Monotonicity of Outer Measure

Containment Transfers Covers in One Direction

Lebesgue outer measure is defined by minimizing the total length of interval covers. If $A\subseteq B$, every cover of $B$ is also a cover of $A$. This inclusion of cover families is the whole mechanism behind monotonicity: minimizing over the larger family of covers cannot give a larger infimum. The direction matters. A cover of $A$ need not cover $B$, so the corresponding statement about cover families goes only one way.

We work with arbitrary sets $A,B\subseteq\mathbb{R}$, whether or not they are Borel or Lebesgue measurable. The quantity in question is Lebesgue outer measure $m^*$, not the Lebesgue measure $m$ restricted to measurable sets. No measurability, boundedness, or finiteness assumption on $A$ or $B$ is needed.

Definition: An interval cover of $E\subseteq\mathbb{R}$ is a finite or countable family of open intervals whose union contains $E$. The length of a bounded open interval $(a,b)$ is $b-a$; an unbounded open interval has length $+\infty$. The empty family is an interval cover of $\varnothing$, with total length $0$. The Lebesgue outer measure of $E$ is $$ m^*(E)=\inf\left\{\sum_j |I_j|:\ E\subseteq\bigcup_j I_j,\ \ I_j\text{ open intervals}\right\}, $$ where the infimum and sums take values in $[0,+\infty]$.

Every set has at least one admissible cover: the single interval $\mathbb{R}$ covers it, with total length $+\infty$. Thus the infimum is taken over a nonempty family. It may be finite or infinite. We will use the defining property of an infimum: if the infimum $c$ is finite, then for each $\varepsilon>0$ some admissible value is less than $c+\varepsilon$. The infimum need not itself be the cost of any cover.

The Cover-Family Inclusion

Lemma: For $E\subseteq\mathbb{R}$, let $\mathcal{C}(E)$ be the family of interval covers of $E$. If $A\subseteq B$, then $\mathcal{C}(B)\subseteq\mathcal{C}(A)$.

Proof. Assume $A\subseteq B$, and take any cover $(I_j)\in\mathcal{C}(B)$. By the definition of a cover, each $x\in B$ belongs to at least one $I_j$. Every $x\in A$ also belongs to $B$, so it belongs to at least one of the same intervals. Therefore $(I_j)$ covers $A$, and hence $(I_j)\in\mathcal{C}(A)$. This proves $\mathcal{C}(B)\subseteq\mathcal{C}(A)$. If $B$ is empty, then $A$ is empty as well, and the same argument holds. $\square$

Since the covers of $B$ are among the covers of $A$, the set of costs over which $m^*(A)$ takes its infimum includes all the costs available for $m^*(B)$, and possibly more. The next theorem makes that comparison precise, including when one of the values is infinite.

Theorem (Monotonicity of Outer Measure): For all $A,B\subseteq\mathbb{R}$, if $A\subseteq B$, then $m^*(A)\le m^*(B)$. The inequality is in the extended nonnegative real numbers and does not require measurability.

Proof. Suppose first that $m^*(B)=+\infty$. Since $m^*(A)\in[0,+\infty]$, it follows directly that $m^*(A)\le+\infty=m^*(B)$.

Now suppose $m^*(B)<+\infty$. Fix $\varepsilon>0$. By the defining property of the infimum, there is an interval cover $(I_j)$ of $B$ whose total length satisfies $$ \sum_j |I_j|<m^*(B)+\varepsilon. $$ Because $A\subseteq B$, the same intervals cover $A$, by the lemma. The definition of $m^*(A)$ therefore gives $$ m^*(A)\le \sum_j |I_j|<m^*(B)+\varepsilon. $$ This holds for every $\varepsilon>0$. If $m^*(A)>m^*(B)$, choosing $\varepsilon=(m^*(A)-m^*(B))/2>0$ would contradict the displayed inequality. Thus $m^*(A)\le m^*(B)$. This argument includes $m^*(B)=0$: the cover can be chosen with total length less than any prescribed positive $\varepsilon$. $\square$

Worked Comparisons

Worked Example: Two Nested Intervals

Let $A=[0,1]$ and $B=(-2,3)$. First check the containment: for every $x\in[0,1]$, we have $-2<0\le x\le1<3$, so $x\in(-2,3)$ and $A\subseteq B$.

The single interval $I=(-2,3)$ covers $B$, and its length is $3-(-2)=5$. It also covers $A$. Thus the definition gives $m^*(B)\le5$ and $m^*(A)\le5$, while the theorem gives the stronger comparison $m^*(A)\le m^*(B)$. These statements do not compute either outer measure exactly.

As a separate cover of $A$, take $J=(-1,2)$. It covers every $x\in[0,1]$, since $-1<0\le x\le1<2$, and its length is $2-(-1)=3$. Hence $m^*(A)\le3$. This does not show $m^*(B)\le3$: indeed, $J$ does not cover $B$, since $-\tfrac32\in(-2,3)$ but $-\tfrac32\notin(-1,2)$. Check: the valid transferred cover is $(-2,3)$, whose length is $3-(-2)=5$, and it covers both sets.

Worked Example: A Disconnected Set Inside an Interval

Let $A=[0,1]\cup[2,3]$ and $B=(-1,4)$. For points in the first interval, $-1<0\le x\le1<4$; for points in the second, $-1<2\le x\le3<4$. Thus $A\subseteq B$.

