Restricting A Measure To A Measurable Set

Measuring Only the Part Inside a Set

A measure on a space assigns sizes to its measurable sets. If we want to count only the portion of each set lying inside a fixed measurable region, the natural operation is to intersect with that region before measuring. This simple construction is useful because it produces another measure, even when the original measure is infinite and even when the chosen region has measure zero.

Throughout the general discussion, let $(X,\mathcal{M},\mu)$ be an arbitrary measure space: $\mathcal{M}$ is a sigma-algebra on $X$, and $\mu:\mathcal{M}\to[0,\infty]$ is a measure. No assumption of finiteness, sigma-finiteness, or completeness is made. Fix $A\in\mathcal{M}$. All equalities below are exact set-function equalities; no almost-everywhere qualification is involved.

Definition: The restriction of $\mu$ to $A$, viewed as a set function on the original sigma-algebra $\mathcal{M}$, is $$ \mu_A(E)=\mu(E\cap A),\qquad E\in\mathcal{M}. $$ It measures only the part of $E$ that lies in $A$. Its values are allowed to be $+\infty$.

The definition requires $A$ to be measurable: then $E\cap A\in\mathcal{M}$, so $\mu(E\cap A)$ is defined for every $E\in\mathcal{M}$. The measure space need not have finite total measure. In particular, $\mu_A(X)=\mu(A)$, which may be infinite.

Why the Restriction Is a Measure

To verify that $\mu_A$ is a measure on $(X,\mathcal{M})$, we check the two defining axioms: it assigns zero to the empty set, and it is countably additive on disjoint measurable sets. Nonnegativity and the range $[0,\infty]$ follow at once from the corresponding properties of $\mu$.

Theorem: For any measure space $(X,\mathcal{M},\mu)$, with no finiteness, sigma-finiteness, or completeness assumption, and any $A\in\mathcal{M}$, the set function $\mu_A(E)=\mu(E\cap A)$ is a measure on $(X,\mathcal{M})$.

Proof. First, $$ \mu_A(\varnothing)=\mu(\varnothing\cap A)=\mu(\varnothing)=0. $$ Now let $E_1,E_2,\ldots$ be pairwise disjoint members of $\mathcal{M}$. The sets $E_n\cap A$ are measurable and pairwise disjoint: if a point belonged to both $E_i\cap A$ and $E_j\cap A$ for $i\ne j$, it would belong to both $E_i$ and $E_j$, contrary to disjointness. Also, $$ \left(\bigcup_{n=1}^{\infty}E_n\right)\cap A =\bigcup_{n=1}^{\infty}(E_n\cap A). $$ Countable additivity of $\mu$, applied to this disjoint sequence, therefore gives $$ \mu_A\left(\bigcup_{n=1}^{\infty}E_n\right) =\mu\left(\bigcup_{n=1}^{\infty}(E_n\cap A)\right) =\sum_{n=1}^{\infty}\mu(E_n\cap A) =\sum_{n=1}^{\infty}\mu_A(E_n). $$ This equality is valid also when the sum is $+\infty$, since countable additivity of $\mu$ takes values in $[0,\infty]$. Thus $\mu_A$ satisfies both measure axioms. $\square$

There is also a common way to describe the restriction when the underlying space itself is changed to $A$. Define the trace sigma-algebra $$ \mathcal{M}_A=\{E\cap A:E\in\mathcal{M}\} $$ on $A$, and define $\mu|_A(B)=\mu(B)$ for $B\in\mathcal{M}_A$. These definitions are valid because every $B\in\mathcal{M}_A$ is a measurable subset of $X$. The trace family is a sigma-algebra on $A$: it contains $\varnothing$ and $A$, complements relative to $A$ satisfy $A\setminus(E\cap A)=E^c\cap A$, and countable unions satisfy $\bigcup_n(E_n\cap A)=(\bigcup_n E_n)\cap A$. The same two-axiom argument proves that $\mu|_A$ is a measure on $(A,\mathcal{M}_A)$. This version records only subsets of $A$; $\mu_A$, in contrast, remains defined on all of $\mathcal{M}$.

Basic Consequences of Restriction

Some useful properties follow directly from the definition. Each statement here concerns the arbitrary measure space $(X,\mathcal{M},\mu)$ above, with $A\in\mathcal{M}$; none requires finite measure or sigma-finiteness.

Total mass: $\mu_A(X)=\mu(X\cap A)=\mu(A)$.
Sets inside the region: if $E\in\mathcal{M}$ and $E\subseteq A$, then $\mu_A(E)=\mu(E)$.
Comparison with the original measure: $\mu_A(E)\leq\mu(E)$, since $E\cap A\subseteq E$ and measures are monotone.
Null values: $\mu_A(E)=0$ exactly when $\mu(E\cap A)=0$, by the definition.

Proposition: For any measure space $(X,\mathcal{M},\mu)$ and any $A,B\in\mathcal{M}$, restricting first to $A$ and then to $B$ gives the same measure as restricting directly to $A\cap B$: $$ (\mu_A)_B=\mu_{A\cap B}. $$

Proof. The measure $\mu_A$ is defined on $\mathcal{M}$, so its restriction to the measurable set $B$ is defined. For every $E\in\mathcal{M}$, the definitions and associativity of intersection give $$ (\mu_A)_B(E)=\mu_A(E\cap B) =\mu((E\cap B)\cap A) =\mu(E\cap(A\cap B)) =\mu_{A\cap B}(E). $$ The two set functions therefore agree on every member of their common domain $\mathcal{M}$, which proves the assertion. $\square$

Worked Example: Restricting Lebesgue Measure to an Interval

Take $(\mathbb{R},\mathcal{L},m)$, Lebesgue measure on the real line, and $A=[1,3]$. This measure space is not finite, since $m(\mathbb{R})=\infty$; it is sigma-finite and complete. Let $E=(-\infty,2]\cup[4,5]$. Compute $\mu_A(E)$, where $\mu_A(E)=m(E\cap A)$.

