Definition of a Measure

The Definition of a Measure

A measurable space $(X,\mathcal{F})$ specifies the sets on which a size assignment is allowed to operate. The formal definition of a measure turns the guiding principle from the previous tutorial into three precise requirements: the domain is $\mathcal{F}$, the values are nonnegative and may be infinite, and the sizes of countably many pairwise disjoint sets add.

Definition: Let $(X,\mathcal{F})$ be a measurable space. A measure on $(X,\mathcal{F})$ is a function $\mu:\mathcal{F}\to[0,\infty]$ such that:

$\mu(\varnothing)=0$; and
for every sequence $A_1,A_2,\ldots$ of pairwise disjoint sets in $\mathcal{F}$, $$ \mu\left(\bigcup_{n=1}^{\infty} A_n\right)=\sum_{n=1}^{\infty}\mu(A_n). $$

The codomain $[0,\infty]$ consists of the nonnegative real numbers together with $+\infty$. The sum on the right is an extended nonnegative sum: it is the limit of its increasing partial sums, or equivalently their supremum. Thus the sum is allowed to be infinite. Because all terms are nonnegative, no cancellation or expression of the form $+\infty-(+\infty)$ is involved.

Each part of the definition matters. The function must be defined on the sigma-algebra $\mathcal{F}$, not on an unspecified collection of subsets. The empty set must have measure zero. Finally, countable additivity is required for every countable disjoint family, not only for pairs or finite families. Finite additivity and monotonicity, established in the previous tutorial, are consequences of these axioms; they are useful facts, but do not replace countable additivity.

Checking the Axioms in Examples

The first examples show how to read the definition directly. To check countable additivity, the key question is whether each point or contribution to the measure is counted exactly once in a disjoint union.

Worked Example: The Zero Measure

Let $(X,\mathcal{F})$ be any measurable space, and define $\mu(A)=0$ for every $A\in\mathcal{F}$. Then $\mu(\varnothing)=0$. If $A_1,A_2,\ldots$ are pairwise disjoint members of $\mathcal{F}$, their union belongs to $\mathcal{F}$, and $$ \mu\left(\bigcup_{n=1}^{\infty}A_n\right)=0 \quad\text{and}\quad \sum_{n=1}^{\infty}\mu(A_n)=\sum_{n=1}^{\infty}0=0. $$ The two sides agree, so $\mu$ is a measure. Its total measure is $\mu(X)=0$. This confirms that the definition does not require a measure to assign positive size to the whole space.

Worked Example: A Weighted Measure on Three Points

Let $X=\{p,q,r\}$, let $\mathcal{F}=\mathcal{P}(X)$, and assign weights $w(p)=2$, $w(q)=0$, and $w(r)=5$. Define $$ \mu(A)=\sum_{x\in A}w(x),\qquad A\subseteq X. $$ The empty set contains no points, so $\mu(\varnothing)=0$. For any pairwise disjoint sequence $A_1,A_2,\ldots$, each of the three points belongs to at most one set $A_n$. Consequently, its weight contributes to the measure of the union exactly when it belongs to one of the sets, and then contributes exactly once. In particular, $$ \mu\left(\bigcup_{n=1}^{\infty}A_n\right) =2\,\mathbf{1}_{\{p\in\bigcup_n A_n\}}+ 5\,\mathbf{1}_{\{r\in\bigcup_n A_n\}} =\sum_{n=1}^{\infty}\mu(A_n). $$ Here $\mathbf{1}_{\{P\}}$ is $1$ when the condition $P$ holds and $0$ otherwise. The point $q$ contributes zero on either side. Thus $\mu$ is a measure. For example, $\mu(\{p,q\})=2$, $\mu(\{r\})=5$, and $\mu(X)=7$.

The same point-by-point check explains why a point mass is a measure. If $\{a\}$ belongs to $\mathcal{F}$, define $\delta_a(A)=1$ when $a\in A$, and $\delta_a(A)=0$ otherwise. In a disjoint sequence, the point $a$ belongs to either no set or exactly one set. Its contribution to the union therefore equals the sum of its contributions to the individual sets.

