Monotonicity And Other Immediate Consequences

Three Basic Consequences of the Measure Axioms

The measure axioms give more than a rule for countable disjoint unions. They also determine how measures behave when one measurable set lies inside another. These elementary consequences are used throughout measure theory, but their statements require care: a measure may take the value \(+\infty\), and subtraction involving \(+\infty\) is not generally defined.

Throughout, \((X,\mathcal{M},\mu)\) is a measure space. Thus \(\mathcal{M}\) is a sigma-algebra of subsets of \(X\), and \(\mu:\mathcal{M}\to[0,\infty]\) satisfies \(\mu(\varnothing)=0\) and countable additivity on pairwise disjoint measurable sets. No assumption that \(\mu\) is finite or sigma-finite is made. All sets appearing below are in \(\mathcal{M}\). We write \(A\setminus B\) for the points in \(A\) that are not in \(B\); this set is measurable because \(\mathcal{M}\) is closed under complements and intersections.

Definition: A measure \(\mu\) on \((X,\mathcal{M})\) assigns a value in \([0,\infty]\) to each measurable set, assigns \(0\) to \(\varnothing\), and is countably additive: if \(E_1,E_2,\ldots\) are pairwise disjoint measurable sets, then \(\mu(\bigcup_{n=1}^{\infty}E_n)=\sum_{n=1}^{\infty}\mu(E_n)\). The sum is a nonnegative extended sum and may equal \(+\infty\).

Under this standard definition, \(\mu(\varnothing)=0\) is one of the axioms. It is useful to see why this normalization matters. If one states only countable additivity and omits the normalization, then countable additivity alone need not force the value at the empty set to be zero: the identically \(+\infty\) set function satisfies countable additivity but assigns \(+\infty\) to \(\varnothing\). However, if a countably additive set function has some measurable set \(E\) with finite value, then the disjoint decomposition \(E=E\cup\varnothing\) gives \(\mu(E)=\mu(E)+\mu(\varnothing)\), which forces \(\mu(\varnothing)=0\). The standard measure axiom avoids this exceptional issue by specifying the value directly.

Finite Additivity and Monotonicity

Countable additivity immediately gives the two-set version for disjoint measurable sets. If \(A,B\in\mathcal{M}\) and \(A\cap B=\varnothing\), apply countable additivity to the sequence \(A,B,\varnothing,\varnothing,\ldots\). The normalization \(\mu(\varnothing)=0\) gives \(\mu(A\cup B)=\mu(A)+\mu(B)\), including when one or both measures are infinite.

Theorem (Monotonicity): For any measure space \((X,\mathcal{M},\mu)\) and measurable sets \(A,B\in\mathcal{M}\), if \(A\subseteq B\), then \(\mu(A)\le\mu(B)\). The measure need not be finite.

Proof. Since \(A\subseteq B\), the sets \(A\) and \(B\setminus A\) are disjoint and \(B=A\cup(B\setminus A)\). Both pieces are measurable. Finite additivity therefore gives \(\mu(B)=\mu(A)+\mu(B\setminus A)\). Since \(\mu(B\setminus A)\ge0\), addition in \([0,\infty]\) implies \(\mu(A)\le\mu(B)\). This also covers the case \(\mu(B)=+\infty\): every value in \([0,\infty]\) is at most \(+\infty\). \(\square\)

Monotonicity says that enlarging a measurable set cannot lower its measure. It does not say that strict inclusion forces a strict increase: the added points may have measure zero. Nor does it require either set to have finite measure.

Worked Example: Comparing Finite Sets with Counting Measure

Let \(X=\{1,2,3,4,5,6\}\), let \(\mathcal{M}\) be the collection of all subsets of \(X\), and let \(\mu\) be counting measure, so \(\mu(E)\) is the number of elements of \(E\). Take \(A=\{1,3,5\}\) and \(B=\{1,2,3,5,6\}\). Each element of \(A\) belongs to \(B\), so \(A\subseteq B\). Counting the elements gives \(\mu(A)=3\) and \(\mu(B)=5\), hence \(3\le5\).

The difference is \(B\setminus A=\{2,6\}\), which has measure \(2\). The disjoint pieces \(A\) and \(B\setminus A\) contain, respectively, \(3\) and \(2\) elements, so their measures add to \(3+2=5=\mu(B)\). Check: this recovers \(\mu(B)=5\) directly from the decomposition and confirms the monotonicity comparison.

Additivity for Nested Sets and the Limits of Subtraction

Theorem (Nested-set decomposition): Let \((X,\mathcal{M},\mu)\) be any measure space, and let \(A,B\in\mathcal{M}\) with \(A\subseteq B\). Then \(\mu(B)=\mu(A)+\mu(B\setminus A)\). This identity is valid whether the measures are finite or infinite.

