Lemma to the Riesz-Fischer Theorem (p=∞)

# Lemma to the Riesz-Fischer Theorem (p=∞)

 Lemma 1: Let $(X, \mathfrak T, \mu)$ be a measure space and let $1 \leq p \leq \infty$. If $(f_n)_{n=1}^{\infty}$ is a sequence of functions in $L^p(X, \mathfrak T, \mu)$ such that there exists a sequence of positive real numbers $(\epsilon_n)_{n=1}^{\infty}$ such that $\displaystyle{\sum_{n=1}^{\infty} \epsilon_n < \infty}$ and $\| f_n - f_{n+1} \|_p \leq \epsilon_n^2$ for all $n \geq 1$, then there exists a function $f \in L^p (X, \mathfrak T, \mu)$ such that: a) $\displaystyle{\lim_{n \to \infty} f_n(x) = f(x)}$ $\mu$-almost everywhere on $X$. b) $\displaystyle{\lim_{n \to \infty} \| f - f_n \|_p = 0}$.

The case when $1 \leq p < \infty$ can be found on the Lemma to the Riesz-Fischer Theorem (1 ≤ p < ∞) page.

(1)
\begin{align} \quad \| f_n - f_{n+m} \|_p &= \| f_n - f_{n+1} + f_{n+1} - f_{n+2} + f_{n+2} - ... + f_{n+m-1} - f_{n+m} \|_p \\ & \leq \| f_n - f_{n+1} \|_p + \| f_{n+1} - f_{n+2} \|_p + ... + \| f_{n+m-1} - f_{n+m} \|_p \\ & \leq \sum_{k=n}^{n+m-1} \| f_k - f_{k+1} \|_p \\ & \leq \sum_{k=n}^{\infty} \| f_k - f_{k+1} \|_p \\ & \leq \sum_{k=n}^{\infty} \epsilon_k^2 \quad (*) \end{align}
• Observe that since $\displaystyle{\sum_{n=1}^{\infty} \epsilon_n < \infty}$ we must have that $\displaystyle{\lim_{n \to \infty} \epsilon_n = 0}$. So there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $\epsilon_n < 1$. So for $n \geq N$ we have that $\epsilon_n^2 < \epsilon_n < 1$ and so $\displaystyle{\lim_{n \to \infty} \epsilon_n^2 = 0}$. From above, we see that as $n \to \infty$, the sum $\displaystyle{\sum_{k=n}^{\infty} \epsilon_k^2 \to 0}$. So from above, we see that for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $m, n \geq N$ then $\| f_n - f_{m+n} \|_p < \epsilon$. Therefore, $(f_n)_{n=1}^{\infty}$ is a pointwise Cauchy sequence of functions.
• Now suppose that $p = \infty$. Let $m, n \in \mathbb{N}$. Then:
(2)
\begin{align} \quad | f_m(x) - f_n(x) | \leq \inf \{ M > 0 : |f_m(x) - f_n(x)| \leq M, \: \mu - \mathrm{a.e. \: on \:} X \} = \| f_m - f_n \|_{\infty} \quad \mu-\mathrm{a.e. \: on \:} X \end{align}
• So there exists a measurable set $E_{m, n} \in \mathfrak T$ with $E_{m,n} \subseteq X$ such that $|f_m(x) - f_n(x)| \leq \| f_m - f_n \|_{\infty}$ for all $x \in E_{m, n}$ where $\mu (X / E_{m, n}) = 0$. Define a new set $E$ by:
(3)
\begin{align} \quad E = \bigcap_{m, n \in \mathbb{N}} E_{m, n} \end{align}
• Then $E_{m, n}$ is a countable intersection of measurable sets and so $E$ is a measurable set. Moreover, we have that $\mu (X \setminus E) = 0$ since:
(4)
\begin{align} \quad \mu (X \setminus E) = \mu \left ( X \setminus \bigcap_{m, n \in \mathbb{N}} E_{m, n} \right ) = \mu \left ( X \cap \left ( \bigcap_{m,n \in \mathbb{N}} E_{m,n} \right)^c \right ) = \mu \left ( X \cap \bigcup_{m, n \in \mathbb{N}} E_{m, n}^c \right ) = \mu \left ( \bigcup_{m, n \in \mathbb{N}} X \cap E_{m,n}^c \right ) = \mu \left ( \bigcup_{m, n \in \mathbb{N}} X / E_{m,n} \right ) \leq \sum_{m, n \in \mathbb{N}} \mu (X \setminus E_{m, n}) = 0 \end{align}
• Therefore $|f_m(x) - f_n(x)| \leq \| f_m - f_n \|_{\infty}$ on all of $E$ for all $m, n \in \mathbb{N}$ and $\mu (X \setminus E) = 0$. So from $(*)$, we have that $(f_n)_{n=1}^{\infty}$ is pointwise Cauchy on $E$, that is, $(f_n(x))_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{C}$ for every $x \in E$. But $\mathbb{C}$ is a complete metric space. So $(f_n(x))_{n=1}^{\infty}$ converges to some complex number for every $x \in E$. We define a function $f : X \to \mathbb{C}$ by:
(5)
\begin{align} \quad f(x) = \left\{\begin{matrix} \lim_{n \to \infty} f_n(x) & \mathrm{if} \: x \in E \\ 0 & \mathrm{if} \: x \in X \setminus E \end{matrix}\right. \end{align}
• Since $(f_n)_{n=1}^{\infty}$ is a Cauchy sequence $\mu$ on $X$ in $L^{\infty} (X, \mathfrak T, \mu)$, the sequence is also bounded on $X$. So there exists an $M \in \mathbb{R}$, $M > 0$ such that $\| f_n \|_{\infty} \leq M$ for all $n \in \mathbb{N}$, $\mu$ on $X$.
• For each $n \in \mathbb{N}$ there exists a measurable set $F_n \in \mathfrak T$ such that $|f_n(x)| \leq \|f_n\|_{\infty}$ for all $x \in F_n$ and $\mu (X \setminus F_n) = 0$. We set:
(6)
\begin{align} \quad F = \bigcap_{n=1}^{\infty} F_n \end{align}
• Therefore $|f_n(x)| \leq \| f_n \|_{\infty} \leq M$ for all $x \in F$, for all $n \in \mathbb{N}$, and $\mu (X \setminus F) = 0$. So for all $x \in E \cap F$ we have that:
(7)
\begin{align} \quad |f(x)| = |\lim_{n \to \infty} f_n(x)| \leq |M| = M \end{align}
• Hence, for all $x \in E \cap F$ we have that there exists an $M \in \mathbb{R}$, $M > 0$ such that $|f(x)| \leq M$ and $\mu (X \setminus E \cap F) = 0$. So $|f(x)| \leq M$ $\mu$-a.e. on $X$ which shows that $f \in L^{\infty} (X, \mathfrak T, \mu)$ and by definition, satisfies (a).
• We now show that $f$ satisfies (b). For all $x \in E$ we have that:
(8)
\begin{align} \quad |f(x) - f_n(x)| = \lim_{m \to \infty} |f_{m+n}(x) - f_n(x)| \leq \lim_{m \to \infty} \| f_{m+n} - f_n \|_{\infty} \leq \sum_{k=n}^{\infty} \epsilon_k^2 \end{align}
• So $\displaystyle{\lim_{n \to \infty} \| f - f_n \|_{\infty} = 0}$.