Why Endpoint Level Sets Need Their Own Description
An extended-real-valued function may take either \(+\infty\) or \(-\infty\), so a real-valued level-set test must account for both endpoints. The key observation is that no finite real number exceeds every positive integer, and no finite real number is below every negative integer. Consequently, countably many ordinary threshold tests identify exactly the points where the function reaches an endpoint.
Let \((X,\mathcal{M},\mu)\) be any measure space; no finiteness, sigma-finiteness, or completeness assumption is needed here. Write \(\overline{\mathbb{R}}=[-\infty,\infty]\). Throughout, \(f:X\to\overline{\mathbb{R}}\) is extended-real-valued. Measurability means that \(f^{-1}(B)\in\mathcal{M}\) for every Borel set \(B\) in \(\overline{\mathbb{R}}\). In particular, for every real \(t\), the sets \(\{x:f(x)>t\}\) and \(\{x:f(x)<t\}\) are measurable: they are preimages of the Borel sets \((t,\infty]\) and \([-\infty,t)\), respectively.
The Countable-Intersection Formulas
Proof. For the first identity, suppose \(f(x)=+\infty\). Then \(f(x)>n\) for every positive integer \(n\), so \(x\) belongs to the intersection on the right. Conversely, suppose \(f(x)>n\) for every positive integer \(n\). If \(f(x)=-\infty\), this fails already for \(n=1\). If \(f(x)\) is finite, choose a positive integer \(n>f(x)\), which exists by the Archimedean property of the real numbers. Then \(f(x)>n\) is false. The only remaining possibility is \(f(x)=+\infty\). This proves the first identity.
For the second identity, if \(f(x)=-\infty\), then \(f(x)<-n\) for every positive integer \(n\). Conversely, suppose \(f(x)<-n\) for every positive integer \(n\). The value \(+\infty\) cannot satisfy even the inequality with \(n=1\). If \(f(x)\) is finite, choose a positive integer \(n>-f(x)\). Then \(-n<f(x)\), so \(f(x)<-n\) is false. Thus \(f(x)=-\infty\), proving the second identity. Each set in each intersection is measurable by the threshold-set observation, and \(\mathcal{M}\) is closed under countable intersections. Hence both endpoint level sets are in \(\mathcal{M}\). \(\square\)
The use of all positive integers is essential to the description: a single threshold only says that the function is sufficiently large or sufficiently negative. The intersection tests whether it passes every such threshold. The thresholds need not be integers specifically; any sequence tending to \(+\infty\) works for the positive endpoint, and its negative works for the negative endpoint. Integers are a convenient choice because they are countable and unbounded in the required directions.
Worked Example: A Positive-Infinity Value at the Endpoint
On \([0,1]\) with the restricted Lebesgue measure space \(([0,1],\mathcal{L}|_{[0,1]},m)\), define \(f(0)=+\infty\) and \(f(x)=1/x\) for \(0<x\le1\). This is an extended-real-valued measurable function: on \((0,1]\) it is continuous, and the exceptional singleton \(\{0\}\) is measurable. Determine its positive- and negative-infinity level sets from the thresholds.
For every positive integer \(n\), \(f(0)=+\infty>n\). At \(x>0\), the strict inequality \(f(x)>n\) is equivalent to \(1/x>n\), which is equivalent to \(x<1/n\). Thus $$ \{x:f(x)>n\}=[0,1/n). $$ The endpoint is excluded because \(f(1/n)=n\), not a value strictly greater than \(n\). Their intersection is \(\{0\}\): zero belongs to every interval, while if \(x>0\), any integer \(n\ge 1/x\) gives \(1/n\le x\), so \(x\notin[0,1/n)\). Therefore \(\{f=+\infty\}=\{0\}\). Its Lebesgue measure is \(0\), as is also seen directly from the measure of a singleton. For every \(n\ge1\), \(f(x)<-n\) holds nowhere, so \(\{f=-\infty\}=\bigcap_{n\ge1}\varnothing=\varnothing\). Check: the original definition assigns \(+\infty\) only at zero and never assigns \(-\infty\), agreeing with both intersections.
Both Endpoints and Finite Values
The two formulas apply independently, so an extended-real-valued function can have either endpoint level set empty, both empty, or one or both nonempty. The endpoint sets cannot overlap because \(+\infty\ne-\infty\). The formulas also give an exact criterion for everywhere finite-valuedness.
Proof. At each point \(x\), the value \(f(x)\) lies in \(\overline{\mathbb{R}}\), whose only values that are not real are \(+\infty\) and \(-\infty\). Therefore \(f\) is finite at every point exactly when neither endpoint level set contains a point, proving the first equivalence. By definition, \(f\) is finite \(\mu\)-almost everywhere exactly when the set of points where it is not finite is contained in a measurable set of measure zero. That set is precisely the union of the two endpoint level sets. The union is measurable by the theorem, so this is equivalent to its measure being zero. No completeness assumption is required. \(\square\)
Worked Example: Two Endpoint Values on a Finite Space
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be its power set, and let \(\mu\) be counting measure. Define \(f(a)=+\infty\), \(f(b)=-\infty\), and \(f(c)=2\). The function is measurable because every subset of \(X\) belongs to \(\mathcal{M}\).
