Constructing an Outer Measure from Open Covers

From Open-Set Content to an Outer Measure

Open-set content assigns a size to open sets using a positive linear functional. The next construction extends this information to every subset of the underlying space: cover a set by countably many open sets, add their contents, and take the infimum over all such covers. The infimum matters. A particular cover supplies an upper bound, but it need not be the least possible total content.

Let $X$ be a locally compact Hausdorff space, and let $C_c(X)$ be the real-valued continuous functions on $X$ with compact support. Suppose $L:C_c(X)\to\mathbb{R}$ is a positive linear functional: $L$ is linear and $L(f)\geq 0$ whenever $f\geq 0$. For an open set $U\subseteq X$, its open-set content is

c(U)=\sup\{L(f): f\in C_c(X),\ 0\leq f\leq 1,\ \operatorname{supp} f\subseteq U\}.

The zero function is admissible for every open $U$, so $c(U)\geq 0$ and $c(\varnothing)=0$. The supremum is allowed to be $+\infty$. The function $c$ is defined on open sets; it is not yet a measure on a sigma-algebra. We now use it to define a set function on the entire power set of $X$.

Definition: For every subset $A\subseteq X$, define $$ \mu^*(A)=\inf\left\{\sum_{n=1}^{\infty}c(U_n):\ U_n\subseteq X\text{ open and }A\subseteq\bigcup_{n=1}^{\infty}U_n\right\}. $$ Here the sum has values in $[0,\infty]$. An empty set may be included in a cover, and an empty cover of $\varnothing$ has sum zero.

The resulting set function $\mu^*$ is an outer measure. An outer measure is a function from all subsets of $X$ to $[0,\infty]$ that is zero on the empty set, monotone, and countably subadditive. These properties do not by themselves say that $\mu^*$ is countably additive on all subsets. Establishing which sets are measurable for $\mu^*$ is a separate step.

Why the Construction Satisfies the Outer-Measure Axioms

Theorem: For any locally compact Hausdorff space $X$ and positive linear functional $L$ as above, the set function $\mu^*$ defined by countable open covers is an outer measure on $X$. No finiteness or sigma-finiteness assumption on $X$ or $c$ is required.

Proof. First, every sum in the definition is nonnegative, so $\mu^*(A)\geq 0$ for all $A\subseteq X$. The empty cover is admissible for $\varnothing$ and has sum zero. Therefore $\mu^*(\varnothing)=0$.

For monotonicity, suppose $A\subseteq B\subseteq X$. Every countable open cover of $B$ is also a cover of $A$. Thus the collection of covers over which the infimum is taken for $A$ includes the collection used for $B$, giving $\mu^*(A)\leq\mu^*(B)$.

For countable subadditivity, let $A_1,A_2,\ldots\subseteq X$. If $\sum_{j=1}^{\infty}\mu^*(A_j)=+\infty$, then $\mu^*(\bigcup_j A_j)\leq\sum_j\mu^*(A_j)$ holds because the right-hand side is $+\infty$. Otherwise the sum is finite, so each $\mu^*(A_j)$ is finite. Fix $\varepsilon>0$. By the definition of an infimum, for each $j\geq 1$ there is a countable open cover $(U_{jk})_{k\geq1}$ of $A_j$ such that

\sum_{k=1}^{\infty}c(U_{jk})<\mu^*(A_j)+\varepsilon 2^{-j}.

All the sets $U_{jk}$, with both indices varying, form a countable open cover of $\bigcup_j A_j$. Consequently,

\mu^*\left(\bigcup_{j=1}^{\infty}A_j\right) \leq \sum_{j=1}^{\infty}\sum_{k=1}^{\infty}c(U_{jk}) \leq \sum_{j=1}^{\infty}\mu^*(A_j)+\varepsilon\sum_{j=1}^{\infty}2^{-j} =\sum_{j=1}^{\infty}\mu^*(A_j)+\varepsilon.

