Egorov’s theorem
Throughout this section, \(m\) denotes Lebesgue measure on \(\mathbb{R}^{d}\), \(m_{e}\) the outer measure, and “a.e.” abbreviates almost everywhere.
Note (Pointwise convergence does not imply uniform convergence)
Recall that there exist sequences of continuous functions \(f_{k} : E \to \mathbb{R}\) such that
\[\lim_{k \to \infty} f_{k} = f\]pointwise on \(E\), but such that \(f_{k}\) does not converge uniformly to \(f\) on compact sets. Egorov’s theorem shows that, for measurable functions on a set of finite measure, pointwise convergence a.e. is indeed uniform outside a set of arbitrarily small measure.
Example (The finite measure hypothesis is essential in Egorov)
Let \(E = \mathbb{R}^{d}\) and
\[f_{k}(x) = \mathbf{1}_{B(0,k)}(x).\]Then \(\lim_{k \to \infty} f_{k} = 1\) pointwise on \(\mathbb{R}^{d}\). Let \(F\) be closed and unbounded; then for every \(k \in \mathbb{N}\) there exists \(x \in F\) such that
\[|1 - f_{k}(x)| = 1,\]since \(F\) has points outside \(B(0,k)\). That is, the convergence cannot be uniform on \(F\). Note moreover that, if \(F\) is closed and \(m(\mathbb{R}^{d} \setminus F) < \infty\), then \(F\) is unbounded. Hence the conclusion of Egorov’s theorem fails for this sequence: no closed set leaving out a set of finite measure admits uniform convergence.
Lemma (Technical lemma preceding Egorov’s theorem)
Let \(E \subseteq \mathbb{R}^{d}\) be measurable with \(m(E) < \infty\) and let \(\{f_{k}\}_{k=1}^{\infty}\) be measurable functions on \(E\) such that
\[\lim_{k \to \infty} f_{k} = f \quad \text{a.e. on } E, \qquad |f(x)| \neq \infty \ \text{a.e. on } E.\]Then, given \(\varepsilon > 0\) and \(\eta > 0\), there exist a closed set \(F \subseteq E\) and \(k_{0} = k_{0}(\varepsilon, \eta) \geq 0\) such that
\[m(E \setminus F) < \eta\]and
\[|f_{k}(x) - f(x)| < \varepsilon \quad \text{for every } x \in F \text{ and } k \geq k_{0}.\]Proof: Define
\[\tilde{E} = \{ x \in E :\ \lim_{k \to \infty} f_{k}(x) = f(x),\ |f(x)| < \infty \}.\]By hypothesis, \(m(E \setminus \tilde{E}) = 0\). For fixed \(\varepsilon > 0\) and \(\eta > 0\), define
\[E_{m} = \{ x \in \tilde{E} :\ k \geq m \implies |f(x) - f_{k}(x)| < \varepsilon \}.\]Then each \(E_{m}\) is measurable, since
\[E_{m} = \tilde{E} \cap \bigcap_{k \geq m} \{ x \in E :\ |f(x) - f_{k}(x)| < \varepsilon \}\]is a countable intersection of measurable sets, and moreover \(E_{m} \subseteq E_{m+1}\), since the condition is imposed on fewer indices \(k\) as \(m\) grows. We have that \(\bigcup_{m=1}^{\infty} E_{m} = \tilde{E}\) (exercise; see the exercise below).
By continuity from below of the measure, we have that
\[m(\tilde{E}) = \lim_{m \to \infty} m(E_{m}).\]Since \(m(\tilde{E}) \leq m(E) < \infty\), we may subtract and obtain
\[\lim_{m \to \infty} m(\tilde{E} \setminus E_{m}) = \lim_{m \to \infty} \big( m(\tilde{E}) - m(E_{m}) \big) = 0.\]Moreover \(E \setminus E_{m} = (E \setminus \tilde{E}) \cup (\tilde{E} \setminus E_{m})\) and \(m(E \setminus \tilde{E}) = 0\), so \(m(E \setminus E_{m}) = m(\tilde{E} \setminus E_{m}) \to 0\). Let then \(k_{0}\) be such that
\[m(E \setminus E_{k_{0}}) < \frac{\eta}{2}.\]Note that if \(x \in E_{k_{0}}\), then, by definition of \(E_{k_{0}}\),
\[|f(x) - f_{k}(x)| < \varepsilon \quad \text{for } k \geq k_{0}.\]Finally, since \(E_{k_{0}}\) is measurable, by inner regularity there exists a closed set \(F\) with \(F \subseteq E_{k_{0}}\) satisfying
\[m(E_{k_{0}} \setminus F) < \frac{\eta}{2}.\]Then
\[m(E \setminus F) \leq m(E \setminus E_{k_{0}}) + m(E_{k_{0}} \setminus F) < \eta,\]| and if \(x \in F \subseteq E_{k_{0}}\) and \(k \geq k_{0}\), we have that $$ | f_{k}(x) - f(x) | < \varepsilon$$. |
Exercise (The union of the \(E_{m}\) recovers \(\tilde{E}\))
With the notation of the preceding proof, show that \(\bigcup_{m=1}^{\infty} E_{m} = \tilde{E}\).
