Hausdorff Outer Measure Is an Outer Measure

From Diameter Covers to an Outer Measure

Hausdorff measure is built by charging a set for the sizes of the pieces in a cover. The diameter restriction matters: as the allowed pieces become smaller, a cover may require more pieces or a larger total cost. The resulting limiting set function is an outer measure on all subsets of a metric space. Its construction does not, by itself, assert that every subset is measurable for that outer measure.

Let $(X,d)$ be a metric space. For a nonempty set $U\subseteq X$, its diameter is $\operatorname{diam}(U)=\sup\{d(x,y):x,y\in U\}$, with $+\infty$ allowed. Fix a finite exponent $s\ge0$ and a scale $\delta>0$. A $\delta$-cover of $A\subseteq X$ is a finite or countable family of nonempty sets $U_i\subseteq X$ such that $A\subseteq\bigcup_i U_i$ and $\operatorname{diam}(U_i)\le\delta$ for every $i$. The empty family covers the empty set. For $s=0$, each nonempty covering set has cost $1$; this specifies the convention at exponent zero without using $0^0$.

Definition: For $A\subseteq X$, define the Hausdorff content at scale $\delta$ by $$ \mathcal{H}^s_\delta(A)=\inf\left\{\sum_i\bigl(\operatorname{diam}(U_i)\bigr)^s:\ (U_i)\text{ is a }\delta\text{-cover of }A\right\},\qquad s>0, $$ and use $\sum_i 1$ in place of the sum of powers when $s=0$. An infimum over no admissible covers is $+\infty$. The Hausdorff outer measure of exponent $s$ is $$ \mathcal{H}^s(A)=\lim_{\delta\downarrow0}\mathcal{H}^s_\delta(A). $$

For $s>0$, a singleton has cost $0^s=0$; for $s=0$, a singleton has cost $1$. At any fixed $\delta$, a smaller permitted scale allows no additional covers, so $\mathcal{H}^s_\delta(A)$ is nondecreasing as $\delta$ decreases. Consequently the displayed limit exists in $[0,\infty]$, though it may be infinite. There is no measure space $(X,\mathcal{M},\mu)$ assumed here: $\mathcal{H}^s$ is defined on the power set of $X$ as an outer measure.

The Outer-Measure Properties at a Fixed Scale

Proposition: For each $\delta>0$ and $s\ge0$, the set function $\mathcal{H}^s_\delta$ is zero on the empty set, monotone, and countably subadditive on subsets of $X$.

Proof. The empty family is a $\delta$-cover of $\varnothing$, with cost zero, so $\mathcal{H}^s_\delta(\varnothing)=0$. Costs are nonnegative. If $A\subseteq B$, every $\delta$-cover of $B$ also covers $A$. Thus the infimum for $A$ is no greater than the infimum for $B$, which proves monotonicity.

Now let $A_1,A_2,\ldots\subseteq X$. If $\sum_{j=1}^\infty\mathcal{H}^s_\delta(A_j)=+\infty$, countable subadditivity holds because its right-hand side is infinite. Otherwise, fix $\varepsilon>0$. For each $j$, the content of $A_j$ is finite, so the definition of infimum gives a $\delta$-cover of $A_j$ whose cost is less than $\mathcal{H}^s_\delta(A_j)+\varepsilon 2^{-j}$. The union of these covers is countable and covers $\bigcup_j A_j$; every set in it still has diameter at most $\delta$. Its total cost is at most $$ \sum_{j=1}^\infty\mathcal{H}^s_\delta(A_j)+\varepsilon\sum_{j=1}^\infty2^{-j} =\sum_{j=1}^\infty\mathcal{H}^s_\delta(A_j)+\varepsilon. $$ The infimum defining the content of the union is no greater than this cost. Since the inequality holds for every $\varepsilon>0$, letting $\varepsilon\downarrow0$ proves $$ \mathcal{H}^s_\delta\left(\bigcup_{j=1}^\infty A_j\right) \le\sum_{j=1}^\infty\mathcal{H}^s_\delta(A_j). $$ This also covers empty sets among the $A_j$, whose empty covers contribute zero. $\square$

Passing to the Limit in the Scale

Taking a limit in the fixed-scale inequality requires care: the scale on the left and on every term on the right must be the same. We first record the elementary limit fact that justifies the countable sum.

