Countable Suprema and Arbitrary Suprema
For a sequence of measurable functions, taking the pointwise supremum preserves measurability. The reason is that each strict upper level set of the supremum is a countable union of strict upper level sets of the individual functions. For an uncountable family, the corresponding union need not be measurable: a sigma-algebra is required to contain countable unions, not unions of arbitrary size.
We first record the countable result. On a measurable space \((X,\mathcal{M})\), a function \(f:X\to[-\infty,\infty]\) is measurable when \(\{x:f(x)>t\}\in\mathcal{M}\) for every \(t\in\mathbb{R}\). In what follows, each \(f_n:X\to\mathbb{R}\) is measurable, and its pointwise supremum is allowed to take the extended-real value \(+\infty\).
Proof. Fix any \(t\in\mathbb{R}\). At a point \(x\), the supremum \(\sup_n f_n(x)\) is strictly greater than \(t\) if and only if at least one \(f_n(x)\) is strictly greater than \(t\). Indeed, if every \(f_n(x)\le t\), then \(t\) is an upper bound and the supremum is at most \(t\); conversely, if some \(f_n(x)>t\), the supremum is at least that value and is strictly greater than \(t\). Thus \(\{x:g(x)>t\}=\bigcup_{n=1}^{\infty}\{x:f_n(x)>t\}\). Each set in this union belongs to \(\mathcal{M}\), and \(\mathcal{M}\) is closed under countable unions. Hence \(\{g>t\}\in\mathcal{M}\) for every real \(t\), which is the measurability condition. \(\square\)
For a family of indicator functions, the connection between a supremum and a union is especially direct. If \(I\) is a nonempty index set and \(E_i\in\mathcal{M}\) for each \(i\in I\), then, pointwise, \(\sup_{i\in I}\chi_{E_i}=\chi_{\cup_{i\in I}E_i}\). At a point in the union at least one indicator equals \(1\), so the supremum is \(1\); at a point outside the union every indicator is \(0\), so the supremum is \(0\). This identity does not say the union is measurable. That depends on whether the particular union belongs to \(\mathcal{M}\).
A Nonmeasurable Set from Rational Translates
We construct a non-Lebesgue-measurable set in \(\mathbb{R}\). On \([0,1]\), define an equivalence relation by \(x\sim y\) if and only if \(x-y\in\mathbb{Q}\). Choose one representative from each equivalence class and let \(V\subseteq[0,1]\) be the set of chosen representatives. This selection uses the axiom of choice. The resulting set is called a Vitali set. We show directly that \(V\) is not Lebesgue measurable.
Proof. Let \(Q_0=\mathbb{Q}\cap[-1,1]\), a countably infinite set. For every \(q\in Q_0\), consider \(V+q=\{v+q:v\in V\}\). If \(V+q_1\) and \(V+q_2\) intersect, there are \(v_1,v_2\in V\) with \(v_1+q_1=v_2+q_2\). Therefore \(v_1-v_2=q_2-q_1\in\mathbb{Q}\), so \(v_1\sim v_2\). Since \(V\) contains just one representative of each equivalence class, \(v_1=v_2\), and then \(q_1=q_2\). Thus the translates for distinct \(q\in Q_0\) are disjoint.
They cover \([0,1]\): for any \(x\in[0,1]\), let \(v\in V\) be the representative equivalent to \(x\). Then \(q=x-v\) is rational and lies in \([-1,1]\), so \(x=v+q\in V+q\). Also, every translate is contained in \([-1,2]\), since \(V\subseteq[0,1]\) and \(q\in[-1,1]\).
Suppose, for contradiction, that \(V\in\mathcal{L}\). Translation preserves Lebesgue measurability and measure, so each \(V+q\) is measurable and has measure \(m(V)\). Since \(V\subseteq[0,1]\), this measure is finite. If \(m(V)=0\), countable subadditivity gives \(m(\bigcup_{q\in Q_0}(V+q))=0\), contradicting that this union contains \([0,1]\), which has measure \(1\). If \(m(V)>0\), countable additivity on the disjoint translates gives the union infinite measure: its measure is the sum of infinitely many equal positive terms. But the union is contained in \([-1,2]\), whose measure is \(3\), a contradiction. Both cases are impossible, so \(V\notin\mathcal{L}\). \(\square\)
Worked Examples: Suprema That Behave Differently
Worked Example: A Measurable Family with a Nonmeasurable Supremum
Work on \((\mathbb{R},\mathcal{L},m)\), where \(m\) is Lebesgue measure. This measure space is complete and sigma-finite, but not finite. For each \(v\in V\), define the real-valued function \(f_v=\chi_{\{v\}}\). Show that every \(f_v\) is measurable, but their pointwise supremum is not.
For each \(v\), the singleton \(\{v\}\) is closed, hence Borel, hence Lebesgue measurable. Therefore its indicator \(f_v\) is measurable. At any \(x\in\mathbb{R}\), if \(x\in V\), then the index \(v=x\) occurs in the family and \(f_x(x)=1\); every indicator is at most \(1\), so the supremum equals \(1\). If \(x\notin V\), then \(x\ne v\) for every \(v\in V\), and every \(f_v(x)=0\), so the supremum equals \(0\). Consequently, \(\sup_{v\in V} f_v(x)=\chi_V(x)\) for every \(x\in\mathbb{R}\).