Theorem (Egorov: a.e. convergence is almost uniform in finite measure)
Let \(E \subseteq \mathbb{R}^{d}\) be measurable, with \(m(E) < \infty\), and let \(\{f_{k}\}_{k=1}^{\infty}\) be measurable functions on \(E\). Suppose that
\[\lim_{k \to \infty} f_{k} = f\]| a.e. on \(E\) and that $$ | f(x) | \neq \infty\(a.e. on\)E\(. Then, given\)\varepsilon > 0\(, there exists a closed set\)F_{\varepsilon} \subseteq E$$ such that: |
- \(m(E \setminus F_{\varepsilon}) < \varepsilon\).
- \(f_{k}\) converges to \(f\) uniformly on \(F_{\varepsilon}\).
Proof: Let \(\varepsilon > 0\). Applying the preceding lemma with \(\varepsilon_{m} = \tfrac{1}{m}\) and \(\eta_{m} = \tfrac{\varepsilon}{2^{m}}\) for each \(m \geq 1\), there exist closed sets \(F_{m} \subseteq E\) and integers \(\kappa_{m}^{\varepsilon}\) such that
\[m(E \setminus F_{m}) < \frac{\varepsilon}{2^{m}}\]and
\[|f(x) - f_{k}(x)| < \frac{1}{m} \quad \text{for every } k \geq \kappa_{m}^{\varepsilon} \text{ and } x \in F_{m}.\]Let
\[F_{\varepsilon} = \bigcap_{m=1}^{\infty} F_{m}.\]Then \(F_{\varepsilon}\) is closed, being an intersection of closed sets, and
\[m(E \setminus F_{\varepsilon}) = m\left( E \cap \bigcup_{m=1}^{\infty} F_{m}^{c} \right) \leq \sum_{m=1}^{\infty} m(E \setminus F_{m}) < \sum_{m=1}^{\infty} \frac{\varepsilon}{2^{m}} = \varepsilon.\]Let us see that the convergence is uniform on \(F_{\varepsilon}\). Given \(\delta > 0\), take \(m \geq 1\) with \(\tfrac{1}{m} < \delta\). If \(x \in F_{\varepsilon} \subseteq F_{m}\) and \(k \geq \kappa_{m}^{\varepsilon}\), then
\[|f(x) - f_{k}(x)| < \frac{1}{m} < \delta.\]Since \(\kappa_{m}^{\varepsilon}\) does not depend on \(x\), we conclude that \(f_{k} \to f\) uniformly on \(F_{\varepsilon}\).
Lusin’s theorem
Definition (Property \(C\) of a function)
A function \(f : E \to \mathbb{R}\) has property \(C\) on \(E\) if, given \(\varepsilon > 0\), there exists a closed set \(F \subseteq E\) such that
- \(m(E \setminus F) < \varepsilon\);
- \(f\) is continuous relative to \(F\), that is, \(f : F \to \mathbb{R}\) is continuous.
Note (Sequential characterisation of relative continuity)
The second condition of the preceding definition is equivalent to the following: if \(\{x_{n}\}_{n=1}^{\infty} \subseteq F\) with
\[\lim_{n \to \infty} x_{n} = x \in F,\]then
\[\lim_{n \to \infty} f(x_{n}) = f(x).\]Lemma (Measurable simple functions have property \(C\))
Let \(\phi\) be a simple, measurable function on \(E\). Then \(\phi\) has property \(C\).
Proof: Write
\[\phi(x) = \sum_{\ell=1}^{m} b_{\ell} \mathbf{1}_{B_{\ell}}(x)\]with the \(B_{\ell} \subseteq E\) measurable, \(B_{i} \cap B_{j} = \emptyset\) and \(b_{i} \neq b_{j}\) if \(i \neq j\), and \(E = \bigcup_{\ell=1}^{m} B_{\ell}\).
