A fundamental result in measure theory is Egorov's Theorem, which says that if a sequence of measurable functions converge point-wise, then restricted to a ``large'' subset of the domain, the sequence of functions in fact converges uniformly. In this post we discuss a version of this theorem for Riemann integrable functions. For simplicity of exposition we will limit ourself to real valued functions defined on the interval $[0,1]$; but the same argument holds for Riemann-Loomis integration of Banach-space-valued functions on (finite) Jordan-measurable subsets of $\mathbb{R}^n$.
"Measure Theory"
One of the key steps in the standard proof of Egorov's Theorem is the following continuity property of the Lebesgue measure:
So we begin by establishing an analogue of this statement. The analogue statement can be phrased in terms of Jordan measurable sets, but for our purposes it is more convenient to argue using closed intervals. To make the discussion easier to follow, we introduce some (not very creative) names of the various objects involved.
Two closed intervals $I,J$ are said to be almost disjoint if $I\cap J$ has cardinality at most 1. (So that the intersection is either empty or a singleton set.)
A subset $E\subseteq [0,1]$ is said to be regular if it can be written as the countable union of pairwise almost disjoint closed intervals. The subset is said to be strongly regular if it can be written as the finite union of pairwise almost disjoint closed intervals.
For an interval $I$ we denote by $|I|$ its length (the difference between its end points). An empty set has $|\emptyset| = 0$. For a non-empty regular subset $E\subseteq [0,1]$, writing $E = \cup I_n$ where $I_n$ are almost disjoint and closed, we let $|E| := \sum |I_n|$. (We note that the sum converges as the partial sums are increasing and uniformly bounded by $1$.)
As a consequence of our definition, we see that if $E\subseteq F \subseteq [0,1]$ are regular sets, then $|E| \leq |F|$. (We are effectively constructing by hand a notion of inner measure.)
Let enumerate the regular sets as $E_i$. Each can be written as a countable union of almost disjoint intervals. Collection all the intervals used to make up all the $E_i$, these intervals are not necessarily almost disjoint, since they may belong to different $E_i$, but the total number is countable, and therefore we can enumerate them as $I_k$. By definition $\bigcup_k I_k = \bigcup_i E_i$. It suffices to show that we can break $I_k$ down into countably many almost disjoint closed intervals.
Our key observation is this: if $F$ is strongly regular, and $J$ a closed interval, then $\overline{J\setminus F}$ is strongly regular. (The complement of $F$ is formed of finitely many open intervals, so its intersection with $J$ is also finitely many intervals.) Additionally, we also have that $F \cup \overline{J\setminus F} = F \cup J$. (A point in $\partial(J\setminus F)$ is either in $\partial J$ or in $\partial F$, so $F$ and $J$ both being closed sets ensure the $\subseteq$ inclusion. For the reverse, we have $F\cup \overline{J\setminus F} \supseteq F \cup (J\setminus F) = F \cup J$ by definition.) By construction, every interval of $\overline{J\setminus F}$ is almost disjoint from every interval of $F$.
Hence we can inductively replace $I_k$ by a finite set of almost disjoint intervals obtained through $\overline{I_k \setminus \bigcup_{\ell = 1}^{k-1} I_\ell}$, such that the replacement at the $k$th step is almost disjoint from all the intervals from the previous steps.
Now, it is no longer the case that a nested sequence of regular sets has a regular intersection. To see an example, consider the usual construction of the Cantor set. At each stage the sets $C_n$ (using the Wikipedia notation) is a finite union of disjoint closed intervals, so is in fact strongly regular. But the Cantor set itself contains no non-degenerate intervals, but also has uncountable cardinality, and so is not "regular" in the sense above.
We will prove the following, much weaker continuity statement.
- If $\lim |E_n| > 0$ then $\bigcap E_n$ is non-empty.
- If $\bigcap E_n = \emptyset$ then $\lim |E_n| = 0$.
First note that when each $E_n$ is strongly regular, the result is purely topological. Strongly regular sets are finite unions of compact sets, and hence are compact. That $\lim |E_n| > 0$ implies that none of the $E_n$ are empty, and hence the result follows from Cantor's intersection theorem.
