Examples of Measures

Building New Measures from Known Ones

A measure is specified by both a measurable space and a rule for assigning sizes to its measurable sets. Once one measure is available, there are several ways to produce others without checking the countable-additivity axiom from the beginning each time. We can focus on part of the original space, transfer the measure to another space through a measurable map, or combine measures. These constructions give examples with different interpretations, including probability measures built from ordinary length.

We will use the familiar Lebesgue measure $\lambda$ on the Borel subsets of $\mathbb{R}$. In particular, an interval with endpoints $a\leq b$ has length $b-a$; whether either endpoint is included does not affect its measure. The constructions below apply more generally to any measure on a measurable space. The trace sigma-algebra and its basic properties were established in the earlier tutorial on measurable spaces.

Restricting a Measure to a Measurable Subset

Suppose we want to measure only the part of a space lying in a particular measurable set $E$. The sets we can measure within $E$ are the members of the trace sigma-algebra $\mathcal{F}|_E$. If $E\in\mathcal{F}$, every such set is also measurable in the original space, so the original measure can be used directly.

Definition: Let $\mu$ be a measure on $(X,\mathcal{F})$, and let $E\in\mathcal{F}$. The restriction of $\mu$ to $E$ is the function $\mu|_E:\mathcal{F}|_E\to[0,\infty]$ defined by $$ (\mu|_E)(C)=\mu(C),\qquad C\in\mathcal{F}|_E. $$

Theorem (Restriction of a Measure): If $\mu$ is a measure on $(X,\mathcal{F})$ and $E\in\mathcal{F}$, then $\mu|_E$ is a measure on $(E,\mathcal{F}|_E)$.

Proof. The empty set belongs to the trace sigma-algebra, and $(\mu|_E)(\varnothing)=\mu(\varnothing)=0$. Now let $C_1,C_2,\ldots$ be pairwise disjoint members of $\mathcal{F}|_E$. Each $C_n$ is in $\mathcal{F}$, and their union is in $\mathcal{F}|_E$. Applying countable additivity of $\mu$ gives $$ (\mu|_E)\left(\bigcup_{n=1}^{\infty}C_n\right) =\mu\left(\bigcup_{n=1}^{\infty}C_n\right) =\sum_{n=1}^{\infty}\mu(C_n) =\sum_{n=1}^{\infty}(\mu|_E)(C_n). $$ Thus the restriction satisfies both requirements in the definition of a measure. $\square$

Worked Example: Length on a Bounded Interval

Take $E=[2,5]$ and restrict Lebesgue measure on the Borel subsets of $\mathbb{R}$ to the trace sigma-algebra on $E$. The total measure of this restricted space is $$ (\lambda|_E)(E)=\lambda([2,5])=5-2=3. $$ It is a finite measure, but not a probability measure because its total measure is not $1$. We can rescale it to obtain a probability measure $P$ on $E$: $$ P(C)=\frac{1}{3}(\lambda|_E)(C),\qquad C\in\mathcal{B}(\mathbb{R})|_E. $$ Indeed, $P(E)=\frac{1}{3}\lambda([2,5])=1$. For example, $C=[2,7/2]$ has length $7/2-2=3/2$, so $$ P([2,7/2])=\frac{1}{3}\cdot\frac{3}{2}=\frac{1}{2}. $$ This is the uniform probability measure on the interval: the probability assigned to a subinterval is its length divided by the total length $3$.

More generally, multiplying a measure by a positive constant gives another measure: if $c>0$, then $A\mapsto c\mu(A)$ is countably additive because multiplication by $c$ preserves sums of nonnegative terms, including infinite sums. The zero measure is the special zero-valued measure described earlier in the course; we do not interpret $0\cdot\infty$ as a product when discussing this scaling.

Transporting a Measure by a Measurable Map

A measure can also be used to assign sizes to sets in a different space. The key is to ask which points in the original space map into a given measurable set. These points form its preimage. Measurability of the map ensures that each such preimage is a set on which the original measure is defined.

