Measurability Through Preimages

Testing Measurability with Threshold Sets

A real-valued function is measurable when the preimage of every Borel set is measurable in its domain. That definition can appear to demand many separate checks. In practice, it is often enough to examine preimages of simple sets that describe whether the function lies above a threshold. This tutorial develops that test and shows how a countable collection of rational thresholds suffices.

Throughout, let \((X,\mathcal{F})\) be a measurable space and let \(f:X\to\mathbb{R}\). Measurability of \(f\) means measurability as a map from \((X,\mathcal{F})\) to \((\mathbb{R},\mathcal{B}(\mathbb{R}))\). For a real number \(a\), the set \(\{x\in X:f(x)>a\}\) records the inputs where the function exceeds \(a\). Such sets are also called strict superlevel sets.

Theorem (Threshold Characterization of Measurability): For \(f:X\to\mathbb{R}\), the following conditions are equivalent:

\(f\) is measurable;
\(\{x\in X:f(x)>q\}\in\mathcal{F}\) for every rational number \(q\);
\(\{x\in X:f(x)>a\}\in\mathcal{F}\) for every real number \(a\).

If any of these conditions holds, then the preimages of all four types of rays \((a,\infty)\), \([a,\infty)\), \((-\infty,a)\), and \((-\infty,a]\) are measurable for every real \(a\).

Proof. Suppose first that \(f\) is measurable. Each ray \((a,\infty)\) is open, and hence Borel. Therefore its preimage \(\{x:f(x)>a\}\) belongs to \(\mathcal{F}\) for every real \(a\). In particular, this holds for every rational \(q\).

Now assume that \(\{x:f(x)>q\}\in\mathcal{F}\) for every rational \(q\). Fix a real \(a\). Density of the rationals gives the identity

\{x:f(x)>a\} = \bigcup_{\substack{q\in\mathbb{Q}\\q>a}}\{x:f(x)>q\}.

To check the identity, if \(f(x)>a\), choose a rational \(q\) strictly between \(a\) and \(f(x)\); then \(f(x)>q\). Conversely, if \(f(x)>q\) for some rational \(q>a\), then \(f(x)>a\). The union is countable, so it belongs to \(\mathcal{F}\). Thus the rational-threshold condition implies the condition for every real threshold.

It remains to show that the real-threshold condition implies measurability. First, complements give

\{x:f(x)\leq a\}=X\setminus\{x:f(x)>a\}\in\mathcal{F}.

For any real \(b\), density of the rationals also gives

\{x:f(x)<b\} = \bigcup_{\substack{q\in\mathbb{Q}\\q<b}}\{x:f(x)\leq q\}.

Indeed, if \(f(x)<b\), there is a rational \(q\) with \(f(x)\leq q<b\). Conversely, \(f(x)\leq q<b\) implies \(f(x)<b\). This is a countable union of measurable sets. The remaining closed ray is obtained by taking a complement:

\{x:f(x)\geq a\}=X\setminus\{x:f(x)<a\}\in\mathcal{F}.

Now let \(O\subseteq\mathbb{R}\) be open. By the rational-interval generators for the Borel sigma-algebra, \(O\) is the union of the rational open intervals contained in it. This is a countable union, and for rational \(p<q\),

f^{-1}((p,q)) = \{x:f(x)>p\}\cap\{x:f(x)<q\}\in\mathcal{F}.

Therefore \(f^{-1}(O)\in\mathcal{F}\) for every open \(O\). The collection of all sets \(B\subseteq\mathbb{R}\) for which \(f^{-1}(B)\in\mathcal{F}\) is a sigma-algebra: preimages preserve complements and countable unions. Since this sigma-algebra contains every open set, it contains \(\mathcal{B}(\mathbb{R})\). Hence \(f\) is measurable. The displayed complement and union identities also establish measurability of the other three types of rays. \(\square\)

Why Rational Thresholds Are Enough

The reduction to rational thresholds is useful because it replaces an uncountable family of tests by a countable one. The key point is not that the function takes rational values. Its values can be arbitrary real numbers. Rather, rational numbers are dense: whenever a value lies strictly above a real threshold, a rational threshold can be placed between them.

The same idea explains why strict and non-strict inequalities can be exchanged. For example, the set where \(f\leq a\) is the complement of the set where \(f>a\). To describe the set where \(f<b\), take the union of the sets where \(f\leq q\) over rational \(q<b\). The countability of the rational numbers is essential: sigma-algebras guarantee closure under countable unions, not arbitrary unions.

Worked Example: Testing the Square Function

Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\), with \(\mathbb{R}\) equipped with its Borel sigma-algebra. For each real threshold \(a\), compute the strict superlevel set:

\{x:x^2>a\} = \begin{cases} \mathbb{R}, & a<0,\\ (-\infty,0)\cup(0,\infty), & a=0,\\ (-\infty,-\sqrt{a})\cup(\sqrt{a},\infty), & a>0. \end{cases}

When \(a<0\), \(x^2\geq 0>a\) for every real \(x\), so the preimage is all of \(\mathbb{R}\). When \(a=0\), the inequality \(x^2>0\) holds precisely when \(x\neq 0\). When \(a>0\), \(x^2>a\) is equivalent to \(|x|>\sqrt{a}\), which means \(x<-\sqrt{a}\) or \(x>\sqrt{a}\). Each resulting set is open and therefore Borel. The threshold characterization proves that \(f\) is measurable.