The geometric series equals $1$. Since this inequality holds for every $\varepsilon>0$, letting $\varepsilon$ decrease to zero proves countable subadditivity. This proves all three outer-measure axioms. $\square$

Notice what the proof does not require: it does not require that the content of a union of open sets be bounded by the sum of their contents. Countable subadditivity here comes directly from putting covers together and taking infima. The content’s other properties may be useful elsewhere, but they are not needed for these axioms.

Interval Calculations for the Lebesgue Integral Functional

For explicit computations, take $X=\mathbb{R}$ with its usual topology and let $L(f)=\int f\,dm$ for $f\in C_c(\mathbb{R})$, where $m$ is Lebesgue measure. In this example the open-set content agrees with Lebesgue measure on every open set. Here is the reason. For any admissible $f$ and open $U$, $0\leq f\leq\chi_U$, so $\int f\,dm\leq m(U)$, giving $c(U)\leq m(U)$. Conversely, an open subset of $\mathbb{R}$ is a countable disjoint union of open intervals. Inside each component, continuous compactly supported functions between zero and one can be chosen to equal one on any prescribed compact subinterval. For finitely many components, add such functions on disjoint supports; their sum remains at most one and is admissible for $U$. Taking compact subintervals whose lengths approach the component lengths, then taking larger finite collections of components, gives $c(U)\geq m(U)$, including when $m(U)=+\infty$. Thus $c(U)=m(U)$.

Worked Example: The Content of an Open Interval

Compute $c((2,5))$ directly from the defining supremum. For any admissible $f$, $f$ vanishes outside $(2,5)$ and satisfies $0\leq f\leq1$. Therefore $\int f\,dm\leq m((2,5))=5-2=3$, so $c((2,5))\leq3$.

For the reverse inequality, choose $0<\delta<3/4$. Define a continuous function $f_\delta$ that is zero outside $[2+\delta,5-\delta]$, increases linearly from zero to one on $[2+\delta,2+2\delta]$, equals one on $[2+2\delta,5-2\delta]$, and decreases linearly to zero on $[5-2\delta,5-\delta]$. Its support is compact and contained in $(2,5)$, and $0\leq f_\delta\leq1$. The plateau has length $3-4\delta$, and the two ramps have area $\delta/2$ each. Hence

L(f_\delta)=\int f_\delta\,dm=(3-4\delta)+\frac{\delta}{2}+\frac{\delta}{2}=3-3\delta.

As $\delta$ decreases to zero, these admissible integrals approach $3$. Thus $c((2,5))\geq3$, and $c((2,5))=3$. Check: the upper bound was the interval length $5-2=3$, and the cutoff integrals $3-3\delta$ approach that same value from below.

Worked Example: The Outer Measure of a Closed Interval

Compute $\mu^*([2,5])$. For every $\delta>0$, the single open interval $(2-\delta,5+\delta)$ covers $[2,5]$, and its content is its length. Therefore

\mu^*([2,5])\leq c((2-\delta,5+\delta))=(5+\delta)-(2-\delta)=3+2\delta.

Since this holds for every $\delta>0$, $\mu^*([2,5])\leq3$. For the lower bound, let $(U_n)$ be any countable open cover of $[2,5]$. Since $[2,5]$ is compact, finitely many members of this cover, say $U_{n_1},\ldots,U_{n_r}$, still cover it. For these open sets, $c(U_{n_i})=m(U_{n_i})$. Monotonicity and finite subadditivity of Lebesgue measure give

3=m([2,5])\leq m\left(\bigcup_{i=1}^{r}U_{n_i}\right) \leq\sum_{i=1}^{r}m(U_{n_i}) =\sum_{i=1}^{r}c(U_{n_i}) \leq\sum_{n=1}^{\infty}c(U_n).