Given \(\varepsilon > 0\), by inner regularity take, for each \(1 \leq \ell \leq m\), a closed set \(F_{\ell} \subseteq B_{\ell}\) such that
\[m(B_{\ell} \setminus F_{\ell}) < \frac{\varepsilon}{m}.\]Then
\[F = \bigcup_{\ell=1}^{m} F_{\ell}\]is closed, being a finite union of closed sets. Moreover, since the \(B_{\ell}\) cover \(E\), we have that \(E \setminus F \subseteq \bigcup_{\ell=1}^{m} (B_{\ell} \setminus F_{\ell})\), and hence
\[m(E \setminus F) \leq \sum_{\ell=1}^{m} m(B_{\ell} \setminus F_{\ell}) < m \cdot \frac{\varepsilon}{m} = \varepsilon.\]It remains to see that \(\phi\) is continuous relative to \(F\). Take \(\{x_{n}\}_{n=1}^{\infty} \subseteq F\) such that
\[\lim_{n \to \infty} x_{n} = y \in F.\]Then there exists \(\ell_{0}\) such that \(y \in F_{\ell_{0}}\). We claim that, given \(\ell \neq \ell_{0}\), the set
\[\{x_{n}\}_{n=1}^{\infty} \cap F_{\ell}\]is finite. Otherwise, there exists a subsequence
\[\{x_{n_{k}}\}_{k=1}^{\infty} \subseteq F_{\ell} \quad \text{with} \quad \lim_{k \to \infty} x_{n_{k}} = y.\]Since \(F_{\ell}\) is closed, this implies that \(y \in F_{\ell}\); but \(F_{\ell} \cap F_{\ell_{0}} \subseteq B_{\ell} \cap B_{\ell_{0}} = \emptyset\), which leads to a contradiction.
Hence, since the \(x_{n}\) are distributed among the \(F_{\ell}\) and only finitely many fall outside \(F_{\ell_{0}}\), there exists \(k_{0}\) such that \(x_{n} \in F_{\ell_{0}}\) for \(n \geq k_{0}\). Then
\[\phi(x_{n}) = b_{\ell_{0}} = \phi(y) \quad \text{for } n \geq k_{0},\]that is,
\[\lim_{n \to \infty} \phi(x_{n}) = \phi(y).\]Theorem (Lusin: measurability is equivalent to property \(C\))
Let \(f : E \to \mathbb{R}\) with \(E\) measurable. Then \(f\) is measurable if and only if \(f\) has property \(C\) on \(E\).
Proof: Measurable implies property \(C\), case \(m(E) < \infty\). Let \(f : E \to \mathbb{R}\) be measurable; then there exists a sequence \(\{f_{k}\}_{k=1}^{\infty}\) of simple, measurable functions such that
\[\lim_{k \to \infty} f_{k}(x) = f(x)\]a.e. on \(E\). Let \(\varepsilon > 0\). By the preceding lemma, for each \(k \geq 1\) there exists a closed set \(F_{k} \subseteq E\) satisfying
\[m(E \setminus F_{k}) < \frac{\varepsilon}{2^{k+1}},\]and such that \(f_{k}\) is continuous relative to \(F_{k}\). Since \(m(E) < \infty\), by Egorov’s theorem there exists a closed set \(F_{0} \subseteq E\) such that
\[m(E \setminus F_{0}) < \frac{\varepsilon}{2} \qquad \text{and} \qquad \lim_{k \to \infty} f_{k} = f \ \text{uniformly on } F_{0}.\]If we take
\[F = F_{0} \cap \bigcap_{k=1}^{\infty} F_{k},\]then \(F\) is closed, and since
\[E \setminus F = (E \setminus F_{0}) \cup \bigcup_{k=1}^{\infty} (E \setminus F_{k}),\]we have that
\[m(E \setminus F) \leq m(E \setminus F_{0}) + \sum_{k=1}^{\infty} m(E \setminus F_{k}) < \frac{\varepsilon}{2} + \sum_{k=1}^{\infty} \frac{\varepsilon}{2^{k+1}} = \varepsilon.\]Each \(f_{k}\) is continuous relative to \(F_{k} \supseteq F\), and hence continuous relative to \(F\). Since
\[\lim_{k \to \infty} f_{k} = f\]uniformly on \(F\), and the uniform limit of continuous functions is continuous, we have that \(f\) is continuous relative to \(F\).