Lemma: Suppose $a_{j,n}\in[0,\infty]$ is nondecreasing in $n$ for each $j$, and $a_j=\lim_{n\to\infty}a_{j,n}$. Then $$ \lim_{n\to\infty}\sum_{j=1}^\infty a_{j,n}=\sum_{j=1}^\infty a_j. $$

Proof. For each $n$, $a_{j,n}\le a_j$, so $\sum_j a_{j,n}\le\sum_j a_j$. This gives the upper bound on the limit. For any fixed positive integer $J$, $$ \lim_{n\to\infty}\sum_{j=1}^\infty a_{j,n} \ge \lim_{n\to\infty}\sum_{j=1}^J a_{j,n} =\sum_{j=1}^J a_j. $$ The last equality holds because this is a finite sum of nondecreasing sequences, with extended nonnegative limits allowed. Taking the supremum over $J$ gives the reverse bound, since $\sum_j a_j=\sup_J\sum_{j=1}^J a_j$. The bounds prove the result, including when the sum is infinite. $\square$

Theorem: For any metric space $(X,d)$ and any finite $s\ge0$, $\mathcal{H}^s$ is an outer measure on $X$: it is nonnegative, is zero on $\varnothing$, is monotone, and is countably subadditive.

Proof. Nonnegativity follows from the nonnegative costs. Since $\mathcal{H}^s_\delta(\varnothing)=0$ at every scale, its limit is also zero. If $A\subseteq B$, fixed-scale monotonicity gives $\mathcal{H}^s_\delta(A)\le\mathcal{H}^s_\delta(B)$ for every $\delta>0$; taking the limits proves monotonicity of $\mathcal{H}^s$.

For countable subadditivity, fix $A_1,A_2,\ldots\subseteq X$, and use the sequence of scales $\delta_n=1/n$. For each $j$, $\mathcal{H}^s_{1/n}(A_j)$ is nondecreasing in $n$ and converges to $\mathcal{H}^s(A_j)$. The limit along $1/n$ is the full limit as $\delta\downarrow0$: given $1/(n+1)\le\delta\le1/n$, monotonicity in scale gives $$ \mathcal{H}^s_{1/n}(A_j)\le\mathcal{H}^s_\delta(A_j)\le\mathcal{H}^s_{1/(n+1)}(A_j). $$ Both bounds have the same limit as $n\to\infty$.

Apply fixed-scale countable subadditivity at the common scale $1/n$. Then take $n\to\infty$. The lemma just proved, with $a_{j,n}=\mathcal{H}^s_{1/n}(A_j)$, justifies the limit of the sum. Therefore $$ \mathcal{H}^s\left(\bigcup_{j=1}^\infty A_j\right) =\lim_{n\to\infty}\mathcal{H}^s_{1/n}\left(\bigcup_{j=1}^\infty A_j\right) \le\lim_{n\to\infty}\sum_{j=1}^\infty\mathcal{H}^s_{1/n}(A_j) =\sum_{j=1}^\infty\mathcal{H}^s(A_j). $$ All limits and sums here involve nonnegative extended real numbers. This proves countable subadditivity and completes the outer-measure axioms. $\square$

Worked Diameter-Cover Calculations

Worked Example: The Empty Set at Every Exponent

Let $A=\varnothing$, let $\delta>0$, and let $s\ge0$. The empty family is an admissible cover and has cost $0$. Every admissible cost is nonnegative, so the infimum is exactly $$ \mathcal{H}^s_\delta(\varnothing)=0. $$ Taking the limit as $\delta\downarrow0$ gives $\mathcal{H}^s(\varnothing)=0$. Check: this computation uses the empty cover, not a cover by nonempty sets; the latter would have nonnegative cost and could not lower the infimum below zero.

