Measure Theory Mastery I

From Almost-Everywhere to Uniform Convergence

The previous tutorial showed how countable unions of measurable level sets can establish measurability for constructions involving sequences of functions. Here we use that same set-based approach to study a stronger question: when does pointwise convergence become uniform after a small part of the domain is removed? On a finite-measure space, almost-everywhere convergence has exactly this useful consequence. The result is Egorov’s theorem.

Let \((X,\mathcal{F},\mu)\) be a measure space. Throughout the main result, assume that \(\mu(X)<\infty\), and let \(f_n:X\to\mathbb{R}\) and \(f:X\to\mathbb{R}\) be measurable. We say \(f_n\) converges uniformly to \(f\) on a set \(A\subseteq X\) if, for every \(\delta>0\), there is an integer \(N\) such that \(|f_n(x)-f(x)|<\delta\) for every \(x\in A\) and every \(n\geq N\). The index \(N\) may depend on \(\delta\), but not on \(x\).

First, it is useful to make explicit that the set where convergence holds is measurable. This set will let us apply continuity from below without making assumptions about points in the null exceptional set.

Proof. For every positive integer \(k\), set

The functions \(f_n-f\) are measurable by the arithmetic closure properties of measurable functions, and the absolute value is a continuous function. Thus each set in the intersection is measurable, and \(A_{m,k}\in\mathcal{F}\) by closure under countable intersections. A sequence converges to \(f(x)\) precisely when, for each \(k\), its terms eventually stay within \(1/k\) of \(f(x)\). Consequently,

The right-hand side is formed from measurable sets using countable unions and intersections, so \(C\in\mathcal{F}\). \(\square\)

Egorov’s Theorem

Proof. Let \(C=\{x:f_n(x)\to f(x)\}\), which is measurable by the Measurability of the Convergence Set Theorem. Since convergence holds almost everywhere, \(\mu(X\setminus C)=0\). Fix a positive integer \(k\). As \(m\) increases, the sets \(A_{m,k}\) defined above increase: requiring the bound for every \(n\geq m+1\) imposes no more conditions than requiring it for every \(n\geq m\). The sets \(C\cap A_{m,k}\) therefore increase with \(m\), and their union is \(C\). Indeed, each point of \(C\) eventually satisfies the bound \(1/k\) for all subsequent indices. By Continuity from Below,

Because \(\mu(C)\leq\mu(X)<\infty\), the difference \(\mu(C)-\mu(C\cap A_{m,k})\) tends to zero. Choose \(m_k\) so that

Define the exceptional set

All the sets in this expression are measurable. By Countable Subadditivity and \(\mu(X\setminus C)=0\),

It remains to verify uniform convergence outside \(E\). If \(x\in X\setminus E\), then \(x\in C\) and \(x\in A_{m_k,k}\) for every \(k\). Hence, for each \(k\),

Given any \(\delta>0\), choose \(k\) with \(1/k<\delta\). The index \(m_k\) then works for every \(x\in X\setminus E\), giving \(|f_n(x)-f(x)|<\delta\) whenever \(n\geq m_k\). Thus convergence is uniform on \(X\setminus E\). \(\square\)

The proof has a useful structure. For each accuracy \(1/k\), convergence at each point of \(C\) gives an eventual index, but that index may vary from point to point. The increasing sets \(C\cap A_{m,k}\) collect points for which a common index \(m\) already works. Continuity from below says that these sets eventually cover all but a small amount of \(C\). A countable choice of accuracies then produces one exceptional set that works for every required accuracy.

A Converse Criterion and the Finite-Measure Hypothesis

There is also a useful converse. If uniform convergence can be obtained outside exceptional sets of arbitrarily small measure, then pointwise convergence must hold almost everywhere. Unlike Egorov’s theorem, this implication does not require \(\mu(X)<\infty\).

Proof. The convergence set \(C=\{x:f_n(x)\to f(x)\}\) is measurable by the theorem proved above. For each positive integer \(j\), choose a measurable set \(E_j\) with \(\mu(E_j)<1/j\) such that convergence is uniform, and therefore pointwise, on \(X\setminus E_j\). Any point where convergence fails must belong to every \(E_j\); thus \(X\setminus C\subseteq E_j\) for each \(j\). By monotonicity,

for every positive integer \(j\). A nonnegative extended real number bounded above by \(1/j\) for every \(j\) must be zero. Hence \(\mu(X\setminus C)=0\), which is almost-everywhere convergence. \(\square\)

The finite-measure condition in Egorov’s theorem cannot simply be dropped. On \(\mathbb{R}\) with Lebesgue measure, define \(f_n=\mathbf{1}_{[n,n+1)}\) and \(f=0\). Each fixed real number belongs to at most one of the intervals \([n,n+1)\), so \(f_n(x)\to0\) at every \(x\). Suppose a measurable set \(E\) of finite measure could be removed to make this convergence uniform. Uniform convergence would give an integer \(N\) such that, for every \(n\geq N\), \(|f_n(x)|<1/2\) for all \(x\notin E\). Since \(f_n(x)=1\) on \([n,n+1)\), this forces \([n,n+1)\subseteq E\) for every \(n\geq N\). These intervals are disjoint and each has measure one, so their union has infinite measure. Monotonicity would then imply \(\mu(E)=\infty\), a contradiction.

On a finite-measure space, Egorov’s theorem says that pointwise convergence holds uniformly after a small amount of the space is removed. It does not say that the same exceptional set works for every \(\varepsilon\), nor does it say that convergence is uniform on the entire space. The conclusion also concerns uniform convergence, not a uniform rate on the discarded set. Keeping these distinctions clear is important when using the theorem in later arguments.