Exercise 1.5.5 (Fast $L^1$ convergence)

Question

Let $(X,\mathcal{B},\mu)$, and suppose that $f,\left(fight)_{n\in\mathbb{N}}:X	o\mathbb{C}$ are measurable functions such that
$$\sum_{n=1}^{\infty}\|f_n-f\|_{L^1} < \infty.$$
Show that
\begin{enumerate}[label=(oman*)]
\item $\left(fight)_{n\in\mathbb{N}}$ converges pointwise almost everywhere to $f$.
\item $\left(fight)_{n\in\mathbb{N}}$ converges almost uniformly to $f$.
\end{enumerate}

Elu Thingol · Accepted Answer

$
ewcommand{\intx}[1]{\int_{X} #1 \mathop{}\!\mathrm{d}{\mu}}$
\begin{enumerate}[label=(oman*)]
\item We have to demonstrate that for some null set $E$ we have for all $x\in E$:
  $$\forall \epsilon >0 \quad \exists N \in \mathbb{N} \quad \forall n \geq N: \quad |f(x)-f_n(x)| \leq \epsilon.$$
  Suppose for the sake of contradiction that this is not the case; i.e., there exists non-null set $F \in \mathcal{B}$ such that for all $x\in F$:
  $$\exists \epsilon > 0 \quad \forall N \in \mathbb{N} \quad \exists n_N \geq N: \quad |f(x)-f_{n_N}(x)| > \epsilon.$$
  We then have
  $$\sum_{n=1}^{\infty}\intx{|f-f_n|} \geq \sum_{n_1}^{\infty}\int_{F} \epsilon \ \mathrm{d}\mu = \sum_{n_1}^{\infty} \epsilon \mu(F) = \infty$$
  and so we arrive at contradiction.
\item Note that from the given condition we can derive the convergence of the sequence $\sum_{n=1}^{N}\left|f_n-fight|$ against $\sum_{n=1}^{\infty}|f_n-f|$ in $L^1$ norm. This follows by applying Corollary 1.4.45 (Tonelli's theorem for sums and integrals) to 
  \begin{align*}
    \intx{\left(\sum_{n=1}^{\infty}\left|f_n-fight| - \sum_{n=1}^{N}\left|f_n-fight|ight)} &= \intx{\left(\sum_{n=N+1}^{\infty}\left|f_n-fight|ight)} = \sum_{n=N+1}^{\infty}\intx{\left|f_n-fight|}\
    &= \sum_{n=1}^{\infty}\intx{\left|f_n-fight|} - \sum_{n=1}^{N}\intx{\left|f_n-fight|} \leq \epsilon
  \end{align*}
  where the last term can be adjusted to be less that $\epsilon$ using the convergence of $\sum_{n=1}^{\infty}\|f_n-f\|_{L^1}$. Now we demonstrate the uniform convergence. Pick an arbitrary $\delta>0$ to test the convergence, and an $\epsilon >0$ to measure the exceptional set. Recall that $L^1$ convergence necessarily implies the convergence in measure; thus, for a $\delta>0$ we have a $N_\delta\in\mathbb{N}$ such that
  $$\mu(E_{N_\delta}) \leq \epsilon 	ext{ for } E:=\left\{ x \in X: \sum_{n=N_\delta+1}^{\infty}\left|f_n-fight| > \delta ight\}.$$
  Usually we cannot derive almost uniform convergence from the convergence in measure (cf. Example 1.5.5); but in our case we have the convergence in measure of series, which is why for all $x\in X/E$ and additionally for all $i\geq N_\delta$ we have
  $$\left|f_i(x)-f(x)ight| \leq \sum_{n=N_\delta+1}^{\infty}\left|f_n-fight| \leq \delta $$
  as desired.      
\end{enumerate}

Exercise 1.5.5 (Fast $L^1$ convergence)

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Exercise 1.5.5 (Fast $L^1$ convergence)

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