Worked Example: A Separated Finite Set at Exponent Zero

In $X=\mathbb{R}$ with the usual metric, take $A=\{0,2,5\}$ and $0<\delta<2$. At exponent $s=0$, every nonempty covering set costs $1$. A set of diameter at most $\delta<2$ cannot contain two points of $A$, because the smallest distance between distinct points of $A$ is $2$. Thus at least three covering sets are required, so $\mathcal{H}^0_\delta(A)\ge3$. The three singleton sets $\{0\},\{2\},\{5\}$ form an admissible cover of cost $1+1+1=3$. Hence $\mathcal{H}^0_\delta(A)=3$, and $\mathcal{H}^0(A)=3$. Check: each singleton has diameter $0\le\delta$, and the three required costs sum to $3$.

For any $s>0$, the same singleton cover has cost $0^s+0^s+0^s=0$. Nonnegativity then gives $\mathcal{H}^s_\delta(A)=0$ at every scale, and $\mathcal{H}^s(A)=0$. This contrast follows from the exponent-zero convention, not from a change in the set or the metric.

Worked Example: A Scale-Uniform Cover of an Interval

In $\mathbb{R}$, let $A=[0,1]$, take $s=1$, and fix any $\delta>0$. Set $N=\lceil1/\delta\rceil$, the least integer at least $1/\delta$. Divide $[0,1]$ into $N$ intervals of equal length $1/N$, assigning shared endpoints to just one of the adjacent intervals. Their union is $[0,1]$, and $$ \frac1N\le\delta,\qquad \sum_{k=1}^N\bigl(\operatorname{diam}(U_k)\bigr)^1 =N\cdot\frac1N=1. $$ Thus $\mathcal{H}^1_\delta([0,1])\le1$ for every $\delta>0$, and taking the limit gives $\mathcal{H}^1([0,1])\le1$. For example, at $\delta=1/4$, $N=\lceil4\rceil=4$, each interval has diameter $1/4$, and the total cost is $4\cdot(1/4)=1$. Check: the cover meets the diameter bound at every scale, not only at $\delta=1/4$; the general calculation uses $N\ge1/\delta$.

A Common Scale Error

A tempting but invalid argument is to cover each $A_j$ using a different scale $\delta_j$, combine those covers, and claim that their union is a $\delta$-cover for one fixed $\delta$. For a concrete failure, let $A_1=\{0\}$, $A_2=\{1\}$, and $\delta=1/4$. A cover of $A_1$ may use $\{0\}$, while a cover of $A_2$ at the larger scale $2$ may use $\{0,1\}$, whose diameter is $1$. Combining these pieces does not give a $1/4$-cover: the piece $\{0,1\}$ violates the required diameter bound.

The correction is to choose near-minimizing covers at the same $\delta$ for every set $A_j$, as in the fixed-scale proof. Only after establishing the inequality at each common scale should one let $\delta\downarrow0$. The lemma ensures that the countable sum has the required limit. Finally, being an outer measure means these four set-function axioms hold; it does not mean that every subset of $X$ is measurable or that $\mathcal{H}^s$ is already a measure on the power set.

Check Your Understanding

Use the diameter-cover definition and identify the scale or exponent condition needed in each argument.

For $s>0$, compute $\mathcal{H}^s(\{x\})$ in an arbitrary metric space.
Why does the empty family establish $\mathcal{H}^s_\delta(\varnothing)=0$, while nonnegativity is still needed?
Explain why a $\delta$-cover of $B$ is also a $\delta$-cover of $A$ when $A\subseteq B$.
In the countable-subadditivity proof, why must the covers of all $A_j$ use the same $\delta$?
State the limit fact used to justify passing from $\sum_j\mathcal{H}^s_{1/n}(A_j)$ to $\sum_j\mathcal{H}^s(A_j)$.