Egoroff's Theorem
Table of Contents

Egoroff's Theorem

Recall from the Littlewood's First Principle page that Littlewood's first principle says that if $E$ is a Lebesgue measurable set with $m(E) < \infty$ then for all $\epsilon > 0$ there exists a finite collection of open intervals $\{ I_1, I_2, ..., I_N \}$ such that:

(1)
\begin{align} \quad m \left ( E \Delta \bigcup_{n=1}^{N} I_n \right ) < \epsilon \end{align}

In other words, a Lebesgue measurable set $E$ with finite measure can be covered almost entirely by a finite union of open intervals.

We will now prove another very important theorem known as Egoroff's theorem which is stated below.

Theorem 1 (Egoroff's Theorem): Let $E$ be a Lebesgue measurable set with $m(E) < \infty$. If $(f_n(x))_{n=1}^{\infty}$ is a sequence of Lebesgue measurable functions that converges pointwise to a function $f(x)$ on $E$ then for all $\epsilon > 0$ there exists a closed set $F \subseteq E$ such that:
1) $m(E \setminus F) < \epsilon$.
2) $(f_n(x))_{n=1}^{\infty}$ converges to $f(x)$ uniformly on $F$.
  • Proof: Let $E$ be a Lebesgue measurable set with $m(E) < \infty$ and let $(f_n(x))_{n=1}^{\infty}$ be a sequence of Lebesgue measurable functions on $E$ that converge pointwise to $f(x)$ on $E$.
  • Let $\epsilon > 0$. For each $n \in \mathbb{N}$ define the set $A_n(\epsilon)$ to be the set of elements in $E$ such that $|f_k(x) - f(x)| < \epsilon$ where $k \in \{ n, n+1, ... \}$. That is:
(2)
\begin{align} \quad A_n(\epsilon) = \{ x \in E : |f_k(x) - f(x)| < \epsilon, \: k \in \{ n, n+1, ... \} \} \end{align}
  • Consider the condition that $(f_n(x))_{n=1}^{\infty}$ converges uniformly to $f(x)$ on any set $A \subseteq E$ means that for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $|f_n(x) - f(x) | < \epsilon$ for all $x \in A$. By the definition of the sets $A_n(\epsilon)$ above, we have that $(f_n(x))_{n=1}^{\infty}$ converges uniformly to $f(x)$ on $A$ if and only if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $x \in A$ then:
(3)
\begin{align} \quad x \in A_N(\epsilon) \end{align}
  • Equivalently, $(f_n(x))_{n=1}^{\infty}$ converges to $f(x)$ on $A$ if and only if for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that:
(4)
\begin{align} \quad A \subseteq A_N(\epsilon) \end{align}
  • Furthermore, we note that for a fixed $\epsilon > 0$, the collection of sets $A_1(\epsilon)$, $A_2(\epsilon)$, …, $A_N(\epsilon)$, … is a ascending sequence of sets:
(5)
\begin{align} \quad A_1(\epsilon) \subseteq A_2(\epsilon) \subseteq ... \subseteq A_N(\epsilon) \subseteq A_{N+1} (\epsilon) \subseteq ... \end{align}
  • Let $\epsilon > 0$. Since $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$ we have that for this given $\epsilon$, for each $x \in E$ there exists an $N(\epsilon, x) \in \mathbb{N}$ such that if $n \geq N_0$ then:
(6)
\begin{align} \quad | f_n(x) - f(x) | < \epsilon \end{align}
  • So for each $x \in E$ there exists an $N(\epsilon, x) \in \mathbb{N}$ such that $x \in A_{N(\epsilon, x)}(\epsilon)$ and hence:
(7)
\begin{align} \quad E = \bigcup_{n=1}^{\infty} A_n(\epsilon) \end{align}
  • For each $N \in \mathbb{N}$ let $\displaystyle{\epsilon_k = \frac{1}{k} > 0}$. For each $n \in \mathbb{N}$ consider the sets $\displaystyle{A_n (\epsilon_k) = A_n \left ( \frac{1}{k} \right )}$. Then:
(8)
\begin{align} \quad E = \bigcup_{n=1}^{\infty} A_n \left ( \frac{1}{k} \right ) \end{align}
  • Additionally since $\left ( A_n \left ( \frac{1}{k} \right ) \right )_{n=1}^{\infty}$ is a sequence of ascending sets that converges to $E$, for each $\displaystyle{\epsilon_k = \frac{\epsilon}{2^{k+1}} > 0}$ there exists an $N_k \in \mathbb{N}$ such that:
(9)
\begin{align} \quad m \left ( E \setminus A_{N_k} \left ( \frac{1}{k} \right ) \right ) < \epsilon_k = \frac{\epsilon}{2^{k+1}} \quad (*) \end{align}
  • Let:
(10)
\begin{align} \quad A = \bigcap_{k=1}^{\infty} A_{N_k} \left ( \frac{1}{k} \right ) \end{align}
  • Consider the Lebesgue measure of $m(E \setminus A)$:
(11)
\begin{align} \quad m (E \setminus A) &= m \left ( E \setminus \bigcap_{k=1}^{\infty} A_{N_k} \left ( \frac{1}{k} \right ) \right ) \\ &= m \left ( \bigcup_{k=1}^{\infty} \left ( E \setminus A_{N_k} \left ( \frac{1}{k} \right ) \right ) \right ) \\ & \leq \sum_{k=1}^{\infty} m \left ( E \setminus A_{N_k} \left ( \frac{1}{k} \right ) \right ) \\ & \overset{(*)} \leq \sum_{k=1}^{\infty} \frac{\epsilon}{2^{k+1}} \\ & \leq \frac{\epsilon}{2} \end{align}
  • So for every $\epsilon > 0$, let $k \in \mathbb{N}$ be such that $\displaystyle{\frac{1}{k} < \epsilon}$. Then:
(12)
\begin{align} \quad A = \bigcap_{k=1}^{\infty} A_{N_k} \left ( \frac{1}{k} \right ) \subseteq A_{N_k} \left ( \frac{1}{k} \right ) \subseteq A_{N_k} (\epsilon) \end{align}
  • So $(f_n(x))_{n=1}^{\infty}$ converges uniformly to $f(x)$ on $A$. Since $A \subseteq E$ and $m(E) < \infty$ we have that $m(A) < \infty$. So there exists a closed set $F \subseteq A$ such that $\displaystyle{m(A \setminus F) < \frac{\epsilon}{2}}$ and hence:
(13)
\begin{align} \quad m(E \setminus F) = m(E \setminus A) + m(A \setminus F) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
  • And since $(f_n(x))_{n=1}^{\infty}$ converges uniformly to $f(x)$ on $A$ we also have that $(f_n(x))_{n=1}^{\infty}$ converges uniformly to $f(x)$ on $F$. $\blacksquare$
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