It suffices therefore to show that in the general, regular case, we can construct an approximation using strongly regular sets. The key operation is as follows: suppose that $E\subseteq E'\subseteq [0,1]$ are regular sets, and take $0 < \delta < |E|$. Then by definition there exists $F'\subseteq E'$ strongly regular, such that $|F'| \geq |E'| - \delta$. Now the closure $\overline{[0,1] \setminus F'}$ is also strongly regular, so that $E'\cap \overline{[0,1]\setminus F'}$ is regular. Furthermore, by construction the intervals of $F'$ and $\overline{[0,1]\setminus F'}$ are almost disjoint, which implies that $|E'\cap \overline{[0,1]\setminus F'}| + |F'| = |E'|$, and hence $|E'\cap \overline{[0,1]\setminus F'}| \leq \delta$. As $|E| > \delta$, this means that $E$ is not a subset of $E'\cap \overline{[0,1]\setminus F'}$. And we can conclude that $|E \cap F| \geq |E| - \delta > 0$.
We now apply this procedure inductively. Denote first by $\varepsilon := \lim |E_n| > 0$. First take $F_1\subseteq E_1$ strongly regular, so that $|E_1| - |F_1| < \frac14 \varepsilon$. We then proceed iteratively to construct a sequence $F_n\subseteq E_n$ such that $|E_n| - |F_n| < (\frac12 - \frac1{2^{n+1}} )\varepsilon$.
Assume $F_{n-1}$ has been constructed, then as $E_n \subseteq E_{n-1}$ and has $|E_n| > \varepsilon$, we have that $|E_n\cap F_{n-1}| > (\frac12 + \frac1{2^n})\varepsilon$, and $|E_n \cap \overline{[0,1]\setminus F_{n-1}}| < (\frac12 -\frac1{2^n})\varepsilon$. So using the above procedure, we can choose a strongly regular subset $F_n\subseteq E_n\cap F_{n-1}$, such that $|F_n| > |E_n\cap F_{n-1}| - \frac1{2^{n+1}}\varepsilon$. This implies that \[ |E_n| - |F_n| = |E_n \cap F_{n-1}| - |F_n| + |E_n\cap \overline{[0,1]\setminus F_{n-1}}| < (\frac12 - \frac1{2^{n+1}})\varepsilon \]
By construction our $F_n$ are nested, and that each has $|F_n| > \frac12 \varepsilon$. So we can apply Cantor's intersection theorem as in the first paragraph to conclude that the intersection $\emptyset \neq \bigcap F_n \subseteq \bigcap E_n$.
Egorov's Theorem
Before stating the proof, let us introduce some notations. By a partition of $[0,1]$ we mean a finite collection $\mathcal{P} = {I_1, \ldots, I_N}$ of closed bounded non-degenerate subintervals of $[0,1]$, that are pairwise almost disjoint, and such that $\bigcup \mathcal{P} = [0,1]$.
Given a closed interval $I$ and a bounded function $g$, the osciallation of $g$ on $I$ is \[ \DeclareMathOperator{osc}{osc} \osc_I(g) = \sup_I g - \inf_I g. \] Darboux's characterization of Riemann integrability states, a bounded function $g$ on $[0,1]$ is Riemann integrable if and only if for every $\epsilon,\mu > 0$ there exists a partition $\mathcal{P}$ such that \[ \sum \{ |I| : I\in \mathcal{P}, \osc_I(g) \geq \epsilon \} < \mu. \]
By considering $g_n = |f_n - f|$, it suffices to prove the theorem for a sequence of non-negative Riemann integrable functions convering pointwise to zero.
Riemann integrability means that we can choose partitions $\mathcal{P}_n$ such that \[ \sum \Big\{ |I| : I\in \mathcal{P}_{n}, \osc_I(g_n) \geq \frac{1}{n} \Big\} < \frac1{2^{3+n}} \delta. \] Write \[ \mathcal{Q}_n = \{ I\in \mathcal{P}_n, \osc_I(g_n) < \frac{1}{n} \Big\}, \quad R = \bigcup_{n = 1}^\infty \Big( \bigcup \big(\mathcal{P}_n\setminus \mathcal{Q}_n\big) \Big). \] It is not too hard to see that $R$ is a regular set, and $|R|< \frac18 \delta$. Next let \[ E_{n,k} = \bigcup \{ I\in \mathcal{Q}_n, \inf_I(g_n) \geq \frac1k \}. \] Notice that $E_{n,k}$ is strongly regular, so $\bigcup_{m\geq n} E_{m,k}$ is regular (by Lemma 3). If a point $x\in E_{n,k}$, then $g_n(x) \geq \frac1k$. So \[ x\in \limsup_n E_{n,k} \iff g_n(x) \text{ is frequently } \geq \frac{1}{k}. \] By assumption we have $g_n(x)$ converges pointwise, so that $\limsup_n E_{n,k} = \emptyset$ for any $k$. But $\limsup_n E_{n,k} = \bigcap_n \bigcup_{m\geq n} E_{m,k}$, so we can apply Theorem 4 to find that \[ \lim_{n} \Big| \bigcup_{m\geq n} E_{m,k} \Big| = 0. \] Thus we can choose a sequence $n_k$ such that, while ensuring $n_k \geq k$, defining \[ H_k := \bigcup_{m \geq n_k} E_{m,k} \] we can guarantee \[ |H_k| \leq \frac1{2^{3+k}}\delta. \] And hence if we define \[ H = \bigcup_k H_k \] we have a regular set with $|H| \leq \frac{1}{8} \delta$.