Definition: A map $f:(X,\mathcal{F})\to(Y,\mathcal{G})$ is measurable if $f^{-1}(B)\in\mathcal{F}$ for every $B\in\mathcal{G}$. Given a measure $\mu$ on $(X,\mathcal{F})$, the pushforward measure of $\mu$ under $f$ is the function $f_{\ast}\mu:\mathcal{G}\to[0,\infty]$ defined by $$ (f_{\ast}\mu)(B)=\mu(f^{-1}(B)),\qquad B\in\mathcal{G}. $$

Theorem (Pushforward of a Measure): If $f:(X,\mathcal{F})\to(Y,\mathcal{G})$ is measurable and $\mu$ is a measure on $(X,\mathcal{F})$, then $f_{\ast}\mu$ is a measure on $(Y,\mathcal{G})$.

Proof. Since $f^{-1}(\varnothing)=\varnothing$, we have $(f_{\ast}\mu)(\varnothing)=\mu(\varnothing)=0$. Let $B_1,B_2,\ldots$ be pairwise disjoint members of $\mathcal{G}$. Their preimages are measurable because $f$ is measurable. They are also pairwise disjoint: if $x$ belonged to both $f^{-1}(B_i)$ and $f^{-1}(B_j)$, then $f(x)\in B_i\cap B_j$, which is impossible for $i\ne j$. Preimages preserve unions, so $f^{-1}(\bigcup_n B_n)=\bigcup_n f^{-1}(B_n)$. Countable additivity of $\mu$ now yields $$ (f_{\ast}\mu)\left(\bigcup_{n=1}^{\infty}B_n\right) =\mu\left(\bigcup_{n=1}^{\infty}f^{-1}(B_n)\right) =\sum_{n=1}^{\infty}\mu(f^{-1}(B_n)) =\sum_{n=1}^{\infty}(f_{\ast}\mu)(B_n). $$ Thus $f_{\ast}\mu$ is a measure. $\square$

Worked Example: A Measure Induced by Squaring

Let $X=Y=[0,1]$, equipped with their Borel sigma-algebras, and let $\mu$ be the restriction of Lebesgue measure to $[0,1]$. Define $f(x)=x^2$. This function is continuous, hence measurable by the continuous-preimages result established earlier in the course. Therefore $\nu=f_{\ast}\mu$ is a measure on the Borel subsets of $[0,1]$.

For $0\leq t\leq1$, the preimage of $[0,t]$ is $[0,\sqrt{t}]$: on $[0,1]$, $x^2\leq t$ holds exactly when $x\leq\sqrt{t}$. Consequently, $$ \nu([0,t])=\mu([0,\sqrt{t}])=\sqrt{t}. $$ More generally, if $0\leq a\leq b\leq1$, then $f^{-1}((a,b])=(\sqrt{a},\sqrt{b}]$, and therefore $$ \nu((a,b])=\sqrt{b}-\sqrt{a}. $$ For instance, $\nu((1/4,1])=1-\sqrt{1/4}=1/2$, while $\nu([0,1/4])=\sqrt{1/4}=1/2$. The total mass is $\nu([0,1])=\mu([0,1])=1$, so $\nu$ is a probability measure. It is not uniform length on the target interval: its value on $[0,t]$ is $\sqrt{t}$, rather than $t$.

This example illustrates the direction of the construction. To find $\nu(B)$, we measure $f^{-1}(B)$ in the original space. In general, it is not correct to define a transported measure by taking the measure of the image $f(B)$: $B$ is a subset of the target space, whereas $f(B)$ would apply $f$ to a subset of its domain. The pushforward formula uses preimages precisely so that it applies to measurable sets in the target.

Adding Measures

Two measures on the same measurable space can be combined by adding their values on each measurable set. The result may assign infinite mass, which is allowed. This operation lets different sources of size contribute independently to one measure.

Theorem (Sum of Two Measures): If $\mu$ and $\nu$ are measures on $(X,\mathcal{F})$, then $\rho(A)=\mu(A)+\nu(A)$ defines a measure on $(X,\mathcal{F})$, where addition is in $[0,\infty]$.

