A Set Together with Its Measurable Structure
The Borel Sigma-Algebras tutorial attached a sigma-algebra to a topological space by generating it from the open sets. Measure theory also needs a framework that does not require a topology: a set of points together with a specified collection of subsets declared measurable. The collection matters. The same underlying set can support different measurable structures, depending on which subsets we choose to include.
A sigma-algebra is the structure that makes this choice mathematically workable. It contains the whole space and is closed under complements and countable unions, and therefore also under countable intersections. These closure properties allow later definitions, such as measures and measurable functions, to use countably many measurable sets without leaving the chosen collection.
The notation keeps two pieces of information visible: \(X\) is the collection of points under consideration, while \(\mathcal{F}\) is the collection of subsets that have been designated measurable. In particular, saying that \(E\subseteq X\) is measurable means \(E\in\mathcal{F}\), not merely that \(E\) has some desirable appearance. Measurability is always relative to a specified sigma-algebra.
The complement in the sigma-algebra axioms is taken relative to the whole set \(X\). Thus if \(E\) is measurable, its complement as a measurable subset of \(X\) is \(X\setminus E\). Changing the underlying set can change what the complement means, so it is important to keep the ambient set in view.
Different Sigma-Algebras on the Same Set
Every set \(X\) has two especially simple sigma-algebras. The largest possible one contains every subset; the smallest possible one contains only the empty set and the whole space. They illustrate how much the measurable structure can vary even when the underlying points do not change.
The power set is a sigma-algebra because complements of subsets of \(X\) are still subsets of \(X\), and any union of subsets of \(X\) is still a subset of \(X\). The trivial sigma-algebra is closed under complements, since the complements of \(\varnothing\) and \(X\) are \(X\) and \(\varnothing\), respectively. A countable union of its members is either empty or \(X\), so it is also closed under countable unions.
Worked Example: Two Structures on a Three-Point Set
Let \(X=\{a,b,c\}\). With the power-set sigma-algebra, every subset is measurable: $$ \mathcal{P}(X)=\{\varnothing,\{a\},\{b\},\{c\},\{a,b\},\{a,c\},\{b,c\},X\}. $$ For example, \(\{a,c\}\) is measurable in \((X,\mathcal{P}(X))\).
Now equip the same \(X\) with the trivial sigma-algebra \(\{\varnothing,X\}\). The set \(\{a,c\}\) is not measurable in this structure, even though it is still a subset of \(X\). The points have not changed; only the chosen sigma-algebra has changed. In the power-set structure there are eight measurable subsets, while in the trivial structure there are two.
When \(X\) is a topological space, its Borel sigma-algebra gives another natural measurable structure. It is generated by the open sets, as established in the Borel Sigma-Algebras tutorial. This makes every open set Borel measurable, but the definition of a measurable space does not require a topology at all.
Worked Example: The Borel Measurable Space on the Real Line
The pair \((\mathbb{R},\mathcal{B}(\mathbb{R}))\) is a measurable space. Its underlying set is the real line, and its measurable sets are precisely the Borel sets. For example, the interval \([2,5]\) is Borel: it is closed, and closed sets belong to the Borel sigma-algebra.
This structure is not the same as \((\mathbb{R},\mathcal{P}(\mathbb{R}))\). In the latter, every subset of the real line is measurable by definition. The Borel structure starts from the topology and applies the sigma-algebra operations; it does not define measurability as “being any subset whatsoever.”
Restricting a Measurable Space to a Subset
Suppose we focus on a subset \(A\subseteq X\). To describe which parts of \(A\) inherit measurability from \((X,\mathcal{F})\), we intersect \(A\) with the measurable sets in \(X\). The resulting collection is called the trace sigma-algebra, or the sigma-algebra induced on \(A\).
The definition captures inherited measurability: a subset \(C\) of \(A\) is measurable in the induced structure when it can be written as \(A\cap E\) for some measurable \(E\) in the original space. Intersecting with \(A\) ensures that the resulting set lies inside the new underlying space. In particular, complements in the trace sigma-algebra are relative to \(A\), not to \(X\).
