Measurability Defined by Borel Preimages

Measurability as a Preimage Condition

A function $f$ assigns a real number to each point of a domain. To call $f$ measurable, we ask whether the points whose function values lie in a specified Borel subset of the real line form a measurable subset of the domain. The direction of this test matters: it is the preimage in the domain that must belong to the domain’s sigma-algebra. The target need not carry a measure.

Let $(X,\mathcal M)$ be a measurable space: $X$ is a set and $\mathcal M$ is a sigma-algebra of subsets of $X$. No measure on $X$ is needed for the definition. In particular, if $(X,\mathcal M,\mu)$ is a measure space, the definition does not require that $\mu$ be finite, sigma-finite, or complete. The target is $\mathbb R$ equipped with its Borel sigma-algebra $\mathcal B_{\mathbb R}$, the sigma-algebra generated by the open subsets of $\mathbb R$. Functions here are real-valued, and the definition is a pointwise condition, not an almost-everywhere one.

Definition: A function $f:X\to\mathbb R$ is measurable (or $\mathcal M/\mathcal B_{\mathbb R}$-measurable) if $$ f^{-1}(B)=\{x\in X:f(x)\in B\}\in\mathcal M $$ for every $B\in\mathcal B_{\mathbb R}$. If $X$ is also equipped with a measure $\mu$, one may say that $f$ is $\mu$-measurable to refer to this same condition on the measure space’s sigma-algebra $\mathcal M$.

The quantifier “for every Borel set $B$” is part of the definition. It does not ask that $B$ itself belong to $\mathcal M$: $B$ is a subset of the target $\mathbb R$, while $f^{-1}(B)$ is a subset of the domain $X$. Nor does it ask for the image $f(E)$ of each measurable $E\subseteq X$ to be Borel. Those are different conditions.

Preimages obey basic set identities. For $B\subseteq\mathbb R$, $f^{-1}(B^c)=(f^{-1}(B))^c$, where the complements are taken in $\mathbb R$ and $X$, respectively. For any sequence $(B_n)$ of subsets of $\mathbb R$, $f^{-1}(\bigcup_n B_n)=\bigcup_n f^{-1}(B_n)$ and $f^{-1}(\bigcap_n B_n)=\bigcap_n f^{-1}(B_n)$. These identities follow pointwise: $f(x)$ belongs to a union exactly when it belongs to at least one member of that union, and belongs to an intersection exactly when it belongs to every member. They explain how the Borel-set test converts target-set membership into domain-set membership.

Constant Maps

Proposition: For any measurable space $(X,\mathcal M)$ and any $c\in\mathbb R$, the constant map $f(x)=c$ is measurable. This includes the empty domain, if $X=\varnothing$.

Proof. Fix an arbitrary Borel set $B\in\mathcal B_{\mathbb R}$. If $c\in B$, every $x\in X$ satisfies $f(x)=c\in B$, so $f^{-1}(B)=X\in\mathcal M$. If $c\notin B$, no $x\in X$ satisfies $f(x)\in B$, so $f^{-1}(B)=\varnothing\in\mathcal M$. These cases cover every Borel set $B$, proving measurability. If $X=\varnothing$, both $X$ and $\varnothing$ are the same empty set, and the conclusion still holds. $\square$

Worked Example: Testing a Constant Map

Let $X=\{a,b\}$, $\mathcal M=\{\varnothing,\{a\},\{b\},X\}$, and let $\mu$ be counting measure, so $\mu(\{a\})=\mu(\{b\})=1$. Define $f:X\to\mathbb R$ by $f(a)=f(b)=2$. This is a finite measure space, but the constant-map argument does not depend on finiteness.

For the Borel set $B=(-1,3)$, $2\in B$, so $f^{-1}(B)=X$, which belongs to $\mathcal M$. For $C=(2,5)$, $2\notin C$, since the left endpoint is excluded; hence $f^{-1}(C)=\varnothing$. For the singleton $D=\{2\}$, which is Borel, $2\in D$, so $f^{-1}(D)=X$. In each case the preimage is in $\mathcal M$, and the proposition covers all other Borel sets as well.

Check: for any Borel $B$, the preimage is $X$ if $2\in B$, and $\varnothing$ if $2\notin B$. Both sets are in $\mathcal M$, as required.

Indicator Functions

For a subset $E\subseteq X$, its indicator function $\chi_E:X\to\mathbb R$ is defined by $\chi_E(x)=1$ for $x\in E$, and $\chi_E(x)=0$ for $x\notin E$. This definition makes sense whether or not $E$ is measurable. The question is whether the resulting function passes the Borel-preimage test.

Proposition: Let $(X,\mathcal M)$ be a measurable space and $E\subseteq X$. The real-valued indicator $\chi_E$ is measurable if and only if $E\in\mathcal M$. No measure or finiteness assumption is required.