Worked Example: A Rational-Irrational Piecewise Function

Define \(g:\mathbb{R}\to\mathbb{R}\) by

g(x)= \begin{cases} x, & x\in\mathbb{Q},\\ x+1, & x\notin\mathbb{Q}. \end{cases}

For any real \(a\), separate the rational and irrational inputs when computing the strict superlevel set:

\{x:g(x)>a\} = \bigl(\mathbb{Q}\cap(a,\infty)\bigr) \cup \bigl((\mathbb{R}\setminus\mathbb{Q})\cap(a-1,\infty)\bigr).

On the rational branch, \(g(x)>a\) means \(x>a\). On the irrational branch, \(g(x)>a\) means \(x+1>a\), or \(x>a-1\). The rationals are a countable union of closed singleton sets, hence Borel; their complement and the two open intervals are Borel as well. The displayed union is therefore Borel for every \(a\). By the threshold characterization, \(g\) is Borel measurable. This argument checks the preimages directly and does not require \(g\) to be continuous.

Worked Example: The Positive-Part Function

Let \(h(x)=\max\{x,0\}\) on \(\mathbb{R}\). For \(a<0\), \(h(x)\geq 0>a\) for all \(x\), so \(\{x:h(x)>a\}=\mathbb{R}\). At \(a=0\), the inequality holds precisely when \(x>0\). For \(a>0\), it holds precisely when \(x>a\). Thus

\{x:h(x)>a\} = \begin{cases} \mathbb{R}, & a<0,\\ (0,\infty), & a=0,\\ (a,\infty), & a>0. \end{cases}

Every set in this list is Borel, so the threshold criterion establishes measurability. Considering the threshold \(a=0\) separately matters: replacing the strict inequality with a non-strict one would incorrectly include \(x=0\) in this particular preimage.

Monotone Functions Are Measurable

The threshold test also gives a useful general result. A nondecreasing function may have jump discontinuities, but each of its strict superlevel sets has a particularly simple shape: if it contains a point, it contains every larger point. Such a set is an interval of one of a few possible forms, and is therefore Borel.

Theorem (Measurability of a Monotone Function): Every nondecreasing function \(f:\mathbb{R}\to\mathbb{R}\) is Borel measurable.

Proof. Fix \(a\in\mathbb{R}\), and let \(E=\{x\in\mathbb{R}:f(x)>a\}\). If \(x\in E\) and \(y>x\), nondecreasingness gives \(f(y)\geq f(x)>a\), so \(y\in E\). Thus \(E\) is an upward-closed set.

If \(E=\varnothing\), it is Borel. Suppose \(E\neq\varnothing\). If \(E\) is unbounded below, then for every \(y\in\mathbb{R}\) there is an \(x\in E\) with \(x<y\). Upward closure implies \(y\in E\), so \(E=\mathbb{R}\). Otherwise \(E\) is bounded below, and \(c=\inf E\) is a real number. No \(x<c\) belongs to \(E\). If \(x>c\), then \(x\) cannot be a lower bound for \(E\), so there exists \(y\in E\) with \(y<x\). Upward closure gives \(x\in E\). Membership of \(c\) itself is the only remaining possibility. Consequently, \(E=(c,\infty)\) or \(E=[c,\infty)\). Both are Borel.

For every real \(a\), the preimage \(\{x:f(x)>a\}\) is therefore Borel. The threshold characterization implies that \(f\) is Borel measurable. \(\square\)

Worked Example: A Monotone Function with a Jump

Define \(u:\mathbb{R}\to\mathbb{R}\) by \(u(x)=0\) for \(x<0\) and \(u(x)=2\) for \(x\geq 0\). This function is nondecreasing: within each of the two regions it is constant, and if \(x<0\leq y\), then \(u(x)=0\leq 2=u(y)\). Its threshold preimages are

\{x:u(x)>a\} = \begin{cases} \mathbb{R}, & a<0,\\ [0,\infty), & 0\leq a<2,\\ \varnothing, & a\geq 2. \end{cases}

For thresholds below zero, both possible values exceed \(a\). For thresholds from zero up to but not including two, only the value two exceeds \(a\), and it occurs exactly on \([0,\infty)\). For thresholds at least two, neither value exceeds \(a\). All three types of sets are Borel. This verifies measurability directly and illustrates how a jump affects the endpoint of a threshold set.

Using the Criterion Carefully

The threshold characterization is particularly helpful when a function has an algebraic formula, a piecewise definition, or monotonicity that makes the sets \(\{f>a\}\) easy to describe. Once those sets are known to be measurable for every rational threshold, there is no need to calculate the preimage of each Borel set separately.

There are two common pitfalls. First, a single threshold is not enough: showing that \(\{f>a\}\) is measurable for one chosen \(a\) says nothing by itself about the other thresholds. Second, the sigma-algebra on the domain still matters. The criterion asks that the threshold preimages belong to the given \(\mathcal{F}\); being Borel subsets of \(\mathbb{R}\) is not sufficient if the domain carries a smaller sigma-algebra. For a domain \((\mathbb{R},\mathcal{L})\), Borel threshold sets do belong to \(\mathcal{L}\), since Borel sets are Lebesgue measurable.

Key takeaway: To establish measurability of a real-valued function, it is enough to check that the preimages of rational strict upper rays are measurable. Density of the rationals recovers all real thresholds, and those thresholds determine the preimages of open sets and hence all Borel sets.

Check Your Understanding

Use the threshold characterization and the examples above to answer the following questions.

Why does measurability of \(f\) imply that \(\{x:f(x)>q\}\) is measurable for every rational \(q\)?
How can the set \(\{x:f(x)>a\}\), for real \(a\), be written as a countable union using rational thresholds?
Explain why the preimage of \((p,q)\) can be expressed using two threshold sets.
For the rational-irrational piecewise function, what condition on \(x\) gives \(g(x)>a\) when \(x\) is irrational?
Why must an upward-closed subset of \(\mathbb{R}\) be Borel?