Worked Example: A Point Mass on the Borel Real Line

On $(\mathbb{R},\mathcal{B}(\mathbb{R}))$, define $\delta_{-2}(A)=1$ if $-2\in A$, and $\delta_{-2}(A)=0$ otherwise. This is defined on Borel sets; in particular, the singleton $\{-2\}$ is Borel. The empty set does not contain $-2$, so $\delta_{-2}(\varnothing)=0$. For pairwise disjoint Borel sets $A_n$, if $-2$ belongs to their union, it belongs to exactly one $A_k$. In that case, $$ \delta_{-2}\left(\bigcup_{n=1}^{\infty}A_n\right)=1 =\sum_{n=1}^{\infty}\delta_{-2}(A_n). $$ If $-2$ belongs to none of the sets, both sides are $0$. Hence $\delta_{-2}$ is a measure. For instance, $\delta_{-2}((-3,-1))=1$, while $\delta_{-2}((0,1))=0$.

Continuity from Below

Countable additivity also controls sequences that are not disjoint. One important result concerns an increasing sequence: when each set is contained in the next, the measure of the union is the limit of the measures. This is called continuity from below. The proof converts the increasing sequence into disjoint increments, where the defining axiom applies.

Theorem (Continuity from Below): Let $\mu$ be a measure on $(X,\mathcal{F})$, and let $A_1\subseteq A_2\subseteq\cdots$ be measurable sets. Then $$ \mu\left(\bigcup_{n=1}^{\infty}A_n\right) =\lim_{n\to\infty}\mu(A_n). $$ The limit is allowed to be $+\infty$.

Proof. Define $D_1=A_1$, and for $n\geq2$, define $D_n=A_n\setminus A_{n-1}$. Relative differences of measurable sets are measurable, so each $D_n\in\mathcal{F}$. These sets are pairwise disjoint: if $i<j$, then $D_i\subseteq A_i\subseteq A_{j-1}$, whereas $D_j$ contains no points of $A_{j-1}$. Also, for each $N$, $$ \bigcup_{n=1}^{N}D_n=A_N, \qquad \bigcup_{n=1}^{\infty}D_n=\bigcup_{n=1}^{\infty}A_n. $$ By finite additivity, $\mu(A_N)=\sum_{n=1}^{N}\mu(D_n)$. As $N$ increases, these partial sums increase to $\sum_{n=1}^{\infty}\mu(D_n)$, by the definition of a nonnegative series. Countable additivity gives $$ \mu\left(\bigcup_{n=1}^{\infty}A_n\right) =\mu\left(\bigcup_{n=1}^{\infty}D_n\right) =\sum_{n=1}^{\infty}\mu(D_n) =\lim_{N\to\infty}\mu(A_N). $$ This proves the result, including when the limit is infinite. $\square$

Worked Example: Increasing Intervals

Let $\mu$ be Lebesgue measure on the Borel subsets of $\mathbb{R}$, and let $A_n=(-1,\,4-1/n)$ for positive integers $n$. These intervals increase, and their union is $(-1,4)$: every $x<4$ with $x>-1$ lies below $4-1/n$ for some sufficiently large $n$, while $4$ lies in none of the intervals. Their measures are $$ \mu(A_n)=\left(4-\frac{1}{n}\right)-(-1)=5-\frac{1}{n}. $$ Thus $\lim_{n\to\infty}\mu(A_n)=5$, which agrees with $\mu((-1,4))=5$. The endpoint $4$ is not included in the union, but adding or removing a single endpoint does not change its Lebesgue measure.

Countable Subadditivity

Countable additivity applies to disjoint sets. For arbitrary measurable sets, the corresponding general rule is an inequality: the measure of a countable union is at most the sum of the measures. This result is countable subadditivity. It follows by replacing the original sets with disjoint pieces, without changing their union.

Theorem (Countable Subadditivity): If $\mu$ is a measure on $(X,\mathcal{F})$ and $A_1,A_2,\ldots\in\mathcal{F}$, then $$ \mu\left(\bigcup_{n=1}^{\infty}A_n\right) \leq \sum_{n=1}^{\infty}\mu(A_n). $$