Proof. The inclusion \(A\subseteq B\) implies \(A\cap(B\setminus A)=\varnothing\) and \(A\cup(B\setminus A)=B\). The two sets in the union are measurable. Applying finite additivity to these disjoint sets yields \(\mu(B)=\mu(A)+\mu(B\setminus A)\). No subtraction has been used, so the equality remains meaningful if a term equals \(+\infty\). \(\square\)

This identity is the safe form of the result. It can be rearranged as \(\mu(B\setminus A)=\mu(B)-\mu(A)\) when \(\mu(A)<\infty\): if \(\mu(B)<\infty\), ordinary subtraction applies; if \(\mu(B)=+\infty\), the identity forces \(\mu(B\setminus A)=+\infty\), which agrees with \(+\infty\) minus a finite number. If \(\mu(A)=+\infty\), then monotonicity gives \(\mu(B)=+\infty\) as well. In that situation, the decomposition still holds, but it does not determine the measure of \(B\setminus A\) by subtracting the two infinite values.

Worked Example: Nested Intervals with Lebesgue Measure

Consider \((\mathbb{R},\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. Let \(A=(0,2)\) and \(B=(0,5)\). These measurable sets satisfy \(A\subseteq B\), and their measures are their lengths: \(m(A)=2\) and \(m(B)=5\).

The relative complement is \(B\setminus A=[2,5)\), whose Lebesgue measure is \(5-2=3\). The disjoint decomposition is \(B=(0,2)\cup[2,5)\), so finite additivity gives \(m(B)=2+3=5\). Check: the half-open interval \([2,5)\) has length \(5-2=3\), and adding this to \(m(A)=2\) reproduces \(m(B)=5\). The endpoint \(2\) is included in the difference, but a singleton has Lebesgue measure zero, so the endpoint convention does not change the computed length.

Worked Example: Nested Sets of Infinite Counting Measure

Let \(X=\mathbb{N}=\{1,2,3,\ldots\}\), with its full sigma-algebra and counting measure \(\mu\). Set \(A=\{2,4,6,\ldots\}\) and \(B=\mathbb{N}\). Then \(A\subseteq B\). Both sets are infinite, so \(\mu(A)=+\infty\) and \(\mu(B)=+\infty\). The relative complement is \(B\setminus A=\{1,3,5,\ldots\}\), also infinite, and therefore \(\mu(B\setminus A)=+\infty\).

The nested-set identity reads \(+\infty=+\infty+(+\infty)\), which is valid in the extended nonnegative numbers. It does not yield a subtraction formula: \(+\infty-(+\infty)\) is undefined, not zero. Check: the even and odd positive integers are disjoint and together contain every positive integer; each part has infinitely many elements, so their measures add to the counting measure of \(B\), namely \(+\infty\).

A Common Error: Subtracting Infinite Measures

A tempting but invalid calculation is to take the preceding example and write \(\mu(B\setminus A)=\mu(B)-\mu(A)=+\infty-(+\infty)=0\). The final equality is not an arithmetic rule: \(+\infty-(+\infty)\) is undefined. It also contradicts the direct count, because \(B\setminus A\) contains all odd positive integers and thus has infinite counting measure. The correction is to retain the valid decomposition \(\mu(B)=\mu(A)+\mu(B\setminus A)\); here it says \(+\infty=+\infty+(+\infty)\), without assigning a value to an undefined difference.

For comparison, if \(A\subseteq B\) and \(\mu(A)\) is finite, the decomposition does determine the difference. For example, if \(\mu(A)=4\) and \(\mu(B)=+\infty\), then \(\mu(B\setminus A)\) cannot be finite: adding two finite nonnegative numbers would give a finite result, contrary to \(\mu(B)=+\infty\). Thus \(\mu(B\setminus A)=+\infty\), in agreement with subtracting the finite value \(4\) from \(+\infty\). The essential distinction is whether the quantity being subtracted is finite, not whether the containing set has finite measure.

Why These Consequences Matter

Monotonicity provides a basic way to bound the measure of a set by comparing it with a larger set whose measure is known. The nested-set identity gives an exact accounting of the portion added when a set is enlarged. It is valid on every measure space because it is only finite additivity applied to two disjoint pieces. Subtraction is more restrictive: it is a rearrangement that is safe when the smaller set has finite measure, but it cannot recover information from an expression involving infinity minus infinity.

Keep the three statements distinct. The normalization \(\mu(\varnothing)=0\) belongs to the standard measure axioms. Monotonicity follows from nonnegativity and the nested-set decomposition. The decomposition itself is always valid for nested measurable sets, while a subtraction formula needs the relevant finite-measure condition.

Check Your Understanding

For each question, state whether finiteness is needed and use the nested-set identity before attempting any subtraction.

In a measure space \((X,\mathcal{M},\mu)\), prove that if \(A\subseteq B\) are measurable, then \(\mu(A)\le\mu(B)\).
If \(A\subseteq B\), \(\mu(A)=3\), and \(\mu(B)=11\), determine \(\mu(B\setminus A)\) and justify the subtraction.
Can one determine \(\mu(B\setminus A)\) from \(\mu(A)=+\infty\) and \(\mu(B)=+\infty\) alone? Explain using the nested-set identity.
For counting measure on \(\mathbb{N}\), take \(A=\{n\in\mathbb{N}:n\ge4\}\) and \(B=\mathbb{N}\). Compute the measures of \(A\), \(B\), and \(B\setminus A\), and verify the additive decomposition.
Why does the standard definition of a measure specify \(\mu(\varnothing)=0\), rather than relying on countable additivity alone?