Measurability Is a Statement About Preimages
A real-valued function is not measurable in isolation: measurability depends on the sigma-algebra assigned to its domain and the sigma-algebra used on its range. This distinction matters on \(\mathbb{R}\), where the Borel sigma-algebra \(\mathcal{B}_{\mathbb{R}}\) is strictly smaller than the Lebesgue sigma-algebra \(\mathcal{L}\). The difference is captured by preimages: which sets are allowed in the domain determines whether all required preimages qualify.
For the real-line examples, \(\mathcal{B}_{\mathbb{R}}\) is the sigma-algebra generated by open subsets of \(\mathbb{R}\), and \(\mathcal{L}\) is the sigma-algebra of Lebesgue-measurable sets. Lebesgue measure \(m\) is not finite on \(\mathbb{R}\). The term Borel-measurable function will mean a real-valued function \(f:(\mathbb{R},\mathcal{B}_{\mathbb{R}})\to(\mathbb{R},\mathcal{B}_{\mathbb{R}})\) that is measurable as a map. The term Lebesgue-measurable function will mean a real-valued function \(f:(\mathbb{R},\mathcal{L})\to(\mathbb{R},\mathcal{B}_{\mathbb{R}})\) that is measurable as a map. Neither term refers to integrability.
For a real-valued map with range sigma-algebra \(\mathcal{B}_{\mathbb{R}}\), it is often enough to check a simpler family of preimages. Open rays generate the Borel sigma-algebra, so one can test whether \(\{x:f(x)>a\}\) belongs to the domain sigma-algebra for every real \(a\).
A Preimage Criterion for Real-Valued Functions
Proof. If \(f\) is measurable, then \((a,\infty)\in\mathcal{B}_{\mathbb{R}}\), so its preimage \(\{x:f(x)>a\}\) belongs to \(\mathcal{M}\). For the reverse direction, suppose all these preimages are in \(\mathcal{M}\). For each real \(q\), we can express the other type of strict ray as \(\{x:f(x)<q\}=\bigcup_{n=1}^{\infty}\{x:f(x)\le q-1/n\}\). Each set in this union is the complement of \(\{x:f(x)>q-1/n\}\), so it belongs to \(\mathcal{M}\). Every open subset of \(\mathbb{R}\) is a countable union of open intervals with rational endpoints. For rational \(p<q\), the preimage of \((p,q)\) is \(\{f>p\}\cap\{f<q\}\), which is in \(\mathcal{M}\). Countable unions preserve membership in \(\mathcal{M}\), so the preimage of every open set belongs to \(\mathcal{M}\). Since the open sets generate \(\mathcal{B}_{\mathbb{R}}\), the preimage of every Borel set belongs to \(\mathcal{M}\). Thus \(f\) is measurable. \(\square\)
The proof also explains why checking open rays works: the criterion checks enough sets to generate the full range sigma-algebra. It does not mean that a map into a larger range sigma-algebra can be tested using only open sets.
Worked Example: Testing the Function \(f(x)=x^2\)
Take \(X=\mathbb{R}\) with either \(\mathcal{M}=\mathcal{B}_{\mathbb{R}}\) or \(\mathcal{M}=\mathcal{L}\), and let \(f(x)=x^2\). We check the criterion by computing \(\{x:f(x)>a\}\) for every real \(a\).
If \(a<0\), then \(x^2\ge 0>a\) for every \(x\), so the preimage is \(\mathbb{R}\). If \(a=0\), then \(x^2>0\) exactly when \(x\ne0\), so the preimage is \((-\infty,0)\cup(0,\infty)\). If \(a>0\), the inequality \(x^2>a\) holds exactly when \(|x|>\sqrt a\), giving \((-\infty,-\sqrt a)\cup(\sqrt a,\infty)\). All three forms are Borel sets, and therefore belong to both \(\mathcal{B}_{\mathbb{R}}\) and \(\mathcal{L}\). The criterion proves that \(x^2\) is Borel-measurable and Lebesgue-measurable. Check: at \(a=4\), the computed preimage is \((-\infty,-2)\cup(2,\infty)\); substituting \(x=0\) gives \(0>4\), false, while substituting \(x=3\) gives \(9>4\), true.
How Changing the Domain Changes Measurability
The Lebesgue sigma-algebra contains every Borel set: \(\mathcal{B}_{\mathbb{R}}\subseteq\mathcal{L}\). In fact, it is the completion of the Borel sigma-algebra with respect to Lebesgue measure; in particular, it contains every subset of every Borel set of measure zero. This inclusion immediately gives one direction of the comparison for functions with Borel range.
Proof. Assume \(f\) is Borel-measurable. For every \(A\in\mathcal{B}_{\mathbb{R}}\), its preimage \(f^{-1}(A)\) belongs to \(\mathcal{B}_{\mathbb{R}}\) by the definition of Borel measurability. Since \(\mathcal{B}_{\mathbb{R}}\subseteq\mathcal{L}\), the same preimage belongs to \(\mathcal{L}\). This holds for every Borel set \(A\) in the range, so \(f\) is Lebesgue-measurable. The argument proves only this direction: it does not show that preimages under a Lebesgue-measurable function are Borel sets. \(\square\)
The strict inclusion \(\mathcal{B}_{\mathbb{R}}\subsetneq\mathcal{L}\) explains why the converse can fail. For example, the standard cardinality argument uses the null Cantor set: every subset of it is Lebesgue-measurable, while some such subsets are not Borel. Thus there are sets in \(\mathcal{L}\) that are not in \(\mathcal{B}_{\mathbb{R}}\). No claim that every Lebesgue-measurable function is Borel-measurable can be made just from the domain inclusion.
