Monotonicity of Measures

Set Inclusion Orders Measure

A measure assigns a nonnegative size to each measurable set. The basic relationship between set inclusion and size is simple: a subset cannot have greater measure than the set containing it. This is the monotonicity property established in “Measures.” Here we examine what that inequality does—and does not—tell us, especially when a larger set adds only a null set or when the sets already have infinite measure.

Throughout, let $(X,\mathcal{F},\mu)$ be a measure space. If $A,B\in\mathcal{F}$ and $A\subseteq B$, monotonicity gives $\mu(A)\leq\mu(B)$, with values understood in $[0,\infty]$. This is an order comparison, not necessarily a strict one. To understand whether the measure actually increases, we look at the part added to $A$, namely $B\setminus A$.

Definition: A measure is monotone if $A\subseteq B$ for measurable sets implies $\mu(A)\leq\mu(B)$. A measure is strictly increasing along a particular inclusion $A\subseteq B$ if $\mu(A)<\mu(B)$. Strict increase along one inclusion is not automatic from strict set inclusion.

The monotonicity theorem is already available as a basic property of measures; it will be used here rather than reproved. The key additional tool is an exact formula for the increment. Because $B\setminus A$ is measurable and disjoint from $A$, the set $B$ is the disjoint union of these two pieces.

Theorem (Increment Formula for Nested Sets): If $A,B\in\mathcal{F}$ and $A\subseteq B$, then $$ \mu(B)=\mu(A)+\mu(B\setminus A). $$

Proof. Since $\mathcal{F}$ is a sigma-algebra, it is closed under relative differences, so $B\setminus A\in\mathcal{F}$. The sets $A$ and $B\setminus A$ are disjoint: an element of $A$ cannot at the same time lie outside $A$. Their union is $B$, since each element of $B$ either belongs to $A$ or does not. Countable additivity, applied to these two disjoint sets (and with empty sets added if needed), therefore gives $$ \mu(B)=\mu(A)+\mu(B\setminus A). $$ This equality uses addition in the extended nonnegative reals, where $a+\infty=\infty$ for every $a\in[0,\infty]$. $\square$

The formula explains monotonicity: the added piece has nonnegative measure, so adjoining it cannot lower the measure. More importantly, it identifies exactly what controls a strict increase. When $\mu(A)$ is finite, a positive measure for the added piece forces the measure to increase; an added piece of measure zero leaves the measure unchanged.

When Does a Strict Inclusion Give a Strict Increase?

Theorem (Strict Increase for a Positive Increment): Let $A,B\in\mathcal{F}$ with $A\subseteq B$ and $\mu(A)<\infty$. Then $$ \mu(A)<\mu(B) \quad\Longleftrightarrow\quad \mu(B\setminus A)>0. $$

Proof. The increment formula gives $\mu(B)=\mu(A)+\mu(B\setminus A)$. If $\mu(B\setminus A)>0$, then adding this positive value to the finite number $\mu(A)$ gives a value strictly greater than $\mu(A)$. This remains true if $\mu(B\setminus A)=\infty$, since then $\mu(B)=\infty>\mu(A)$. Conversely, if $\mu(B\setminus A)=0$, the formula gives $\mu(B)=\mu(A)+0=\mu(A)$. Thus strict inequality holds exactly when the increment has positive measure. $\square$

The finiteness hypothesis is essential to this equivalence. If $\mu(A)=\infty$, then every measurable superset $B$ also has infinite measure by monotonicity, regardless of the measure of $B\setminus A$. Even adding a piece of positive measure—or an infinite-measure piece—does not make the numerical value larger than infinity.

Worked Example: Nested Intervals with a Positive Increment

Let $\mu$ be Lebesgue measure on the Borel subsets of $\mathbb{R}$, and take $A=[-2,1]$ and $B=[-2,4]$. Then $A\subseteq B$, and the added set is $B\setminus A=(1,4]$. The endpoint convention does not affect the interval lengths, so $$ \mu(A)=1-(-2)=3,\qquad \mu(B\setminus A)=4-1=3. $$ The increment formula gives $$ \mu(B)=3+3=6, $$ which agrees with the direct calculation $4-(-2)=6$. The inclusion is strict and the added set has positive measure, so the measure strictly increases.

Worked Example: Strict Inclusion with No Increase

Again use Lebesgue measure, and let $A=[2,5]$ and $B=[2,5]\cup\{8\}$. These are Borel sets and $A\subsetneq B$, since $8\in B$ but $8\notin A$. The added set is the singleton $B\setminus A=\{8\}$, whose Lebesgue measure is zero. Thus $$ \mu(A)=5-2=3,\qquad \mu(B\setminus A)=0, $$ and the increment formula gives $$ \mu(B)=3+0=3. $$ So strict inclusion alone does not imply a strict increase in measure. What matters is whether the added portion has positive measure.