The supremum is not measurable: its strict upper level set at \(t=\tfrac12\) is \(\{x:\chi_V(x)>\tfrac12\}=V\), which is not in \(\mathcal{L}\). Check: the values are exactly \(1\) on \(V\) and \(0\) off \(V\), so the level-set test detects precisely the nonmeasurable set used in the construction.
Worked Example: A Countable Supremum for the Rationals
On the same measure space, enumerate the rational numbers as \((q_n)_{n\ge1}\) and put \(h_n=\chi_{\{q_n\}}\). Compute the supremum and verify its measurability.
Each singleton \(\{q_n\}\) is Borel, so each \(h_n\) is real-valued and measurable. For \(x\in\mathbb{R}\), if \(x=q_k\) for some \(k\), then \(h_k(x)=1\), and the supremum is \(1\). If \(x\notin\mathbb{Q}\), then \(h_n(x)=0\) for every \(n\), and the supremum is \(0\). Thus \(\sup_n h_n=\chi_{\mathbb{Q}}\).
For \(t\in\mathbb{R}\), the strict upper level set is \(\{\sup_nh_n>t\}=\mathbb{R}\) when \(t<0\), \(\mathbb{Q}\) when \(0\le t<1\), and \(\varnothing\) when \(t\ge1\). These are all Lebesgue measurable: \(\mathbb{Q}\) is countable and therefore a countable union of Borel singletons. Hence the supremum is measurable. Check: the level-set computation agrees with the countable-supremum proposition, including the endpoint thresholds \(t=0\) and \(t=1\).
Worked Example: An Uncountable Supremum That Is Measurable
For each \(a\in[0,1]\), define \(u_a=\chi_{(a-1,a+1)}\) on \((\mathbb{R},\mathcal{L},m)\). Each interval is open and hence Borel. Compute the supremum over the uncountable index set \([0,1]\).
The union of the intervals \((a-1,a+1)\), for \(a\in[0,1]\), is exactly \((-1,2)\). Every point in one of the intervals lies strictly between \(-1\) and \(2\), since \(a-1\ge-1\) and \(a+1\le2\), with the interval endpoints excluded. Conversely, if \(-1<x<2\), there is an \(a\in[0,1]\) with \(|x-a|<1\): for \(-1<x\le0\), choose \(a=0\); for \(0<x<1\), choose \(a=x\); and for \(1\le x<2\), choose \(a=1\). In each case \(x\in(a-1,a+1)\). Therefore \(\sup_{a\in[0,1]}u_a=\chi_{(-1,2)}\).
This supremum is measurable because \((-1,2)\) is Borel. Check: at \(x=-1\) and \(x=2\), every relevant interval excludes the endpoint, so the supremum is \(0\); at every \(x\in(-1,2)\), the choices above give a function with value \(1\). Uncountability alone therefore does not cause failure.
The Precise Pitfall
A common error is to argue, “Every set \(E_i\) is measurable, so their union is measurable.” A sigma-algebra guarantees closure under countable unions; it does not guarantee closure under arbitrary unions. The Vitali example makes the distinction explicit: every singleton \(\{v\}\) is measurable, but the union \(\bigcup_{v\in V}\{v\}=V\) is not.
There is also a level-set detail worth keeping straight. For a countable supremum, the identity uses a strict inequality: \(\{\sup_n f_n>t\}=\bigcup_n\{f_n>t\}\). Replacing \(>\) by \(\ge\) is not generally valid. For example, on the one-point space \(X=\{x\}\) with its full sigma-algebra, let \(f_n(x)=1-\frac1n\). Then \(\sup_n f_n(x)=1\), so \(\{\sup_n f_n\ge1\}=X\), whereas \(\bigcup_n\{f_n\ge1\}=\varnothing\). The supremum equals the threshold without any term attaining it.
For an uncountable family, the safe diagnostic is to compute its strict upper level sets directly. If any such set is nonmeasurable, the supremum is not measurable. If all are measurable, the supremum is measurable, even if the family is uncountable. The examples above show both outcomes and separate the actual issue—the measurability of the relevant unions—from the mere size of the index set.
Check Your Understanding
Use level sets and the distinction between countable and arbitrary unions to answer each question.
- For measurable sets \(E_n\), prove directly that \(\sup_n\chi_{E_n}=\chi_{\cup_nE_n}\) is measurable.
- Why does the family \(f_v=\chi_{\{v\}}\), indexed by the Vitali set, consist of measurable functions even though its supremum is not measurable?
- Give an example of an uncountable family of measurable functions whose supremum is measurable, and verify the supremum pointwise.
- For the sequence \(f_n(x)=1-\frac1n\) on a one-point measurable space, compare \(\{\sup_n f_n\ge1\}\) with \(\bigcup_n\{f_n\ge1\}\).
- State the countable-supremum level-set identity, and explain why the sigma-algebra axiom applies to its right-hand side.