Measurability with Infinite Values
Limits of real-valued measurable functions need not be finite: a pointwise limit can equal \(+\infty\) or \(-\infty\). To include such values, the codomain must be enlarged, and measurability must be tested against sets in that enlarged codomain. In particular, checking preimages of ordinary open subsets of \(\mathbb{R}\) alone is not enough.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space; no assumption of finiteness, sigma-finiteness, or completeness is needed for the definitions or results below. The functions in this tutorial take values in the extended real line \(\overline{\mathbb{R}}=[-\infty,+\infty]\), with the order \(-\infty<x<+\infty\) for every \(x\in\mathbb{R}\). Equip \(\overline{\mathbb{R}}\) with its order topology. Its open sets include intervals \((a,b)\) and the endpoint rays \([-\infty,b)\) and \((a,+\infty]\), with endpoints allowed in \(\overline{\mathbb{R}}\). Write \(\mathcal{B}_{\overline{\mathbb{R}}}\) for the Borel sigma-algebra generated by these open sets.
This is the usual definition of a measurable map into a measurable space, applied to the target \((\overline{\mathbb{R}},\mathcal{B}_{\overline{\mathbb{R}}})\). For a real-valued function, measurability as a map into \(\mathbb{R}\) is equivalent to measurability as a map into \(\overline{\mathbb{R}}\): Borel sets of either space restrict to Borel sets of the other on \(\mathbb{R}\). For an extended-real-valued function, however, the two endpoints are genuine target values and must be included in the tests.
Open Sets Give a Complete Test
Proof. If \(f\) is measurable, then the preimage of every Borel set belongs to \(\mathcal{M}\), and every open set is Borel, so the open-set condition holds. Conversely, suppose that the preimage of every open subset of \(\overline{\mathbb{R}}\) belongs to \(\mathcal{M}\). Define \(\mathcal{C}=\{B\in\mathcal{B}_{\overline{\mathbb{R}}}: f^{-1}(B)\in\mathcal{M}\}\). Preimages preserve complements and countable unions: \(f^{-1}(B^c)=(f^{-1}(B))^c\) and \(f^{-1}(\bigcup_n B_n)=\bigcup_n f^{-1}(B_n)\). Since \(\mathcal{M}\) is a sigma-algebra, \(\mathcal{C}\) is a sigma-algebra. It contains every open set by assumption, and therefore contains the sigma-algebra generated by the open sets, namely \(\mathcal{B}_{\overline{\mathbb{R}}}\). Thus \(f\) is measurable. \(\square\)
The proposition allows a useful strategy: describe the preimage of an arbitrary Borel target set, or verify the preimages of open target sets. When a formula is continuous on the part of the domain where its values are finite, it is often enough to handle that part by continuity and then account separately for any points assigned an infinite value.
Worked Example: Infinite Values on the Rationals and Irrationals
On \((\mathbb{R},\mathcal{L},m)\), define \(f(x)=+\infty\) for rational \(x\), and \(f(x)=-\infty\) for irrational \(x\). Here \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. Verify measurability and test an open set that distinguishes the two endpoint values.
Let \(B\in\mathcal{B}_{\overline{\mathbb{R}}}\). There are four possibilities. If \(B\) contains neither endpoint, then \(f^{-1}(B)=\varnothing\). If \(B\) contains \(+\infty\) but not \(-\infty\), then \(f^{-1}(B)=\mathbb{Q}\). If it contains \(-\infty\) but not \(+\infty\), then \(f^{-1}(B)=\mathbb{R}\setminus\mathbb{Q}\). If it contains both, then \(f^{-1}(B)=\mathbb{R}\). Each set is Lebesgue measurable: \(\mathbb{Q}\) is countable and hence Borel, its complement is Borel, and \(\varnothing,\mathbb{R}\in\mathcal{L}\). Thus \(f\) is measurable.