I claim that $g_n$ converges uniformly on $[0,1]\setminus (H\cup R)$. Let $\epsilon > 0$. Choose $k$ such that $\frac{2}{k} < \epsilon$. It suffices to show that for every $x\in [0,1]\setminus (H\cup R)$, and every $m \geq n_k$, we have $g_m(x) < \epsilon$. Let $\mathcal{P}_m$ be as at the start of this proof. Our definition means that $x$ belongs to $I$, one of the intervals of $\mathcal{P}_m$ that resides in $\mathcal{Q}_m$. Additionaly $x\not\in E_{m,k}$. So we have that $\inf_I(g_m) < \frac1k$, and $\sup_I(g_m) < \inf_I(g_m) + \osc_I(g_m) < \frac1k + \frac1m \leq \frac2k$. And hence $g_m(x) < \epsilon$ as claimed.
A modification of the proof also provides
Following the previous proof and define $g_n$. Our goal is now to prove that for every $\epsilon > 0$ there exists $N$ such that for all $m \geq N$, $\int g_m < \epsilon$. We may assume without loss of generality that $g_n$ are uniformly bounded by $1$.
Set $\delta = \epsilon$ as in the argument of the previous proof and construct the sets $\mathcal{P}_n$ and $\mathcal{Q}_n$. Then we have that the Riemann integral is bounded by the upper Darboux sum; on the sets where the oscillation is large we bound the supremum by $1$. This yields \[ \int g_n \leq \frac{1}{2^{3+n}} \epsilon + \sum_{I\in \mathcal{Q}_n} |I| \cdot \sup_I(g_n). \] Next identify the strongly regular sets $E_{n,k}$ as before. Our previous proof guarantees that for some $n_k \geq k$, we have that for every $m \geq n_k$ that $|E_{m,k}| \leq \frac{1}{2^{3+k}} \epsilon$.
Now take $k$ such that $\frac{4}{k} < \epsilon$. And set $N = n_k$. For $n \geq N$, we can estimate the Darboux sum. Those intervals in $\mathcal{Q}_n$ where $\inf_I(g_n) < \frac{1}{k}$ has total length no more than one. On that set the value of $g_n$ is no more than $\inf_I(g_n) + \osc_I(g_n) < \frac{2}{k} < \frac12 \epsilon$. On those intervals in $\mathcal{Q}$ where $\inf_I(g_n) \geq \frac{1}{k}$, we apply the uniform bound of $1$, but observe that the total length is less than $\frac18\epsilon$. So altogether we have shown that $\int g_n \leq \frac34\epsilon < \epsilon$ as desired, whenever $n \geq N$.
Extensions
The statement of "Egorov's theorem" above requires that the limiting function be Riemann integrable. It turns out that we can formula a "Cauchy" version of the theorem that does not require assumptions on the limiting function.
Form the doubly-indexed sequence of Riemann integrable functions $h_{n,m}(x) = |f_n(x) - f_m(x)|$. We can consider this as a net, where the indexing set is $\mathbb{N}^2$ with the ordering \[ (n,m) \preceq (n',m') \iff n\leq n' \land m\leq m' \] Then the fact that $f_n$ converges pointwise implies that it is pointwise Cauchy, and hence the net $h_{n,m}$ converges pointwise to zero.
Our proof of Theorem 5 is stable under replacing sequential convergence with net convergence, provided that our indexing set is countable, since we will be taking unions over tail sets of the indexing set. (We do not need to modify Theorem 4; where prevoiusly in Theorem 5 we took $\bigcup_{m\geq n} E_{m,k}$ we will look at the sets $\bigcup_{m,n\geq N} E_{n,m.k}$. Thanks to the monotonicity looking at this "subnet" of the diagonal elements is enough.)
Similarly, Arzela's Bounded Convergence Theorem also does not require that the limiting function is Riemann integrable: as long as the $f_n$ are Riemann integrable, uniformly bounded, and converges pointwise, then the sequence of integrals $\int f_n$ converges.
