Exercise 1.5.2 (Easy convergence implications)

Question

Let $(X,\mathcal{B},\mu)$ be a measure space, and let $\left(fight)_{n\in\mathbb{N}}$ be a sequence of measurable functions from $X$ to $\mathbb{C}$, and $f:X	o\mathbb{C}$ be another measurable function.
\begin{enumerate}[label=(oman*)]
\item If $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ uniformly, then $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ pointwise.
\item If $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ uniformly, then $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ in $L^\infty$-norm. Conversely, If $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ in $L^\infty$-norm, then $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ uniformly  outside of a null set.
\item If $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ in $L^\infty$ norm, then $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ almost uniformly.
\item If $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ almost uniformly, then $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ almost everywhere pointwise.
\item If $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ pointwise, then $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ almost everywhere pointwise.
\item If $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ in $L^1$ norm, then $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ in measure.
\item If $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ almost uniformly, then $\left(fight)_{n\in\mathbb{N}}$ converges to $f$ in measure.
\end{enumerate}

Elu Thingol · Accepted Answer

$ ewcommand{\intx}[1]{\int_{X} #1 \mathop{}\!\mathrm{d}{\mu}}$ The following diagram summarizes the statements of the theorem: \begin{figure} \def\shape_1{rectangle} \begin{tikzpicture}[scale=2] ode[\shape_1,draw, ultra thick, fill=blue,align=center] at (0, 0) (U) {$d_\infty$ convergence\ (uniform convergence)}; ode[\shape_1,draw, ultra thick, fill=blue,align=center] at (1, -1) (Linf) {a.e. $d_\infty$-convergence\$$L^\infty$ convergence)}; ode[\shape_1,draw, ultra thick, fill=blue,align=center] at (3.5, -1) (aU) {almost uniform convergence}; ode[\shape_1,draw, ultra thick, fill=blue,align=center] at (6, -1) (M) {convergence in measure}; ode[\shape_1,draw, ultra thick, fill=blue,align=center] at (5, 0) (P) {pointwise convergence}; ode[\shape_1,draw, ultra thick, fill=blue,align=center] at (7.5, 0) (aeP) {a.e. pointwise convergence}; ode[\shape_1,draw, ultra thick, fill=blue,align=center] at (3, -2) (L1) {$L^1$ convergence}; \draw[->] (U) -- (Linf); \draw[->] (U) -- (P); \draw[->] (P) -- (aeP); \draw[->] (Linf) -- (aU); \draw[->] (aU) -- (M); \draw[->] (aU) -- (aeP); \draw[->] (L1) -- (M); \end{tikzpicture} \caption{Relation between different convergence modes} \end{figure} \begin{enumerate}[label=( oman*)] \item Trivial. \item Trivial. \item Trivial. \item Let $\left(E ight)_{n\in\mathbb{N}}$ be a sequence of measurable sets in $X$ such that for each $n\in\mathbb{N}$ we have $\mu(E_n)<1/n$ and $\left(f ight)_{n\in\mathbb{N}}$ converges uniformly to $f$ on $X/E_n$. Now pick an arbitrary $x\in X$. We have two cases. In the first case $x \in \bigcap_{n \in \mathbb{N}} E_n$, in which case $\lim_{n o\infty}f_n(x)$ is not necessarily $f(x)$. But this is not harmful, since $\mu(\bigcap_{n \in \mathbb{N}} E_n) = 0$. In the second case $x otin \bigcap_{n \in \mathbb{N}} E_n$, which implies $\lim_{n o\infty}f_n(x)=f(x)$ from the uniform convergence. \item Trivial. \item Suppose that we can find a $N \in \mathbb{N}$ such that for all $n> N$ we have $$\intx{\left| f_n - f ight|} < \delta $$ By Markov's inequality we have $$\epsilon \cdot \mu\left(\{x\in X: |f_n-f| \geq \epsilon \} ight) \leq \intx{\left| f_n - f ight|} < \delta$$ \item Pick an arbitrary $\delta>0$ to test the convergence of the measure. Since $\left(f ight)_{n\in\mathbb{N}}$ converges to $f$ almost uniformly, for this $\delta$ we can find a set $E: \mu(E)< \delta$ such that $\left(f ight)_{n\in\mathbb{N}}$ converges uniformly to $f$ on $X/E$. Accordingly, we can find a $N \in \mathbb{N}$ such that for all $x\in X/E$ and for all $n>N$ we have $|f_n-f|< \epsilon$. Thus, for all $n>N$ we have \begin{align*} \mu\left(\{x\in X: |f_n-f| \geq \epsilon \} ight) &= \mu\left(\{x\in E: |f_n-f| \geq \epsilon \} ight) + \mu\left(\{x\in X/E: |f_n-f| \geq \epsilon \} ight)\ &\leq \delta + 0 = \delta \end{align*} \end{enumerate}

Exercise 1.5.2 (Easy convergence implications)

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Exercise 1.5.2 (Easy convergence implications)

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