For the specific open set \(U=(0,+\infty]\), the endpoint \(+\infty\) belongs to \(U\), while \(-\infty\) does not. Therefore \(f^{-1}(U)=\mathbb{Q}\), and \(m(f^{-1}(U))=0\). The strict lower endpoint matters: \(0\notin U\), though \(f\) never takes the value \(0\) in this example. Check: for \(V=[-\infty,0)\), the preimage is \(\mathbb{R}\setminus\mathbb{Q}\), because \(-\infty\in V\) and \(+\infty\notin V\).
Worked Example: A Reciprocal with a Positive-Infinite Value
On \((\mathbb{R},\mathcal{L},m)\), define \(f(0)=+\infty\) and \(f(x)=1/x\) for \(x\ne0\). Verify measurability, then compute the preimage of \(B=(-\infty,-1)\cup\{+\infty\}\), a Borel subset of \(\overline{\mathbb{R}}\).
For any Borel set \(C\subseteq\overline{\mathbb{R}}\), its preimage has the form \(f^{-1}(C)=\{x\ne0:1/x\in C\cap\mathbb{R}\}\cup\{0\}\) if \(+\infty\in C\), and has no \(\{0\}\) term if \(+\infty\notin C\). The set \(C\cap\mathbb{R}\) is Borel in \(\mathbb{R}\). The reciprocal function is continuous on the open set \(\mathbb{R}\setminus\{0\}\), so its preimage of \(C\cap\mathbb{R}\), taken within that open set, is Borel there and hence Borel in \(\mathbb{R}\). Adding or omitting the Borel singleton \(\{0\}\) preserves Borel measurability. Since every Borel subset of \(\mathbb{R}\) belongs to \(\mathcal{L}\), \(f\) is measurable.
Now calculate the requested preimage. For \(x\ne0\), the condition \(1/x<-1\) holds exactly when \(-1<x<0\): on negative \(x\), multiplying by \(x\) reverses the inequality, giving \(1>-x\), or \(x>-1\); positive \(x\) cannot satisfy it. The additional target value \(+\infty\) is attained at \(x=0\). Hence \(f^{-1}(B)=(-1,0)\cup\{0\}=(-1,0]\). Check: \(x=-1\) gives \(-1\), which is excluded by the strict inequality, and \(x=0\) gives \(+\infty\), which is included. The interval \((-1,0]\) is Borel and has Lebesgue measure \(1\).
A Real-Valued Coordinate for the Extended Line
The extended line can also be represented by an ordinary bounded interval. Define \(T:\overline{\mathbb{R}}\to[-\pi/2,\pi/2]\) by \(T(t)=\arctan(t)\) for finite \(t\), \(T(-\infty)=-\pi/2\), and \(T(+\infty)=\pi/2\). This map is a homeomorphism: it is a strictly increasing bijection, and its endpoint behavior follows from \(\arctan(t)\to-\pi/2\) as \(t\to-\infty\) and \(\arctan(t)\to\pi/2\) as \(t\to+\infty\).
Proof. Suppose first that \(f\) is measurable. For every Borel set \(D\subseteq\mathbb{R}\), \(T^{-1}(D)\) is Borel in \(\overline{\mathbb{R}}\), because \(T\) is continuous and \(D\) is Borel. Therefore \((T\circ f)^{-1}(D)=f^{-1}(T^{-1}(D))\in\mathcal{M}\), so \(T\circ f\) is measurable. Conversely, suppose \(T\circ f\) is measurable. For \(B\in\mathcal{B}_{\overline{\mathbb{R}}}\), the set \(T(B)\) is Borel in the closed interval \([-\pi/2,\pi/2]\), since \(T^{-1}\) is continuous. A Borel subset of this closed interval is the intersection of the interval with a Borel subset \(D\) of \(\mathbb{R}\). Because \(T\circ f\) always takes values in the interval, \((T\circ f)^{-1}(T(B))=(T\circ f)^{-1}(D)\). This set belongs to \(\mathcal{M}\) by measurability of \(T\circ f\). Since \(T\) is one-to-one, this preimage is exactly \(f^{-1}(B)\). Thus \(f\) is measurable. \(\square\)
Worked Example: A Logarithm Attaining Negative Infinity
Let \(X=[0,\infty)\) with its relative Lebesgue sigma-algebra and Lebesgue measure restricted to \(X\). Define \(f(0)=-\infty\) and \(f(x)=\log x\) for \(x>0\). Verify measurability and compute the preimage of \(U=[-\infty,0)\).