Proof. Define $B_1=A_1$ and, for $n\geq2$, define $$ B_n=A_n\setminus\bigcup_{j=1}^{n-1}A_j. $$ Finite unions and relative differences of measurable sets are measurable, so $B_n\in\mathcal{F}$. The sets $B_n$ are pairwise disjoint, and every point in $\bigcup_n A_n$ belongs to the first $A_n$ in the sequence that contains it. Therefore $\bigcup_n B_n=\bigcup_n A_n$. By countable additivity and the monotonicity theorem from the previous tutorial, $$ \mu\left(\bigcup_{n=1}^{\infty}A_n\right) =\sum_{n=1}^{\infty}\mu(B_n) \leq\sum_{n=1}^{\infty}\mu(A_n), $$ since $B_n\subseteq A_n$ for every $n$. This proves countable subadditivity. $\square$

The inequality is useful even when the sets overlap heavily. It does not assert equality: adding the measures of overlapping sets can count the same portion more than once. If the sets are pairwise disjoint, the defining axiom gives equality instead.

Continuity from Above and Its Hypothesis

There is a corresponding limit rule for decreasing sequences, but it requires a finiteness assumption. Without that assumption, subtracting from the measure of the first set can involve the undefined expression $+\infty-(+\infty)$. The next theorem states the safe version.

Theorem (Continuity from Above for Finite Measure): Let $\mu$ be a measure, and let $A_1\supseteq A_2\supseteq\cdots$ be measurable sets with $\mu(A_1)<\infty$. Then $$ \mu\left(\bigcap_{n=1}^{\infty}A_n\right) =\lim_{n\to\infty}\mu(A_n). $$

Proof. Set $C_n=A_1\setminus A_n$. Since the $A_n$ decrease, the $C_n$ increase. Moreover, $$ \bigcup_{n=1}^{\infty}C_n=A_1\setminus\bigcap_{n=1}^{\infty}A_n. $$ Continuity from below therefore gives $$ \mu\left(A_1\setminus\bigcap_{n=1}^{\infty}A_n\right) =\lim_{n\to\infty}\mu(C_n). $$ For every $n$, the disjoint union $A_1=A_n\cup C_n$ gives $\mu(A_1)=\mu(A_n)+\mu(C_n)$. Since $\mu(A_1)$ is finite, both terms on the right are finite, and subtraction is legitimate: $$ \mu(A_n)=\mu(A_1)-\mu(C_n). $$ Taking limits and using the preceding identity yields $$ \lim_{n\to\infty}\mu(A_n) =\mu(A_1)-\mu\left(A_1\setminus\bigcap_{n=1}^{\infty}A_n\right) =\mu\left(\bigcap_{n=1}^{\infty}A_n\right). $$ The last equality follows by decomposing $A_1$ into the intersection and its relative complement; both have finite measure. $\square$

What the Definition Does—and Does Not—Require

A measure is not required to be a probability measure, to be finite on every set, or to assign positive size to every nonempty set. When $\mu(X)=1$, it is called a probability measure. When $\mu(X)<\infty$, it is called a finite measure. These are additional descriptions of particular measures, not extra axioms in the general definition. The zero measure and the point mass above are both valid measures, despite having very different values.

The distinction between finite and infinite values is especially important when applying limit rules. Continuity from below needs no finiteness assumption; its proof uses increasing sums of nonnegative quantities. Continuity from above, in the form proved here, depends on $\mu(A_1)<\infty$. For example, with counting measure on $\mathbb{N}$, the decreasing sets $A_n=\{n,n+1,n+2,\ldots\}$ all have infinite measure, while their intersection is empty and has measure zero. Thus the measures do not converge to the measure of the intersection. This does not contradict continuity from above: the first set has infinite measure.

The definition provides a common structure for size assignments, while the measurable space and the particular measure determine what is being measured. Countable additivity is the central axiom; continuity from below and countable subadditivity show how it can be used beyond disjoint unions. The next tutorial examines further examples of measures and how their different constructions fit these axioms.

Check Your Understanding

Use the definition and results in this tutorial to answer the following questions.

What are the domain, range, and two axioms in the definition of a measure?
Why is the sum in countable additivity well-defined even when it is infinite?
In the proof of continuity from below, how are the sets $D_n$ constructed, and why are they disjoint?
How does disjointification lead to countable subadditivity for a sequence of sets that may overlap?
Where is the hypothesis $\mu(A_1)<\infty$ used in continuity from above?
Why does the decreasing sequence $A_n=\{n,n+1,\ldots\}$ under counting measure not violate the stated continuity-from-above theorem?

What You'll Learn