Intersect each piece of $E$ with $A$. We have $(-\infty,2]\cap[1,3]=[1,2]$ and $[4,5]\cap[1,3]=\varnothing$. Hence $E\cap A=[1,2]$. The interval has Lebesgue measure $2-1=1$, so $$ \mu_A(E)=m([1,2])=1. $$ As a separate check of the total mass, $\mu_A(\mathbb{R})=m(\mathbb{R}\cap[1,3])=m([1,3])=3-1=2$. The first answer is consistent with $E\cap A\subseteq A$, since $1\leq2$.

Worked Example: Restricting Counting Measure

Let $X=\mathbb{N}=\{1,2,\ldots\}$ with its power-set sigma-algebra and counting measure $\#$, so $\#(S)$ is the number of elements of $S$, possibly $+\infty$. This space is sigma-finite and complete but not finite. Set $A=\{2,4,6\}$ and $E=\{1,2,3,4,5\}$. The restricted measure is $\#_A(E)=\#(E\cap A)$.

The intersection is $E\cap A=\{2,4\}$, which has two elements, so $\#_A(E)=2$. The total mass is $\#_A(\mathbb{N})=\#(A)=3$. To check the calculation directly, the only elements of $E$ counted by the restriction are 2 and 4; 1, 3, and 5 are excluded because they are not in $A$. Thus the intersection count is exactly 2, while the full restricted mass is exactly 3.

Worked Example: Restricting a Dirac Measure

Let $(\mathbb{R},\mathcal{B}_{\mathbb{R}},\delta_0)$ be the real line with its Borel sigma-algebra and the Dirac measure at zero, defined by $\delta_0(C)=1$ if $0\in C$ and $\delta_0(C)=0$ if $0\notin C$. This is a finite, hence sigma-finite, measure space; it is not complete on the Borel sigma-algebra. Restrict it to $A=\{0\}$.

For $E=[-1,1]$, the intersection $E\cap A=\{0\}$ contains zero. Therefore $(\delta_0)_A(E)=\delta_0(\{0\})=1$. For $F=(0,2)$, the intersection $F\cap A=\varnothing$, so $(\delta_0)_A(F)=\delta_0(\varnothing)=0$. The total mass is $(\delta_0)_A(\mathbb{R})=\delta_0(\{0\})=1$. These values agree with the defining membership test for $\delta_0$: the restricted measure counts a set precisely when its intersection with $A$ contains the atom at zero.

Common Mistake: Subtracting Measures Instead of Intersecting

The restriction is defined by measuring an intersection, not by subtracting the measure of $A$ from the measure of $E$. For example, in $(\mathbb{R},\mathcal{L},m)$, take $A=[0,2]$ and $E=[1,3]$. Both intervals have measure $2$, so the incorrect calculation $m(E)-m(A)$ gives $2-2=0$. But $E\cap A=[1,2]$, whose measure is $2-1=1$. The correct value is therefore $$ m_A(E)=m(E\cap A)=m([1,2])=1. $$ Subtraction does not in general measure the portion of $E$ inside $A$, and it can even produce a negative number when $m(A)>m(E)$. Intersect first, then apply the original measure.

One further point is that the restriction is a measure on the stated sigma-algebra, but completeness should not be silently inferred. In the general construction above, no completeness assumption was used. In particular, when $\mu_A$ is viewed on $(X,\mathcal{M})$, every subset of $X\setminus A$ has restricted measure zero if that subset is measurable; such subsets need not all belong to $\mathcal{M}$. On the other hand, the total-mass formula shows immediately that $\mu_A$ is finite exactly when $\mu(A)<\infty$. If $\mu$ is sigma-finite, then $\mu_A$ is sigma-finite as well: for a cover $X=\bigcup_n C_n$ with $\mu(C_n)<\infty$, the sets $C_n$ still cover $X$, and $\mu_A(C_n)=\mu(C_n\cap A)\leq\mu(C_n)<\infty$.

Check Your Understanding

For each question, use the definition $\mu_A(E)=\mu(E\cap A)$, and keep track of the underlying sigma-algebra.

On $(\mathbb{R},\mathcal{L},m)$, let $A=[-2,1]$ and $E=[0,4]$. Find $m_A(E)$ and $m_A(\mathbb{R})$.
Prove directly from the definition that if $E\subseteq A$ and $E\in\mathcal{M}$, then $\mu_A(E)=\mu(E)$.
For counting measure on $\mathbb{N}$, let $A$ be the odd positive integers. What is $\#_A(\{2,3,4,5\})$?
If $\mu(A)=0$, what is $\mu_A(E)$ for an arbitrary $E\in\mathcal{M}$? Justify your answer using monotonicity.
Why does the countable-additivity proof remain valid if $\sum_n\mu_A(E_n)=+\infty$?

Restricting A Measure To A Measurable Set

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