This lower bound holds for every countable open cover, so taking the infimum gives $\mu^*([2,5])\geq3$. Therefore $\mu^*([2,5])=3$. Check: the result matches the interval length $5-2=3$; the upper bound $3+2\delta$ tends to $3$, and no open cover can have total content below $3$.

Worked Example: A Countable Dense Set Has Outer Measure Zero

Let $\mathbb{Q}\cap[0,1]=\{q_1,q_2,\ldots\}$, with each rational listed at least once. Fix any $\varepsilon>0$. For each $n\geq1$, take the open interval centered at $q_n$ with radius $\varepsilon/2^{n+1}$:

U_n=\left(q_n-\frac{\varepsilon}{2^{n+1}},q_n+\frac{\varepsilon}{2^{n+1}}\right).

The center $q_n$ belongs to $U_n$, so these intervals cover every point of $\mathbb{Q}\cap[0,1]$. Each interval has length $\varepsilon/2^n$, and hence

\sum_{n=1}^{\infty}c(U_n) =\sum_{n=1}^{\infty}\frac{\varepsilon}{2^n} =\varepsilon.

It follows that $\mu^*(\mathbb{Q}\cap[0,1])\leq\varepsilon$ for every $\varepsilon>0$. Since outer measures are nonnegative, $\mu^*(\mathbb{Q}\cap[0,1])=0$. Check: the total covering length is $\varepsilon\sum_{n=1}^{\infty}2^{-n}=\varepsilon$, which can be made arbitrarily small. Countability is essential to this particular construction because it lets us assign a summable sequence of interval lengths.

A Common Mistake: Confusing a Cover with the Infimum

Consider the two open intervals $U_1=(-1,3/5)$ and $U_2=(2/5,2)$. Their union covers $[0,1]$, and each has length $8/5$, so this cover has total content $16/5$. It would be wrong to conclude that $\mu^*([0,1])=16/5$: the definition takes the infimum over all covers, and this is only one cover. In fact, for every $\delta>0$, the single interval $(-\delta,1+\delta)$ covers $[0,1]$ and has content $1+2\delta$, giving $\mu^*([0,1])\leq1$. The lower bound from the interval-cover argument gives $\mu^*([0,1])\geq m([0,1])=1$, so the correct value is $1$.

A related domain mistake is to insert $c([0,1])$ into the definition. The content $c$ is defined for open sets, whereas $[0,1]$ is closed. The outer measure $\mu^*([0,1])$ is defined by open covers; it is not obtained by applying $c$ directly to the closed interval.

What the Construction Establishes

The construction produces an outer measure on every subset of $X$, without first choosing a sigma-algebra of measurable sets. Its value records the smallest total open-set content available from countable covers. The theorem proves exactly the outer-measure axioms; it does not yet establish that every subset satisfies the criterion needed for countable additivity. Nor does the fact that open sets supplied the covers, by itself, make every subset measurable. Those questions require a further measurability argument.

For the Lebesgue integral functional on $\mathbb{R}$, the open-set content is the length of open sets, so the construction recovers the expected outer sizes of intervals and assigns outer measure zero to countable sets. In the general locally compact Hausdorff setting, the same cover-and-infimum procedure is what extends the functional’s open-set information to a set function on all subsets.

Check Your Understanding

Use the definition of $\mu^*$ and the proof of its axioms to answer these questions.

Why does the empty cover imply $\mu^*(\varnothing)=0$, rather than merely an upper bound on this value?
In the countable subadditivity proof, why is it enough to consider the case $\sum_j\mu^*(A_j)<\infty$ after treating the infinite case?
For the Lebesgue integral functional, compute the content of $(4,9)$ using the upper bound and compactly supported cutoffs.
Give an open cover of $[0,1]$ whose total content is $1+2\delta$, where $\delta>0$, and explain why this proves an upper bound rather than an equality by itself.
Cover an enumeration $(r_n)$ of a countable subset of $\mathbb{R}$ by intervals of lengths $\varepsilon/2^n$. What does this show about its outer measure?