Proof. First suppose $E\in\mathcal M$. Fix any Borel set $B\subseteq\mathbb R$. The only values of $\chi_E$ are $0$ and $1$, so there are four possibilities. If both $0$ and $1$ belong to $B$, then $\chi_E^{-1}(B)=X$. If $1\in B$ and $0\notin B$, then $\chi_E^{-1}(B)=E$. If $0\in B$ and $1\notin B$, then $\chi_E^{-1}(B)=E^c$. If neither $0$ nor $1$ belongs to $B$, then $\chi_E^{-1}(B)=\varnothing$. Since $X,\varnothing\in\mathcal M$, $E\in\mathcal M$ by assumption, and $E^c\in\mathcal M$ because $\mathcal M$ is a sigma-algebra, every listed preimage is measurable. Thus $\chi_E$ is measurable.

Conversely, suppose $\chi_E$ is measurable. The singleton $\{1\}$ is a closed, hence Borel, subset of $\mathbb R$. By the definition of the indicator, $\chi_E(x)\in\{1\}$ holds exactly when $x\in E$. Therefore $\chi_E^{-1}(\{1\})=E$. Measurability of $\chi_E$ requires this preimage to belong to $\mathcal M$, so $E\in\mathcal M$. Both implications hold, proving the equivalence. $\square$

Worked Example: An Indicator on a Finite Measured Space

Take $X=\{a,b,c,d\}$, let $\mathcal M$ be the set of all subsets of $X$, and let $\mu$ be counting measure. Set $E=\{a,c\}$. Since every subset of $X$ lies in $\mathcal M$, $E$ is measurable. The indicator has $\chi_E(a)=1$, $\chi_E(b)=0$, $\chi_E(c)=1$, and $\chi_E(d)=0$.

For $B=[1,2]$, the value $1$ belongs to $B$ and $0$ does not, so $\chi_E^{-1}(B)=\{a,c\}=E$. For $C=[-1,0]$, $0\in C$ and $1\notin C$, so $\chi_E^{-1}(C)=\{b,d\}=E^c$. For $D=(-1,2)$, both $0$ and $1$ belong to $D$, so $\chi_E^{-1}(D)=X$. For $F=(2,3)$, neither value belongs to $F$, so $\chi_E^{-1}(F)=\varnothing$. All four preimages belong to $\mathcal M$.

Check: the four cases depend only on whether $B$ contains $0$ and $1$, not on its other points. The computed preimages match the proposition’s four-case description, so every Borel preimage is measurable.

Worked Example: A Finite-Range Function That Is Not Measurable

Let $X=\{u,v\}$ and $\mathcal M=\{\varnothing,X\}$, a sigma-algebra on $X$. Define $E=\{u\}$, which is not in $\mathcal M$, and form $f=\chi_E$. Then $f(u)=1$ and $f(v)=0$. For the Borel set $B=\{1\}$, the preimage is $f^{-1}(B)=\{u\}$, which is not in $\mathcal M$. Thus $f$ is not measurable.

Every other Borel preimage can also be checked from the two values: it is $X$ if $B$ contains both $0$ and $1$, $\{u\}$ if it contains $1$ but not $0$, $\{v\}$ if it contains $0$ but not $1$, and $\varnothing$ if it contains neither. The middle two sets are not in $\mathcal M$, so the definition fails. This example is a direct check of the criterion: $\chi_E$ is measurable exactly when $E$ is measurable.

What the Definition Detects

The constant-map and indicator tests highlight two different features of the definition. A constant map has only one value, so each Borel set pulls back to one of the two guaranteed members $\varnothing$ or $X$ of a sigma-algebra. An indicator has two possible values, and the Borel set $\{1\}$ pulls back precisely to the set used to define it. If that set is absent from the domain sigma-algebra, the function fails the test.

Common mistake: finite range does not imply measurability. In the two-point example, $f$ takes only the values $0$ and $1$, but $f^{-1}(\{1\})=\{u\}\notin\mathcal M$. The incorrect argument “the range is finite, therefore the function is measurable” ignores the required domain preimages. The correction is to test those preimages: for an indicator, the preimage of the Borel singleton $\{1\}$ is exactly $E$, so measurability forces $E\in\mathcal M$. Conversely, once $E\in\mathcal M$, the four-case calculation proves that every Borel preimage is measurable.

This preimage viewpoint is foundational: it specifies what measurability means before one asks how measurable functions behave under further constructions or how they are used in integration. For the examples here, no integral, convergence assumption, or measure-theoretic size calculation is involved. The decisive information is the sigma-algebra on the domain and the Borel sets in the target.

Check Your Understanding

Answer each question by applying the Borel-preimage definition, and distinguish carefully between sets in the target and sets in the domain.

For a constant map $f(x)=c$, what are the two possible forms of $f^{-1}(B)$, and when does each occur?
For an indicator $\chi_E$, compute the preimage of a Borel set $B$ in each of the four cases determined by whether $0$ and $1$ belong to $B$.
Why does measurability of $\chi_E$ imply $E\in\mathcal M$? Name the Borel set whose preimage gives $E$.
On $X=\{u,v\}$ with $\mathcal M=\{\varnothing,X\}$, decide whether the indicator of $\{v\}$ is measurable and justify your answer with a Borel preimage.
Does a measure on $X$, or a finiteness assumption on that measure, enter the definition of a measurable map into $\mathbb R$? Explain.

Measurability Defined by Borel Preimages

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