The interval $(-1,4)$ covers $B$, and its length is $4-(-1)=5$. It therefore covers $A$ too, so $m^*(A)\le m^*(B)\le5$. There is also a two-interval cover of $A$: $$ [0,1]\subset(-\tfrac12,\tfrac32),\qquad [2,3]\subset(\tfrac32,\tfrac72). $$ The first interval has length $\tfrac32-(-\tfrac12)=2$, and the second has length $\tfrac72-\tfrac32=2$. Their total length is $2+2=4$, giving $m^*(A)\le4$. This separate cover is not needed for monotonicity; it illustrates that different admissible covers can give different upper bounds. Check: the intervals cover the two pieces respectively, and the computed total is $4$, while the cover of $B$ has length $5$.

Worked Example: Using an Epsilon-Optimal Cover

Let $A=(-\tfrac34,\tfrac14)$ and $B=(-1,\tfrac12)$. If $x\in A$, then $-1<-\tfrac34<x<\tfrac14<\tfrac12$, so $x\in B$. Hence $A\subseteq B$. The interval $B$ itself is a cover of $B$, of length $\tfrac12-(-1)=\tfrac32$. In particular, $m^*(B)\le\tfrac32<+\infty$.

Take $\varepsilon=\tfrac1{10}$. The infimum property supplies a cover of $B$ with total length $L<m^*(B)+\tfrac1{10}$. Since $m^*(B)\le\tfrac32$, this cover also satisfies $L<\tfrac32+\tfrac1{10}=\tfrac85$. It covers $A$, so $m^*(A)\le L<\tfrac85$. More generally, taking covers with cost less than $m^*(B)+\varepsilon$ for every $\varepsilon>0$ gives $m^*(A)\le m^*(B)$, as in the theorem. Check: $\tfrac32+\tfrac1{10}=\tfrac{15}{10}+\tfrac1{10}=\tfrac{16}{10}=\tfrac85$; the bound follows from the stated cover cost, not from assuming that an optimal cover exists.

Worked Example: The Empty Set

Let $A=\varnothing$ and $B=(0,1)$. Then $A\subseteq B$. By convention the empty family covers $\varnothing$ and has total length $0$. All cover costs are nonnegative, so their infimum is at least $0$; the empty cover attains $0$. Therefore $m^*(A)=0$.

The interval $(0,1)$ covers $B$ and has length $1-0=1$. Consequently $0=m^*(A)\le m^*(B)\le1$. Check: the lower bound is the monotonicity theorem, and the upper bound comes from the displayed interval cover. This example does not require a prior calculation of $m^*((0,1))$.

Nested Sets and a Common Direction Error

Corollary: If $A_1\subseteq A_2\subseteq\cdots$ are subsets of $\mathbb{R}$, then $m^*(A_1)\le m^*(A_2)\le\cdots$. The values may be finite or infinite; the conclusion is only that this numerical sequence is nondecreasing.

Proof. For each positive integer $n$, the assumed containment $A_n\subseteq A_{n+1}$ allows application of the monotonicity theorem with $A=A_n$ and $B=A_{n+1}$. It gives $m^*(A_n)\le m^*(A_{n+1})$. This holds for every $n$, which proves the claim. $\square$

Common mistake: reversing the cover implication. A tempting but incorrect argument says that if $A\subseteq B$, then every cover of $A$ is a cover of $B$. The example $A=[0,1]$, $B=(-2,3)$ disproves this: $(-1,2)$ covers all of $A$, since $-1<0\le x\le1<2$, but does not cover $B$, since $-\tfrac32\in B$ and $-\tfrac32\notin(-1,2)$. The correct implication is that every cover of $B$ covers $A$. Therefore $\mathcal{C}(B)\subseteq\mathcal{C}(A)$, and the infimum over $\mathcal{C}(A)$ is no greater than the infimum over $\mathcal{C}(B)$.

Monotonicity gives an ordering, not an exact evaluation: an explicit cover provides an upper bound, and a smaller set cannot have greater outer measure than a containing set. Neither statement says that a least-cost cover exists. The epsilon argument is what lets the proof work even when the infimum is approached but not attained. Also, the nested-set corollary does not assert that the outer measure of a union equals the limit of the outer measures; it establishes only the term-by-term ordering proved above.

Check Your Understanding

For each question, identify the containment and the direction in which an interval cover transfers.

If $A\subseteq B$, which family is contained in which: $\mathcal{C}(A)\subseteq\mathcal{C}(B)$ or $\mathcal{C}(B)\subseteq\mathcal{C}(A)$? Explain from the definition of a cover.
Let $A=[1,2]$ and $B=(0,3)$. Verify $A\subseteq B$, compute the length of the interval $B$, and state the resulting upper bound for $m^*(A)$.
Why does an interval cover of a set $A$ not necessarily give any cover of a larger set $B\supseteq A$? Give a concrete example.
In the epsilon proof, why can one choose a cover of $B$ of cost less than $m^*(B)+\varepsilon$ when $m^*(B)$ is finite?
If $A_n\subseteq A_{n+1}$ for every $n$, what ordering follows for $m^*(A_n)$, and what additional claim does monotonicity alone not establish?

What You'll Learn