Proof. First, \(A\in\mathcal{F}|_A\), since \(A=A\cap X\) and \(X\in\mathcal{F}\). Now let \(C\in\mathcal{F}|_A\). By definition, \(C=A\cap E\) for some \(E\in\mathcal{F}\). Its complement relative to \(A\) is $$ A\setminus C=A\setminus(A\cap E)=A\cap(X\setminus E). $$ Since \(\mathcal{F}\) is a sigma-algebra, \(X\setminus E\in\mathcal{F}\). Thus \(A\setminus C\in\mathcal{F}|_A\).
Finally, let \(C_n\in\mathcal{F}|_A\) for every positive integer \(n\). For each \(n\), choose \(E_n\in\mathcal{F}\) such that \(C_n=A\cap E_n\). Then $$ \bigcup_{n=1}^{\infty}C_n =\bigcup_{n=1}^{\infty}(A\cap E_n) =A\cap\left(\bigcup_{n=1}^{\infty}E_n\right). $$ Because \(\mathcal{F}\) is closed under countable unions, \(\bigcup_{n=1}^{\infty}E_n\in\mathcal{F}\). The displayed equality therefore expresses \(\bigcup_{n=1}^{\infty}C_n\) as a member of \(\mathcal{F}|_A\). We have shown that the trace contains \(A\), is closed under complements relative to \(A\), and is closed under countable unions. Hence it is a sigma-algebra on \(A\). \(\square\)
Worked Example: A Trace on a Two-Point Subset
Let \(X=\{a,b,c,d\}\), let \(\mathcal{F}=\mathcal{P}(X)\), and take \(A=\{a,c\}\). The trace sigma-algebra is $$ \mathcal{F}|_A=\{A\cap E:E\subseteq X\}=\mathcal{P}(A) =\{\varnothing,\{a\},\{c\},\{a,c\}\}. $$ To check that every subset of \(A\) occurs, take any \(C\subseteq A\) and choose \(E=C\), which is also a subset of \(X\). Then \(A\cap E=A\cap C=C\). Conversely, every set \(A\cap E\) is a subset of \(A\). Thus the trace is exactly \(\mathcal{P}(A)\).
By contrast, let \(\mathcal{F}=\{\varnothing,X\}\) on the same four-point set. Then $$ \mathcal{F}|_A=\{A\cap\varnothing,A\cap X\}=\{\varnothing,A\}. $$ The induced structure on \(A\) is trivial. This comparison shows that taking a subset does not by itself determine how many of its subsets are measurable; the original sigma-algebra also matters.
Restriction in Stages
A subset may itself contain a smaller subset of interest. It is useful that restricting first to the larger subset and then to the smaller one gives the same measurable structure as restricting directly. This compatibility follows from associativity of intersection.
Proof. By the definition of trace, a member of \((\mathcal{F}|_A)|_B\) has the form \(B\cap C\), where \(C\in\mathcal{F}|_A\). There is some \(E\in\mathcal{F}\) with \(C=A\cap E\). Therefore $$ B\cap C=B\cap(A\cap E)=(B\cap A)\cap E=B\cap E, $$ because \(B\subseteq A\). This shows that every member of \((\mathcal{F}|_A)|_B\) belongs to \(\mathcal{F}|_B\).
