From Box Covers to an Outer Measure
A box-cover construction assigns a size to every subset of \(\mathbb{R}^n\), whether or not that subset is measurable. The idea is to cover a set by boxes, add their volumes, and take the infimum over all such covers. The infimum matters: a cover with a small total volume gives an upper bound, but it need not be the best cover, and there need not be a cover whose cost equals the infimum. The resulting set function is an outer measure. Here we define it and prove its three basic properties: its value at the empty set, monotonicity, and countable subadditivity.
Fix a positive integer \(n\). A half-open box in \(\mathbb{R}^n\) is a set \(Q=\prod_{i=1}^n[a_i,b_i)\), where \(a_i,b_i\in\mathbb{R}\) and \(a_i<b_i\) for each \(i\). Its volume is the product of its side lengths: \(\operatorname{vol}(Q)=\prod_{i=1}^n(b_i-a_i)\). The half-open convention includes each left endpoint and excludes each right endpoint. In particular, a point whose coordinate is \(b_i\) is not in this box.
The infimum is taken in \([0,\infty]\). There is at least one countable box cover of every \(E\subseteq\mathbb{R}^n\): the boxes \(\prod_{i=1}^n[z_i,z_i+1)\), with \((z_1,\ldots,z_n)\in\mathbb{Z}^n\), cover all of \(\mathbb{R}^n\). Their total cost is infinite, but they ensure the collection of admissible covers is nonempty. This construction defines \(m^*\) on all subsets, not just Borel or Lebesgue measurable sets. At this stage, \(m^*\) is an outer measure, not a measure on a sigma-algebra.
Three Basic Properties
- \(m^*(\varnothing)=0\).
- If \(E\subseteq F\), then \(m^*(E)\le m^*(F)\).
- \(m^*(\bigcup_{j=1}^{\infty}E_j)\le\sum_{j=1}^{\infty}m^*(E_j)\), with the right-hand side allowed to be \(+\infty\).
Proof. For the first property, the empty family covers \(\varnothing\) with cost \(0\), so \(m^*(\varnothing)\le 0\). Every cover has nonnegative cost, and therefore its infimum is nonnegative. Thus \(m^*(\varnothing)=0\).
For monotonicity, suppose \(E\subseteq F\). Every box cover of \(F\) also covers \(E\). Consequently, the covers available in the infimum defining \(m^*(E)\) include all covers available for \(m^*(F)\). Taking infima gives \(m^*(E)\le m^*(F)\). This reasoning also covers the case \(E=\varnothing\) and the case \(m^*(F)=+\infty\).
For countable subadditivity, write \(S=\sum_{j=1}^{\infty}m^*(E_j)\). If \(S=+\infty\), the claimed inequality holds because every outer measure is at most \(+\infty\). Suppose instead that \(S<+\infty\). Then each \(m^*(E_j)\) is finite. Fix \(\varepsilon>0\). By the definition of infimum, for each \(j\ge1\) there is a countable box cover \((Q_{j,k})_k\) of \(E_j\) such that $$ \sum_k\operatorname{vol}(Q_{j,k})<m^*(E_j)+\frac{\varepsilon}{2^j}. $$ The strict inequality is available because \(m^*(E_j)\) is finite and \(\varepsilon/2^j>0\). If \(E_j=\varnothing\), the empty cover is allowed and also satisfies this inequality.
The boxes from all these covers together form a countable family: they are indexed by pairs of positive integers, or by a subset of such pairs if some covers are finite. Their union contains every \(E_j\), and hence contains \(\bigcup_jE_j\). Their total cost, obtained by summing nonnegative terms, satisfies $$ \sum_{j=1}^{\infty}\sum_k\operatorname{vol}(Q_{j,k}) <\sum_{j=1}^{\infty}m^*(E_j)+\sum_{j=1}^{\infty}\frac{\varepsilon}{2^j} =S+\varepsilon, $$ since \(\sum_{j=1}^{\infty}2^{-j}=1\). It follows from the infimum defining outer measure that \(m^*(\bigcup_jE_j)\le S+\varepsilon\). This holds for every \(\varepsilon>0\). If \(m^*(\bigcup_jE_j)>S\), choosing \(\varepsilon\) smaller than the positive difference would contradict that inequality. Therefore \(m^*(\bigcup_jE_j)\le S\), as required. \(\square\)
The key step in the last part is not that each set has a cover of exactly minimal cost; an infimum need not be attained. Instead, each set gets a cover within a chosen positive error of its own infimum. The errors must have finite total, which is why the allocation \(\varepsilon/2^j\) is useful.
Worked Consequences of the Definition
Worked Example: A Singleton Has Outer Measure Zero
Fix \(x=(x_1,\ldots,x_n)\in\mathbb{R}^n\). For each positive integer \(k\), the box \(Q_k=\prod_{i=1}^n[x_i,x_i+1/k)\) contains \(x\): every coordinate \(x_i\) is included as a left endpoint. Its volume is $$ \operatorname{vol}(Q_k)=\prod_{i=1}^n\frac1k=\frac1{k^n}. $$ Thus \(m^*(\{x\})\le 1/k^n\) for every \(k\). For any \(\varepsilon>0\), choose an integer \(k>\varepsilon^{-1/n}\). Then \(1/k^n<\varepsilon\), so \(m^*(\{x\})\le\varepsilon\) for every positive \(\varepsilon\). Since outer measure is nonnegative, \(m^*(\{x\})=0\).
