Why Test Open Sets?
Measurability of a real-valued function is defined by requiring that preimages of Borel sets be measurable. Since there are many Borel sets, it is useful to have a smaller test family. Open sets provide such a test, and rational open intervals reduce it further to a countable family. The countability is essential: a sigma-algebra is closed under countable unions, not arbitrary unions.
Throughout, let \((X,\mathcal{M},\mu)\) be any measure space, with no assumption that \(\mu\) is finite, sigma-finite, or complete. The function \(f:X\to\mathbb{R}\) is real-valued, not extended-real-valued. For \(A\subseteq\mathbb{R}\), write \(f^{-1}(A)=\{x\in X:f(x)\in A\}\). A real-valued function is \(\mathcal{M}\)-measurable if \(f^{-1}(B)\in\mathcal{M}\) for every Borel set \(B\in\mathcal{B}_{\mathbb{R}}\). These conditions concern sets and do not depend on the particular measure \(\mu\).
To verify the basis property, take \(y\in U\), where \(U\) is open. There is an \(\varepsilon>0\) such that \((y-\varepsilon,y+\varepsilon)\subseteq U\). By density of \(\mathbb{Q}\) in \(\mathbb{R}\), choose rationals \(p,q\) with \(y-\varepsilon<p<y<q<y+\varepsilon\). Then \(y\in(p,q)\subseteq U\). Thus every point of \(U\) belongs to a rational open interval contained in \(U\), as required. If \(U=\varnothing\), it is the union of no intervals.
The Open-Set and Rational-Interval Criteria
- \(f\) is \(\mathcal{M}\)-measurable, meaning \(f^{-1}(B)\in\mathcal{M}\) for every \(B\in\mathcal{B}_{\mathbb{R}}\).
- \(f^{-1}(U)\in\mathcal{M}\) for every open set \(U\subseteq\mathbb{R}\).
- \(f^{-1}((p,q))\in\mathcal{M}\) for every \(p,q\in\mathbb{Q}\) with \(p<q\).
Proof. If \(f\) is measurable, then the preimage of every Borel set is in \(\mathcal{M}\). Every open set, and in particular every interval \((p,q)\), is Borel. This proves the first condition implies each of the other two.
Suppose next that \(f^{-1}((p,q))\in\mathcal{M}\) for every rational pair \(p<q\). Let \(U\subseteq\mathbb{R}\) be open. By the basis property, \(U\) is the union of all rational intervals contained in it. Taking preimages gives $$ f^{-1}(U)=\bigcup_{\substack{p,q\in\mathbb{Q}\\p<q,\ (p,q)\subseteq U}} f^{-1}((p,q)). $$ There are at most countably many pairs \((p,q)\) in this union. Each set on the right belongs to \(\mathcal{M}\) by assumption, so the union belongs to \(\mathcal{M}\). This proves the second condition.
Finally, suppose \(f^{-1}(U)\in\mathcal{M}\) for every open \(U\subseteq\mathbb{R}\). Define $$ \mathcal{C}=\{B\subseteq\mathbb{R}: f^{-1}(B)\in\mathcal{M}\}. $$ This is a sigma-algebra: \(f^{-1}(\mathbb{R})=X\), \(f^{-1}(B^c)=(f^{-1}(B))^c\), and \(f^{-1}(\bigcup_{n=1}^{\infty}B_n)=\bigcup_{n=1}^{\infty}f^{-1}(B_n)\). Thus \(\mathcal{C}\) contains \(\mathbb{R}\), is closed under complements, and is closed under countable unions. By assumption it contains every open set. Since \(\mathcal{B}_{\mathbb{R}}=\sigma(\{\text{open subsets of }\mathbb{R}\})\), it contains every Borel set. Hence \(f\) is measurable. \(\square\)
The theorem gives two practical routes. One can check preimages of every open set directly, or check only the countable family of rational intervals. The latter often makes the proof shorter: once the interval preimages are known to be measurable, the theorem supplies the open-set test.
Worked Example: The Identity Map and an Irrational Endpoint
Take \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{L}\), and \(\mu=m\), Lebesgue measure on \(\mathbb{R}\). Let \(f(x)=x\) and \(U=(0,\sqrt{2})\). The rational-interval basis gives $$ f^{-1}(U)=U=\bigcup_{\substack{p,q\in\mathbb{Q}\\0<p<q<\sqrt{2}}}(p,q). $$ Indeed, every interval in the union is contained in \((0,\sqrt{2})\); conversely, for each \(x\in(0,\sqrt{2})\), density of the rationals gives \(p,q\in\mathbb{Q}\) with \(0<p<x<q<\sqrt{2}\). The union is countable, so it is Lebesgue measurable. Its measure is \(m((0,\sqrt{2}))=\sqrt{2}-0=\sqrt{2}\). Check: the endpoints are excluded, but adding or removing either endpoint does not change Lebesgue measure, so the interval length remains \(\sqrt{2}\).
Worked Example: An Open-Set Test for the Floor Function
On \((\mathbb{R},\mathcal{L},m)\), define \(f(x)=\lfloor x\rfloor\), the greatest integer less than or equal to \(x\). For any open set \(U\subseteq\mathbb{R}\), $$ f^{-1}(U)=\bigcup_{\substack{n\in\mathbb{Z}\\n\in U}}[n,n+1). $$ To verify this identity, \(f(x)=n\) holds exactly when \(n\le x<n+1\). Therefore \(f(x)\in U\) holds exactly when \(x\in[n,n+1)\) for some integer \(n\in U\). Each half-open interval \([n,n+1)\) is Lebesgue measurable, and the union is countable because \(\mathbb{Z}\) is countable. The open-set test proves that \(f\) is measurable.