At threshold \(n=1\), \(\{f>1\}=\{a,c\}\), since both \(+\infty\) and \(2\) strictly exceed \(1\). At \(n=2\), \(\{f>2\}=\{a\}\), since \(2>2\) is false. For every \(n\ge2\), \(\{f>n\}=\{a\}\). Hence the intersection of all positive-threshold sets is \(\{a\}\). For every \(n\ge1\), \(\{f<-n\}=\{b\}\): \(-\infty<-n\), while \(2<-n\) is false. Thus the negative-threshold intersection is \(\{b\}\). Counting measure gives \(\mu(\{a\})=1\) and \(\mu(\{b\})=1\). Check: direct inspection of the three assigned values gives the same endpoint sets, and \(c\) is excluded from both because its value is finite.
Worked Example: An Endpoint Set of Measure Zero
On \((\mathbb{R},\mathcal{L},m)\), let \(f(x)=-\infty\) for \(x\in\mathbb{Q}\) and \(f(x)=0\) for \(x\notin\mathbb{Q}\). This function is measurable: for any Borel set \(B\subseteq\overline{\mathbb{R}}\), its preimage is one of \(\varnothing,\mathbb{Q},\mathbb{R}\setminus\mathbb{Q},\mathbb{R}\), all of which are Lebesgue measurable.
For every positive integer \(n\), the strict lower-threshold set \(\{f<-n\}\) is exactly \(\mathbb{Q}\), because \(-\infty<-n\), whereas \(0<-n\) is false. Thus \(\bigcap_{n\ge1}\{f<-n\}=\mathbb{Q}=\{f=-\infty\}\). The strict upper-threshold set \(\{f>n\}\) is empty for every \(n\), so \(\{f=+\infty\}=\varnothing\). Since \(\mathbb{Q}\) is countable, it has Lebesgue measure zero: if its elements are enumerated as \(q_1,q_2,\ldots\), then for every \(\varepsilon>0\) it can be covered by intervals of respective lengths \(\varepsilon 2^{-k}\), whose total length is \(\varepsilon\). Therefore \(f\) is finite almost everywhere but not finite everywhere. Check: on every irrational point \(f=0\), and on every rational point \(f=-\infty\); this matches the computed endpoint sets exactly.
A Common Mistake: Taking a Union
The positive-infinity level set is an intersection, not a union, of upper-threshold sets. A union asks whether the function exceeds at least one integer; an intersection asks whether it exceeds every positive integer. Only the latter forces the value to be \(+\infty\). The same distinction applies to lower thresholds and \(-\infty\).
For a concrete incorrect calculation, let \(f(x)=3\) at every point of a nonempty measurable space \((X,\mathcal{M},\mu)\). Then \(\{f>1\}=X\) and \(\{f>2\}=X\), so \(\bigcup_{n\ge1}\{f>n\}=X\). Claiming this union equals \(\{f=+\infty\}\) would incorrectly label every point as an infinite value. In fact, \(\{f>3\}=\varnothing\), so the intersection \(\bigcap_{n\ge1}\{f>n\}\) is empty, as required because \(f\) is finite everywhere. For the negative endpoint, the corresponding mistaken union also fails: with the constant function \(f=-3\), the union \(\bigcup_{n\ge1}\{f<-n\}\) contains \(X\) (already at \(n=4\) it does not; rather, the correct test is \(n=1,2\), for which \(-3<-1\) and \(-3<-2\), so the union is \(X\)), although \(\{f=-\infty\}=\varnothing\). The intersection is empty because \(\{f<-4\}=\varnothing\).
These descriptions are useful whenever a proof needs to know whether an extended-real-valued measurable function actually takes endpoint values. They establish measurability of the exceptional sets directly from measurable threshold sets, and they separate two different claims: being finite everywhere means both endpoint sets are empty, while being finite almost everywhere means their union has measure zero. A null endpoint set may still be nonempty, as the rational example shows.
Check Your Understanding
For each question, use strict inequalities in the threshold sets and distinguish pointwise conclusions from almost-everywhere conclusions.
- Write the countable-intersection formula for \(\{f=+\infty\}\) and justify why a finite value cannot belong to that intersection.
- On \([0,1]\), define \(g(0)=-\infty\) and \(g(x)=-1/x\) for \(x>0\). Find \(\{g<-n\}\) for each positive integer \(n\), then compute their intersection.
- If \(f\) is extended-real-valued and measurable on \((X,\mathcal{M},\mu)\), what condition on its two endpoint level sets is equivalent to \(f\) being finite everywhere?
- For a constant function with value \(4\), compare \(\bigcup_{n\ge1}\{f>n\}\) with \(\bigcap_{n\ge1}\{f>n\}\). Which one gives the positive-infinity level set?
- Can an extended-real-valued measurable function fail to be finite at some points while still being finite \(\mu\)-almost everywhere? State the condition on the union of its endpoint level sets.