For the reverse inclusion, let \(D\in\mathcal{F}|_B\). Then \(D=B\cap E\) for some \(E\in\mathcal{F}\). The set \(A\cap E\) belongs to \(\mathcal{F}|_A\), so $$ D=B\cap E=B\cap(A\cap E) $$ belongs to \((\mathcal{F}|_A)|_B\). We have proved both inclusions, and hence the two trace sigma-algebras are equal. \(\square\)
Worked Example: Restricting Borel Sets to Nested Intervals
Consider the Borel measurable space \((\mathbb{R},\mathcal{B}(\mathbb{R}))\), with \(A=[-3,3]\) and \(B=[-1,1]\). Since \(B\subseteq A\subseteq\mathbb{R}\), transitivity gives $$ \bigl(\mathcal{B}(\mathbb{R})|_{[-3,3]}\bigr)|_{[-1,1]} =\mathcal{B}(\mathbb{R})|_{[-1,1]}. $$ For instance, the measurable set \(E=(-2,0)\in\mathcal{B}(\mathbb{R})\) yields $$ [-1,1]\cap E=[-1,0). $$ Restricting in stages gives the same result: first \([-3,3]\cap E=(-2,0)\), and then \([-1,1]\cap(-2,0)=[-1,0)\). The theorem applies to every Borel \(E\), not just this interval.
When the Subset Is Already Measurable
If \(A\in\mathcal{F}\), the trace admits a particularly simple description. In this case, the measurable subsets of \(A\) are exactly those sets that are already measurable in \(X\) and lie inside \(A\). The condition \(A\in\mathcal{F}\) is essential for this equivalence.
Proof. Suppose \(C\in\mathcal{F}|_A\). Then \(C=A\cap E\) for some \(E\in\mathcal{F}\). Since \(A,E\in\mathcal{F}\) and a sigma-algebra is closed under finite intersections, \(C\in\mathcal{F}\). Also \(C\subseteq A\). This proves $$ \mathcal{F}|_A\subseteq\{C\in\mathcal{F}:C\subseteq A\}. $$ Conversely, suppose \(C\in\mathcal{F}\) and \(C\subseteq A\). Then \(C=A\cap C\), so the definition of the trace gives \(C\in\mathcal{F}|_A\). This proves the reverse inclusion and the claimed equality. \(\square\)
The measurability hypothesis on \(A\) cannot simply be omitted. For example, if \(\mathcal{F}=\{\varnothing,X\}\) and \(A\) is a nonempty proper subset of \(X\), then \(A\notin\mathcal{F}\). The trace is still a valid sigma-algebra on \(A\), but the formula \(\{C\in\mathcal{F}:C\subseteq A\}\) gives only \(\{\varnothing\}\), which does not contain \(A\) and is therefore not a sigma-algebra on \(A\). The trace definition works for every subset; the simplified description requires a measurable subset.
Why the Measurable-Space Viewpoint Matters
A measurable space separates the underlying points from the collection of sets on which later analysis will be defined. A topology supplies one way to choose that collection, through the Borel sigma-algebra, but it is not the only way. The power-set and trivial sigma-algebras show the range of possible choices, while trace sigma-algebras let us pass to a subset without changing the ambient notion of inherited measurability.
A common pitfall is to call a subset “measurable” without specifying the structure in which it is being considered. A set can be measurable in one sigma-algebra and not in another, as the three-point example demonstrated. Another pitfall is to take complements relative to \(X\) after passing to \(A\). In the trace space, the whole space is \(A\), so complements must be formed as \(A\setminus C\). The trace construction ensures that these relative complements remain measurable.
These ideas prepare the ground for measure theory: a measure will assign sizes to sets in a sigma-algebra, and a measurable space specifies which sets are eligible for that assignment. Keeping the pair \((X,\mathcal{F})\) explicit makes those later definitions precise, even when the underlying set has no topology or geometric interpretation.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What are the two components of a measurable space, and what does it mean for a subset to be measurable in that space?
- On \(X=\{a,b,c\}\), is \(\{a\}\) measurable for the trivial sigma-algebra? Is it measurable for the power-set sigma-algebra?
- Why are complements in \(\mathcal{F}|_A\) taken relative to \(A\), rather than to \(X\)?
- If \(B\subseteq A\subseteq X\), what does the Transitivity of the Trace Theorem say about restricting first to \(A\) and then to \(B\)?
- Which additional hypothesis is needed for \(\mathcal{F}|_A=\{C\in\mathcal{F}:C\subseteq A\}\), and where is it used in the proof?