Check: in \(\mathbb{R}^2\), taking \(k=2\) gives a square of side \(1/2\) and area \((1/2)(1/2)=1/4\); taking \(k=4\) gives area \((1/4)(1/4)=1/16\). These costs decrease to zero, consistent with the infimum being zero.
Worked Example: Covering the Rational Points in the Unit Interval
Let \(E=\mathbb{Q}\cap[0,1]\), where \(\mathbb{Q}\) is the set of rational numbers, and enumerate its points without repetition as \(q_1,q_2,\ldots\). For \(\varepsilon>0\), cover \(q_j\) by the interval \(I_j=[q_j,q_j+\varepsilon/2^j)\). Its left endpoint is included, so \(q_j\in I_j\), including when \(q_j=1\). The intervals may extend past \(1\); that causes no problem for a cover. Their total length is $$ \sum_{j=1}^{\infty}\frac{\varepsilon}{2^j} =\varepsilon\sum_{j=1}^{\infty}\frac1{2^j} =\varepsilon. $$ For example, when \(\varepsilon=1/4\), the first four lengths are \(1/8,1/16,1/32,1/64\); their sum is \(15/64\), and the full sum is \(1/4=16/64\). Thus \(m^*(E)\le\varepsilon\) for every \(\varepsilon>0\), which implies \(m^*(E)=0\).
Check: the conclusion uses both parts of the definition: the displayed intervals are an admissible countable cover, and their total cost can be made smaller than any prescribed positive number. A single cover of cost \(1/4\) alone would show only \(m^*(E)\le1/4\), not that the outer measure is zero.
Worked Example: A Cover of the Closed Unit Square
In \(\mathbb{R}^2\), let \(E=[0,1]\times[0,1]\). For each positive integer \(k\), use the single box \(Q_k=[0,1+1/k)\times[0,1+1/k)\). Every coordinate of every point in \(E\) is at least \(0\) and at most \(1\), and \(1<1+1/k\), so \(E\subseteq Q_k\). The box has side length \(1+1/k\), giving the upper bound $$ m^*(E)\le \operatorname{vol}(Q_k) =\left(1+\frac1k\right)^2 =1+\frac2k+\frac1{k^2}. $$ For \(k=1\) this cost is \(4\); for \(k=2\) it is \(9/4\); and for \(k=4\) it is \(25/16\). As \(k\to\infty\), the cost tends to \(1\). Hence for every \(\varepsilon>0\), a sufficiently large \(k\) gives a cover of cost less than \(1+\varepsilon\), and therefore \(m^*(E)\le1\).
Check: the right endpoint \(1+1/k\) is excluded from \(Q_k\), but the points with coordinate \(1\) are still included because \(1<1+1/k\). This calculation supplies an upper bound only; a cover construction by itself does not establish a matching lower bound.
Why the Error Budget Matters
A common mistake in proving countable subadditivity is to give every set the same error allowance. For a concrete illustration, take \(E_j=\{j\}\subseteq\mathbb{R}\), for \(j\ge1\). Covering each \(E_j\) by \([j,j+\varepsilon)\) gives a valid cover of the union, but its total length is \(\sum_{j=1}^{\infty}\varepsilon=+\infty\) for every fixed \(\varepsilon>0\). It is incorrect to treat that infinite total as a useful bound of \(+\varepsilon\).
The correction is to give the \(j\)-th set the length \(\varepsilon/2^j\). Then \([j,j+\varepsilon/2^j)\) contains \(j\), and the combined cost is $$ \sum_{j=1}^{\infty}\frac{\varepsilon}{2^j}=\varepsilon. $$ Since this works for every \(\varepsilon>0\), the union has outer measure zero. The same allocation principle is what makes the general countable-subadditivity proof work.
These three properties are the defining structural features needed of an outer measure: a zero value at the empty set, monotonicity, and countable subadditivity. They do not say that the outer measure is additive on disjoint sets, nor do they make every subset measurable. Outer measure is deliberately defined first on all subsets; a separate measurability condition is needed before restricting it to obtain a measure.
Check Your Understanding
For each question, distinguish a cover's cost from the infimum over all covers.
- Why is the empty family an admissible cover of \(\varnothing\), and why does nonnegativity of all cover costs matter for the value of \(m^*(\varnothing)\)?
- If \(E\subseteq F\), explain why every box cover of \(F\) is also a box cover of \(E\). Which direction does the resulting inequality between outer measures take?
- In the proof of countable subadditivity, why is a single fixed error \(\varepsilon\) for every \(E_j\) not a suitable error allocation?
- For a point \(x\in\mathbb{R}^n\), compute the volume of \(\prod_{i=1}^n[x_i,x_i+1/k)\) and explain how that gives \(m^*(\{x\})=0\).
- Why does a box cover of \([0,1]^2\) with cost approaching \(1\) establish an upper bound but not, by itself, establish equality?