For any \(B\in\mathcal{B}_{\overline{\mathbb{R}}}\), the preimage is \(\{x>0:\log x\in B\cap\mathbb{R}\}\), together with \(\{0\}\) if \(-\infty\in B\). The set \(B\cap\mathbb{R}\) is Borel in \(\mathbb{R}\). Since \(\log\) is continuous on \((0,\infty)\), the first set is Borel relative to \((0,\infty)\), and therefore Borel relative to \(X\). The singleton \(\{0\}\) is also Borel in \(X\). The full preimage is thus Borel in \(X\) in either case, so \(f\) is measurable.
The set \(U=[-\infty,0)\) contains \(-\infty\), excludes \(0\), and includes finite negative real numbers. For \(x>0\), \(\log x<0\) exactly when \(0<x<1\). The assigned value at \(x=0\) also lies in \(U\). Consequently \(f^{-1}(U)=\{0\}\cup(0,1)=[0,1)\), which has measure \(1\). Check: \(f(0)=-\infty\in U\), \(f(1)=0\notin U\), and \(f(2)=\log 2>0\notin U\); these endpoint and finite-value checks agree with \([0,1)\).
A Common Mistake: Testing Only Finite Values
Here is a concrete failure of the real-valued open-set test when it is used with the wrong codomain. Let \(X=\{a,b\}\), let \(\mathcal{M}=\{\varnothing,X\}\), and let \(\mu\) be the probability measure with \(\mu(X)=1\). The singleton \(\{a\}\) is not in \(\mathcal{M}\). Define \(f(a)=+\infty\) and \(f(b)=-\infty\).
A tempting but incorrect calculation is: for every ordinary open set \(V\subseteq\mathbb{R}\), \(f^{-1}(V)=\varnothing\), so \(f\) is measurable. That calculation checks only finite target values. In the extended-real topology, \(U=(0,+\infty]\) is open, and \(f^{-1}(U)=\{a\}\notin\mathcal{M}\). Thus \(f\) is not measurable. The correction is to test Borel sets in \(\overline{\mathbb{R}}\), not merely Borel sets in \(\mathbb{R}\). This is precisely where the endpoint values affect the definition.
The extended-real definition is essential whenever limits or formulas produce infinities, but it does not by itself assert that any integral exists. Measurability is a condition on preimages; integrability requires additional conditions and is a separate question. Keeping the target space explicit prevents an endpoint from being silently ignored.
Check Your Understanding
In each question, use the Borel sigma-algebra of the extended real line as the target sigma-algebra.
- On \((\mathbb{R},\mathcal{L},m)\), define \(g(x)=+\infty\) on \(\mathbb{Z}\) and \(g(x)=0\) elsewhere. Verify measurability by considering the preimage of an arbitrary Borel subset of \(\overline{\mathbb{R}}\).
- For the reciprocal example, find the preimage of \((-\infty,0)\). Does it contain \(x=0\)? Explain using the assigned value there.
- Show directly that the preimage of an open set under a measurable extended-real-valued function is measurable. Which part of the definition is being used?
- For the logarithm example, compute the preimage of \([-\infty,-1]\), checking separately \(x=0\) and \(x>0\).
- Explain why the two-point function in the common-mistake example passes every test using open subsets of \(\mathbb{R}\) but fails an open-set test in \(\overline{\mathbb{R}}\).