Proof. Since both measures assign zero to the empty set, $\rho(\varnothing)=0+0=0$. Let $A_1,A_2,\ldots$ be pairwise disjoint members of $\mathcal{F}$. Countable additivity of $\mu$ and $\nu$ gives $$ \rho\left(\bigcup_{n=1}^{\infty}A_n\right) =\sum_{n=1}^{\infty}\mu(A_n)+\sum_{n=1}^{\infty}\nu(A_n). $$ For nonnegative series, the sum of the two series equals the series of termwise sums. This follows by taking limits of the increasing partial sums: for every $N$, the sum of the first $N$ terms of the termwise series is $\sum_{n=1}^{N}\mu(A_n)+\sum_{n=1}^{N}\nu(A_n)$, and these partial sums increase to the displayed extended sum. Hence $$ \rho\left(\bigcup_{n=1}^{\infty}A_n\right) =\sum_{n=1}^{\infty}\bigl(\mu(A_n)+\nu(A_n)\bigr) =\sum_{n=1}^{\infty}\rho(A_n). $$ The argument remains valid if either series is infinite, because all terms are nonnegative. Thus $\rho$ is a measure. $\square$

Worked Example: A Mixture of Two Probability Measures

On $[0,1]$, let $\mu$ be Lebesgue measure restricted to the interval, and let $\nu$ be the pushforward measure induced by $f(x)=x^2$ from the preceding example. Each has total measure $1$. Define $$ \rho(B)=\frac{1}{2}\mu(B)+\frac{1}{2}\nu(B),\qquad B\in\mathcal{B}([0,1]). $$ The sum theorem and positive scaling show that $\rho$ is a measure. Its total measure is $$ \rho([0,1])=\frac{1}{2}\mu([0,1])+\frac{1}{2}\nu([0,1]) =\frac{1}{2}+\frac{1}{2}=1, $$ so it is a probability measure. For $B=[0,1/4]$, the two component measures are $\mu(B)=1/4$ and $\nu(B)=\sqrt{1/4}=1/2$. Therefore $$ \rho([0,1/4])=\frac{1}{2}\cdot\frac{1}{4}+\frac{1}{2}\cdot\frac{1}{2} =\frac{1}{8}+\frac{1}{4}=\frac{3}{8}. $$ This construction balances the two measures: half the mass comes from uniform length and half from the distribution induced by squaring.

Choosing the Right Construction

These examples share a useful pattern: the measure axioms are preserved when the construction respects empty sets and disjoint unions. Restriction keeps the original values on a smaller measurable space. A pushforward evaluates target sets through their preimages. Addition combines two valid assignments on the same measurable space. In each proof, countable additivity of the original measure supplies the essential step.

The hypotheses matter. Restricting to $E$ requires that $E$ be measurable, so that the trace sigma-algebra consists of sets on which the original measure can be evaluated. A pushforward requires a measurable map; without it, some target preimages may not be measurable, and the proposed value may not be defined. For sums, both measures must have the same measurable domain so their values can be added for each set.

Finally, a measure need not have total mass $1$. The restriction of length to $[2,5]$ has total mass $3$, while normalizing it produces a probability measure. Pushforward can change how mass is distributed without changing total mass when the target contains the image: the preimage of the whole target is the entire domain. These constructions provide reusable ways to obtain measures from familiar ones, rather than treating every example as an isolated verification.

Check Your Understanding

Use the definitions and proofs above to answer the following questions.

Why must the subset used for restriction be measurable, and which sigma-algebra is used on that subset?
For a measurable map $f$, how does the preimage of a disjoint sequence of target sets behave?
If $f(x)=x^2$ on $[0,1]$, calculate the pushforward measure of $(1/9,4/9]$.
What is the total measure of the restriction of Lebesgue measure to $[3,8]$, and what constant would normalize it to a probability measure?
Why does the sum of two measures remain countably additive even when some of its values are infinite?

What You'll Learn