For a specific check, take \(U=(2/5,11/5)\). The integers in this open interval are exactly \(1\) and \(2\): \(2/5<1<11/5\), \(2/5<2<11/5\), while \(0<2/5\) and \(3>11/5\). Hence $$ f^{-1}(U)=[1,2)\cup[2,3)=[1,3). $$ The equality follows because the first interval includes \(1\) and excludes \(2\), while the second includes \(2\) and excludes \(3\). Directly, \(\lfloor x\rfloor\) is either \(1\) or \(2\) precisely for \(1\le x<3\). The result is measurable, and its measure is \(3-1=2\).
The Closed-Set Criterion
Proof. If \(f\) is measurable, then every closed set \(F\) is Borel, so \(f^{-1}(F)\in\mathcal{M}\). Conversely, suppose every closed-set preimage belongs to \(\mathcal{M}\). If \(U\subseteq\mathbb{R}\) is open, then \(U^c\) is closed and $$ f^{-1}(U)=f^{-1}((U^c)^c)=\big(f^{-1}(U^c)\big)^c. $$ The set \(f^{-1}(U^c)\) is measurable by assumption, so its complement is in \(\mathcal{M}\). Thus all open-set preimages are measurable, and the open-set test implies that \(f\) is measurable. \(\square\)
This criterion is equivalent to the open-set criterion, not a separate notion of measurability. It can be convenient when a function is naturally described by the values it takes in closed sets. As with the open-set test, no condition on the measure’s size or completeness is needed.
Worked Example: A Closed-Set Test for an Affine Function
Let \(X=\mathbb{R}\) with Lebesgue measure and define \(f(x)=3x-2\). For a closed set \(F\subseteq\mathbb{R}\), $$ f^{-1}(F)=\left\{\frac{y+2}{3}:y\in F\right\}. $$ The map \(S(y)=(y+2)/3\) takes open intervals \((a,b)\) to open intervals \(((a+2)/3,(b+2)/3)\); because every open subset of \(\mathbb{R}\) is a union of open intervals, \(S\) takes open sets to open sets. It is a bijection, so \(S(F)^c=S(F^c)\). Since \(F^c\) is open, \(S(F^c)\) is open and \(S(F)\) is closed. Therefore \(f^{-1}(F)\) is closed, hence Lebesgue measurable, for every closed \(F\). The closed-set criterion proves that \(f\) is measurable.
For \(F=[1,5]\), the preimage can also be computed directly: $$ 1\le 3x-2\le 5 \quad\Longleftrightarrow\quad 3\le 3x\le 7 \quad\Longleftrightarrow\quad 1\le x\le \frac{7}{3}. $$ Thus \(f^{-1}([1,5])=[1,7/3]\), a closed interval. Check: \(f(1)=3(1)-2=1\) and \(f(7/3)=3(7/3)-2=7-2=5\), so both endpoints belong, as required. Its Lebesgue measure is \(7/3-1=4/3\).
Why Countability Matters
A common mistake is to represent an open set as a union of neighborhoods chosen around all of its points, take preimages, and then conclude that the preimage is measurable because it is a union of measurable sets. The conclusion does not follow from that reasoning: a sigma-algebra is guaranteed to be closed under countable unions, not arbitrary unions.
For a concrete instance of the faulty step, let \(f(x)=x\) on \((\mathbb{R},\mathcal{L},m)\) and \(U=(0,1)\). For each \(x\in(0,1)\), set \(\delta_x=\tfrac12\min\{x,1-x\}\). Then \((x-\delta_x,x+\delta_x)\subseteq(0,1)\), and the union of these intervals over all \(x\in(0,1)\) is \((0,1)\). The preimage of each interval under the identity is the same interval. The erroneous calculation is to invoke closure of \(\mathcal{L}\) under this potentially uncountable union. That closure has not been established and is not a sigma-algebra axiom.
The correction is to use the rational basis. In this example, $$ (0,1)=\bigcup_{\substack{p,q\in\mathbb{Q}\\0<p<q<1}}(p,q), $$ a countable union of measurable intervals. More generally, the rational-interval theorem shows exactly how to replace point-by-point neighborhoods by a countable family. It is this reduction, rather than any assumption about \(\mu\), that makes the open-set test work.
Check Your Understanding
Use the definitions and criteria above to justify each answer, paying attention to whether unions are countable.
- Why does measurability of \(f^{-1}((p,q))\) for rational \(p<q\) imply measurability of \(f^{-1}(U)\) for every open \(U\subseteq\mathbb{R}\)?
- For \(f(x)=\lfloor x\rfloor\), find \(f^{-1}((-1/2,3/2))\), stating explicitly which integers belong to the target interval.
- Prove that if \(f^{-1}(F)\in\mathcal{M}\) for every closed \(F\subseteq\mathbb{R}\), then \(f^{-1}(U)\in\mathcal{M}\) for every open \(U\subseteq\mathbb{R}\).
- For \(f(x)=2x+1\) on \(\mathbb{R}\), compute \(f^{-1}([3,7])\) and check both endpoints by substitution.
- What closure property is missing from the claim that an arbitrary union of measurable sets is measurable? How does the rational-interval basis avoid needing it?