Equal Measures and Null Differences

For nested sets with a finite-measure smaller set, the increment formula gives a complete test for equality. The larger set has the same measure precisely when the difference between the sets is a null set. Here a measurable set $N$ is called a null set if $\mu(N)=0$. A null difference need not be empty: the preceding example added one point, but other null sets can be infinite.

Corollary (Equality for Nested Sets of Finite Measure): If $A\subseteq B$, $A,B\in\mathcal{F}$, and $\mu(A)<\infty$, then $$ \mu(A)=\mu(B) \quad\Longleftrightarrow\quad \mu(B\setminus A)=0. $$

Proof. If $\mu(B\setminus A)=0$, the increment formula immediately gives $\mu(B)=\mu(A)+0=\mu(A)$. For the converse, suppose $\mu(B)=\mu(A)<\infty$. The increment formula says $$ \mu(A)=\mu(A)+\mu(B\setminus A). $$ If $\mu(B\setminus A)>0$, the right side would be strictly greater than the finite number $\mu(A)$; if $\mu(B\setminus A)=\infty$, the right side would be infinite. Both contradict the displayed equality to a finite number. The only possibility is $\mu(B\setminus A)=0$. $\square$

The finite-measure assumption again prevents an invalid inference. If $\mu(A)=\mu(B)=\infty$, equal measure does not imply a null difference. The increment formula still holds, but $\infty+\mu(B\setminus A)=\infty$ for every possible value of the increment. In that case the equality of the two measures gives no information about the size of the added part.

Worked Example: Equal Infinite Measures with a Nonzero Difference

Take $X=\mathbb{Z}$ with the power-set sigma-algebra and counting measure, which assigns to a set its number of elements, with value $\infty$ for an infinite set. Let $A=\{1,2,3,\ldots\}$ and $B=\mathbb{Z}$. Then $A\subsetneq B$, and both sets are infinite, so $$ \mu(A)=\infty,\qquad \mu(B)=\infty. $$ However, $B\setminus A=\{\ldots,-2,-1,0\}$ is also infinite, so $\mu(B\setminus A)=\infty$. The increment formula reads $$ \mu(B)=\mu(A)+\mu(B\setminus A)=\infty+\infty=\infty. $$ The equal measures do not mean the difference is null. They simply show that adding more points to an already infinite counting-measure set does not change its measure value.

Reading Monotonicity Carefully

Monotonicity is often used to compare a complicated set with a simpler subset or superset. The direction of the inequality follows the direction of inclusion: if $A\subseteq B$, then $\mu(A)\leq\mu(B)$. Reversing that direction is a common error. For example, knowing that a set contains another does not make its measure smaller.

There are also limits to what the inequality says. If $\mu(A)=\mu(B)$, it does not follow in general that $A=B$. Under the finite-measure hypothesis for the smaller set, equality says only that the difference $B\setminus A$ is null. And if both measures are infinite, equality may hold even when the difference has infinite measure. Thus set equality, equality of measures, and equality up to a null difference are distinct statements.

Check inclusion.
For measurable $A\subseteq B$, monotonicity gives $\mu(A)\leq\mu(B)$.

Identify the increment.
Use $B\setminus A$ and the formula $\mu(B)=\mu(A)+\mu(B\setminus A)$.

Check whether the smaller measure is finite.
If $\mu(A)<\infty$, the increment determines whether the inequality is strict and whether equality means a null difference.

Treat infinite values separately.
If $\mu(A)=\infty$, every measurable superset has measure $\infty$, and equality alone says nothing about the difference.

The increment formula is useful because it replaces a vague comparison by a decomposition into disjoint measurable pieces. It also guards against treating infinity as an ordinary real number: one may add nonnegative extended values, but one cannot cancel an infinite term or infer the size of an increment from $\infty=\infty+\text{increment}$.

In applications, the distinction between set inclusion and measure increase is especially important when adding exceptional points or other null sets. A set may become strictly larger without gaining any measure. Conversely, when the added part has positive measure and the original set has finite measure, the increase is guaranteed. These conclusions follow directly from the decomposition, rather than from the visual or set-theoretic size of the sets alone.

Check Your Understanding

Use the increment formula and the monotonicity results to answer the following questions.

If $A\subseteq B$, what inequality does monotonicity give for $\mu(A)$ and $\mu(B)$?
Write the increment formula for nested measurable sets and explain which two sets form the disjoint decomposition.
Suppose $\mu(A)<\infty$ and $A\subseteq B$. What condition on $B\setminus A$ is equivalent to $\mu(A)<\mu(B)$?
For nested sets with finite $\mu(A)$, what does $\mu(A)=\mu(B)$ tell us about $B\setminus A$?
Why can equal infinite measures fail to imply that the difference of nested sets has measure zero?

What You'll Learn