empty or a singleton. For $n\geq1$, a singleton has $\mathcal{H}^n$-measure zero: it can be covered by itself, whose diameter is zero. Thus the image has zero $n$-dimensional Hausdorff measure. The proposition is stated for $a>0$ because its proof uses the inverse dilation.

Worked Example: Dilating a Measurable Set in Three Dimensions

Suppose $E\in\mathcal{L}$ in $\mathbb{R}^3$ and $m(E)=5/4$. The unnormalized proportionality factor in dimension $3$ is $\omega_3/2^3=(4\pi/3)/8=\pi/6$. Hence

\mathcal{H}^3(E)=\frac{\pi}{6}\cdot\frac54=\frac{5\pi}{24}.

Now let $F=2E+b$, where $b\in\mathbb{R}^3$. The dilation proposition gives $\mathcal{H}^3(F)=2^3\mathcal{H}^3(E)=8(5\pi/24)=5\pi/3$. Independently, Lebesgue measure scales by the determinant of the dilation, so $m(F)=2^3(5/4)=10$, and the proportionality theorem gives $\mathcal{H}^3(F)=(\pi/6)10=5\pi/3$. Check: both calculations yield $5\pi/3$, and their ratio to the original value is $(5\pi/3)/(5\pi/24)=8=2^3$.

Null Sets and the Importance of Normalization

For subsets of $\mathbb{R}^n$, the theorem transfers statements about $n$-dimensional Lebesgue volume directly to $n$-dimensional Hausdorff measure. For example, if $E\in\mathcal{L}$ is null, then its normalized Hausdorff measure is zero. Every subset $A\subseteq E$, whether or not $A$ was initially known to be Lebesgue measurable, has Hausdorff outer measure zero by monotonicity. Such a subset is measurable for the completed Hausdorff measure.

Worked Example: A Line Segment Has Zero Planar Area

Consider $E=[0,1]\times\{0\}\subseteq\mathbb{R}^2$. For any $r>0$, $E$ is contained in the rectangle $[0,1]\times[-r,r]$, whose area is $2r$. Since $m(E)\leq2r$ for every $r>0$ and $m(E)\geq0$, letting $r\downarrow0$ gives $m(E)=0$. The proportionality theorem then gives $\mathcal{H}^2(E)=(\pi/4)0=0$, and $\mathcal{H}^2_{\mathrm{norm}}(E)=0$. Check: the normalized value is $(4/\pi)\mathcal{H}^2(E)=(4/\pi)0=0=m(E)$. This is a statement about planar, or two-dimensional, measure; it does not assert that the segment has zero one-dimensional Hausdorff measure.

A common error is to identify the unnormalized $\mathcal{H}^n$ with Lebesgue measure without checking the convention. In the plane, for a measurable set of area $4$, the incorrect calculation would be $\mathcal{H}^2(E)=4$. Under the diameter-power convention used here, the correction is

\mathcal{H}^2(E)=\frac{\pi}{4}\cdot4=\pi, \qquad \mathcal{H}^2_{\mathrm{norm}}(E)=\frac{4}{\pi}\cdot\pi=4.

The value $4$ belongs to Lebesgue measure and to the normalized Hausdorff measure; the unnormalized Hausdorff measure is $\pi$. If a different definition builds the factor $2^n/\omega_n$ into Hausdorff measure from the start, then its notation may denote what is called $\mathcal{H}^n_{\mathrm{norm}}$ here. Always check the definition before comparing constants.

The distinction between dimension and measure matters as well. The proportionality theorem is about the measure $\mathcal{H}^n$ in the ambient dimension $n$; it is not a formula for Hausdorff dimension, which is determined by the behavior of $\mathcal{H}^s$ as the exponent $s$ varies. Its practical use is precise: for Lebesgue measurable subsets of $\mathbb{R}^n$, it converts $n$-dimensional Hausdorff measure to volume by one fixed factor, and the normalized convention makes that factor exactly one.

Check Your Understanding

In each question, distinguish the unnormalized $\mathcal{H}^n$ from $\mathcal{H}^n_{\mathrm{norm}}$.

In $\mathbb{R}^2$, if $E$ is Lebesgue measurable and $m(E)=8$, what are $\mathcal{H}^2(E)$ and $\mathcal{H}^2_{\mathrm{norm}}(E)$?
Explain why the proportionality theorem implies that every Lebesgue measurable null set in $\mathbb{R}^n$ has zero $\mathcal{H}^n$-measure.
Use the scaling proposition to express $\mathcal{H}^n(3E+b)$ in terms of $\mathcal{H}^n(E)$, including when the latter is infinite.
In $\mathbb{R}^3$, what normalization factor multiplies the diameter-power $\mathcal{H}^3$ to produce Lebesgue measure?
Why does zero $\mathcal{H}^2$-measure of a line segment in the plane not imply zero $\mathcal{H}^1$-measure?

Hausdorff Measure And Lebesgue Measure

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