Measurable implies property \(C\), case \(m(E) = \infty\). Set
\[E_{k} = E \cap \{ x \in \mathbb{R}^{d} :\ k-1 \leq |x| < k \}, \quad k \geq 1.\]The \(E_{k}\) are measurable, pairwise disjoint, cover \(E\) and \(m(E_{k}) \leq m(B(0,k)) < \infty\). By the previous case, for each \(k\) there exists a closed set \(F_{k} \subseteq E_{k}\) such that \(f\) is continuous relative to \(F_{k}\) and
\[m(E_{k} \setminus F_{k}) \leq \frac{\varepsilon}{2^{k}}.\]Take
\[F = \bigcup_{k=1}^{\infty} F_{k}.\]Since \(E \setminus F \subseteq \bigcup_{k=1}^{\infty} (E_{k} \setminus F_{k})\), we have that \(m(E \setminus F) \leq \sum_{k=1}^{\infty} \varepsilon/2^{k} = \varepsilon\).
Let us see that \(F\) is closed. Let \(\{x_{n}\}_{n=1}^{\infty} \subseteq F\) with \(\lim_{n \to \infty} x_{n} = y \in \mathbb{R}^{d}\). Since the sequence converges, it is bounded: there exists \(k_{0}\) such that
\[n \geq k_{0} \implies |x_{n}| \leq |y| + 1.\]| Let \(M\) be an integer with $$M \geq | y | + 2\(. If\)x_{n} \in F_{k} \subseteq E_{k}\(with\)n \geq k_{0}\(, then\)k - 1 \leq | x_{n} | \leq | y | + 1\(, that is\)k \leq | y | + 2 \leq M$$. Hence |
which is a finite union of closed sets and therefore closed. Then \(y \in \bigcup_{k=1}^{M} F_{k} \subseteq F\), and \(F\) is closed.
Let us see that \(f\) is continuous relative to \(F\). Let \(\{x_{n}\}_{n=1}^{\infty} \subseteq F\) with \(\lim_{n \to \infty} x_{n} = y \in F\). By the above, there exists \(k_{0}\) such that \(x_{n} \in \bigcup_{k=1}^{M} F_{k}\) for \(n \geq k_{0}\). Since the \(F_{k}\) are closed and pairwise disjoint, by the argument in the proof of the preceding lemma there exist \(k_{1}\) and \(\ell_{1}\) such that
\[n \geq k_{1} \implies x_{n} \in F_{\ell_{1}},\]with \(y \in F_{\ell_{1}}\). Since \(f\) is continuous relative to \(F_{\ell_{1}}\), we conclude that \(\lim_{n \to \infty} f(x_{n}) = f(y)\).
Property \(C\) implies measurable. If \(f\) has property \(C\), then for each \(k \geq 1\) there exists a closed set \(F_{k} \subseteq E\) satisfying:
- \(m(E \setminus F_{k}) < \frac{1}{k}\);
- \(f\) is continuous relative to \(F_{k}\).
Let
\[H = \bigcup_{k=1}^{\infty} F_{k},\]then \(H \subseteq E\) and \(m(E \setminus H) = 0\) (exercise; see the exercise below). Write \(Z = E \setminus H\). Finally, for each \(a \in \mathbb{R}\),
\[\begin{aligned} \{ x \in E :\ f(x) > a \} &= \{ x \in H :\ f(x) > a \} \cup \{ x \in Z :\ f(x) > a \} \\ &= \bigcup_{k=1}^{\infty} \{ x \in F_{k} :\ f(x) > a \} \cup \{ x \in Z :\ f(x) > a \}. \end{aligned}\]Each \(\{ x \in F_{k} :\ f(x) > a \}\) is measurable: since \(f\) is continuous relative to \(F_{k}\), this set is open relative to \(F_{k}\), that is, of the form \(F_{k} \cap G\) with \(G\) open in \(\mathbb{R}^{d}\), which is measurable. On the other hand, \(\{ x \in Z :\ f(x) > a \} \subseteq Z\) has outer measure zero and is therefore measurable. Then \(\{ f > a \}\) is measurable for every \(a \in \mathbb{R}\), that is, \(f\) is measurable.
Exercise (The complement of the union of the \(F_{k}\) is null)
With the notation of the preceding proof, show that if \(H = \bigcup_{k=1}^{\infty} F_{k}\) with \(m(E \setminus F_{k}) < \tfrac{1}{k}\) for every \(k \geq 1\), then \(m(E \setminus H) = 0\).