Worked Example: The Identity Map with Two Domain Sigma-Algebras
Let \(f:\mathbb{R}\to\mathbb{R}\) be the identity map \(f(x)=x\), with range sigma-algebra \(\mathcal{B}_{\mathbb{R}}\). First take the domain sigma-algebra to be \(\mathcal{B}_{\mathbb{R}}\). For every Borel set \(A\), \(f^{-1}(A)=A\), which is Borel; hence \(f\) is Borel-measurable. Next take the domain sigma-algebra to be \(\mathcal{L}\). The same calculation gives \(f^{-1}(A)=A\in\mathcal{L}\) for every Borel \(A\); hence it is also Lebesgue-measurable.
Now keep the identity map and the Borel range sigma-algebra, but shrink the domain sigma-algebra to \(\mathcal{M}_0=\{\varnothing,\mathbb{R}\}\). This is a sigma-algebra. The open range set \((0,\infty)\) has preimage \((0,\infty)\), which is neither \(\varnothing\) nor \(\mathbb{R}\), so it is not in \(\mathcal{M}_0\). The identity is not measurable for this domain sigma-algebra. Check: the preimage calculation is unchanged in all three cases; only the domain collection in which the preimage must lie has changed.
More generally, if \(\mathcal{M}_1\subseteq\mathcal{M}_2\) are sigma-algebras on the same set \(X\), measurability into a fixed range sigma-algebra with domain \(\mathcal{M}_1\) implies measurability with domain \(\mathcal{M}_2\). Each required preimage already lies in \(\mathcal{M}_1\), hence in \(\mathcal{M}_2\). The reverse implication need not hold. The range has a corresponding, opposite inclusion rule: if \(\mathcal{N}_1\subseteq\mathcal{N}_2\), measurability into \(\mathcal{N}_2\) implies measurability into \(\mathcal{N}_1\), because the smaller range sigma-algebra imposes fewer preimage conditions.
Worked Example: A Sigma-Algebra Tailored to a Function
Let \(H=(-\infty,0]\), let \(X=\mathbb{R}\), and equip \(X\) with the sigma-algebra \(\mathcal{M}=\{\varnothing,H,H^c,\mathbb{R}\}\). Define the real-valued function \(g\) by \(g(x)=0\) for \(x\in H\), and \(g(x)=1\) for \(x\in H^c\). We check its preimage for an arbitrary Borel set \(A\subseteq\mathbb{R}\).
If neither \(0\) nor \(1\) belongs to \(A\), then \(g^{-1}(A)=\varnothing\). If both belong to \(A\), then \(g^{-1}(A)=\mathbb{R}\). If \(0\in A\) and \(1\notin A\), then \(g^{-1}(A)=H\). If \(1\in A\) and \(0\notin A\), then \(g^{-1}(A)=H^c\). Every possible preimage is in \(\mathcal{M}\), so \(g:(X,\mathcal{M})\to(\mathbb{R},\mathcal{B}_{\mathbb{R}})\) is measurable. Check: for \(A=(1/2,3/2)\), only \(1\) belongs to \(A\), and the preimage is \(H^c=(0,\infty)\), which is in \(\mathcal{M}\). For \(A=(-1/2,1/2)\), only \(0\) belongs to \(A\), and the preimage is \(H\).
When the Range Sigma-Algebra Changes
There is another distinction to keep in view. For \(f:(\mathbb{R},\mathcal{B}_{\mathbb{R}})\to(\mathbb{R},\mathcal{L})\), measurability requires \(f^{-1}(A)\in\mathcal{B}_{\mathbb{R}}\) for every \(A\in\mathcal{L}\), including sets that are not Borel. Checking only that preimages of open sets are Borel establishes measurability into \(\mathcal{B}_{\mathbb{R}}\), not into the larger sigma-algebra \(\mathcal{L}\).
For a concrete test, suppose \(A\in\mathcal{L}\setminus\mathcal{B}_{\mathbb{R}}\) and consider the identity map with Borel domain and Lebesgue range. Its preimage of \(A\) is \(f^{-1}(A)=A\), which is not in the Borel domain sigma-algebra. Consequently, that identity map is not measurable into \((\mathbb{R},\mathcal{L})\), even though its preimages of open sets are open. The failure is not in the computation of open preimages; it is in checking too small a family of range sets.
Common mistake. A mistaken argument says: “The identity has open preimages of open sets, so it is measurable from \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\) into \((\mathbb{R},\mathcal{L})\).” The calculation \(f^{-1}(U)=U\) is correct for each open \(U\), but it does not check every Lebesgue-measurable range set. Taking \(A\in\mathcal{L}\setminus\mathcal{B}_{\mathbb{R}}\) gives \(f^{-1}(A)=A\notin\mathcal{B}_{\mathbb{R}}\), so the claimed measurability fails.
When checking any measurability claim, first write down the domain sigma-algebra and the range sigma-algebra. Then test preimages of a generating family when the range sets are generated by that family; otherwise, check the full required collection. This keeps Borel measurability, Lebesgue measurability, and measurability for other domain structures distinct.
Check Your Understanding
For each question, state the domain and range sigma-algebras before deciding whether the map is measurable.
- Use the open-ray criterion to verify that \(f(x)=x^2\) is measurable from \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\) to \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\).
- Explain why a Borel-measurable real-valued function on \(\mathbb{R}\) is Lebesgue-measurable.
- For the identity map on \(\mathbb{R}\), test measurability from the domain sigma-algebra \(\{\varnothing,\mathbb{R}\}\) into \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\) by using the open set \((0,\infty)\).
- In the tailored-sigma-algebra example, compute \(g^{-1}(A)\) when \(A\) contains both \(0\) and \(1\), and when it contains neither.
- Why does checking preimages of open sets alone not establish measurability into \((\mathbb{R},\mathcal{L})\)?