Exercise 1.5.3 (Convergence for step functions)

Question

Let $(X,\mathcal{B},\mu)$ be a measurable set. Let $\left(fight)_{n\in\mathbb{N}}$ be a sequence of functions defined by
$$f_n:X	o [0,+\infty), \qquad f_n = A_n \cdot 1_{E_n} $$
for measurable sets $\left(Eight)_{n\in\mathbb{N}}$ of non-zero measure $\mu(E_n)>0$ and some positive reals $\left(Aight)_{n\in\mathbb{N}}$ which either converge to zero ($\lim_{n	o\infty}A_n = 0$) or are bounded away from zero ($\exists c>0$ such that $\forall n\in\mathbb{N}: A_n>C$).
ewline

Establish the following claims:
\begin{enumerate}[label=(oman*)]
\item $\left(fight)_{n\in\mathbb{N}}$ converges uniformly to zero if and only if $A_n	o 0$ as $n	o\infty$.
\item $\left(fight)_{n\in\mathbb{N}}$ converges in $L^\infty$ norm to zero if and only if $A_n	o 0$ as $n	o\infty$.
\item $\left(fight)_{n\in\mathbb{N}}$ converges almost uniformly to zero if and only if $A_n	o 0$ as $n	o\infty$ or $\mu(E_N^*)	o 0$ as $N	o\infty$.
\item $\left(fight)_{n\in\mathbb{N}}$ converges pointwise to zero if and only if $A_n	o 0$ as $n	o\infty$ or $\bigcap_{N=1}^{\infty}E_N^* = \emptyset$.
\item $\left(fight)_{n\in\mathbb{N}}$ converges pointwise almost everywhere to zero if and only if $\left(Aight)_{n\in\mathbb{N}}	o 0$ as $n	o\infty$ or $\bigcap_{N=1}^{\infty}E_N^*$ is a null set.
\item $\left(fight)_{n\in\mathbb{N}}$ converges in measure to zero if and only if $\left(Aight)_{n\in\mathbb{N}}	o 0$ as $n	o\infty$ or $\mu(E_n)	o 0$ as $n	o\infty$.
\item $\left(fight)_{n\in\mathbb{N}}$ converges in $L^1$ norm to zero if and only if $A_n\mu(E_n)	o 0$ as $n 	o \infty$.
\end{enumerate}

Elu Thingol · Accepted Answer

$ ewcommand{\intx}[1]{\int_{X} #1 \mathop{}\!\mathrm{d}{\mu}}$ $ ewcommand{\seq}[1]{\left( #1_n ight)_{{n} \in {\mathbb{N}}}}$ \begin{enumerate}[label=( oman*)] \item First, suppose that $\left(f ight)_{n\in\mathbb{N}}$ uniformly converges to zero, i.e., for every $x\in X$ we have $\forall \epsilon>0$ there exists a $N\in\mathbb{N}$ such that for all $n\geq N$ it is true that $A_n\cdot 1_{E_n}(x) \leq \epsilon$. In particular, set $\epsilon<1$, and choose an $x\in E_n eq \emptyset$. Then $A_n\cdot 1_{E_n}(x) = A_n$ should also be less that $\epsilon$, and so $\left(A ight)_{n\in\mathbb{N}} o 0$.\ Conversely, suppose that $\left(A ight)_{n\in\mathbb{N}} o 0$. Then for any $x\in X$ and any $\epsilon>0$ we either have $A_n\cdot 1_{E_n}(x) \leq A_n$ for which we have a $N\in\mathbb{N}$ such that $\forall n \geq N$ it is $A_n \leq \epsilon$, and so is $A_n\cdot 1_{E_n}(x) \leq \epsilon$. \item Suppose that $\left(f ight)_{n\in\mathbb{N}}$ converges to 0 almost everywhere uniformly. In particular, it converges uniformly on $X/E$ for some null set $E\subset X$, and so $A_n o 0$ by the previous part. ewline Conversely, if $A_n o 0$ as $n o\infty$, then $\left(f ight)_{n\in\mathbb{N}}$ converges to 0 uniformly by part (i) of this exercise, and thus, in particular it converges uniformly almost everywhere to 0. \item Suppose that $\left(f ight)_{n\in\mathbb{N}}$ converges to $0$ almost uniformly, i.e., for every $\epsilon>0$ there exists a set $E\in\mathcal{B}$ such that $\mu(E)\leq \epsilon$ and $(f_n estriction_{X/E})_{n\in\mathbb{N}}$ converges to $0$ uniformly. Again, by part (i) we trivially have $\left(A ight)_{n\in\mathbb{N}} o 0$ as a consequence of choosing any arbitrary $\epsilon$. ewline Conversely, suppose that $\left(A ight)_{n\in\mathbb{N}} o 0$. Then $\left(f ight)_{n\in\mathbb{N}}$ converges to $0$ uniformly, in particular, almost uniformly. \item Suppose that $\left(f ight)_{n\in\mathbb{N}}$ converges to 0 pointwise, i.e., for all $x \in X$ we have $A_n\cdot 1_{E_n}(x) o 0$. Notice that we have two cases. (1) $\seq{{1_{E}}}$ converges to $0$ for all $x\in X$. In the case of the indicator function, this is equivalent to saying that for all $x\in X$ it is true that $1_{E_n}(x) = 0$ for all $n\geq N$ after some $N\in\mathbb{N}$. In other words, for each $x\in X$ there exists a $N \in \mathbb{N}$ such that $x otin \bigcup_{n\geq N} E_n$. But this would imply that $\bigcap_{N=1}^{\infty}\bigcup_{n\geq N} E_n = \bigcap_{N=1}^\infty E^*_n = \emptyset$, as desired. Thus, we at least have the vanishing of the support. (2) In the second case, suppose that there exists a $x\in X$ for which $\seq{{1_{E}}}$ does not converge to zero. In the case of the indicator function, this is equivalent to saying that for infinitely many $n\in\mathbb{N}$ we have $1_{E_n}(x) = 1$. But for any $\epsilon>0$ we can find a $N\in\mathbb{N}$ such that for all $n\geq N$: $A_n\cdot 1_{E_n} \leq \epsilon$. This is only possible when $A_n \leq \epsilon$ for all $n \geq N$, and so we get $\left(A ight)_{n\in\mathbb{N}} o 0$. ewline Now suppose conversely that either $\left(A ight)_{n\in\mathbb{N}} o 0$ or $\bigcap_{N=1}^{\infty}E_N^* = \emptyset$. In the first case, $\left(f ight)_{n\in\mathbb{N}}$ converges to 0 uniformly by part (i), and so it converges to zero pointwise. In the latter case, we have $$\emptyset = \bigcap_{N=1}^{\infty}\bigcup_{n\geq N} E_n \supset \bigcap_{N=1}^{\infty} E_n = \emptyset,$$ and so $\lim_{n o\infty}1_{E_n}(x) = 0$ for any $x\in X$. From this, we easily conclude that $f_n(x) = A_n1_{E_n}(x) o 0$ for any $x \in X$. \item Suppose that $f$ converges pointwise almost everywhere to 0, i.e., for a null set $F\in\mathcal{B}$ we have $(f_n estriction_{X/E})_{n\in\mathbb{N}}$ converges pointwise everywhere to $0$ on $X/E$. By the previous part, this has one (also maybe both) of the two implications. (1) $\left(A ight)_{n\in\mathbb{N}} o 0$ as $n o\infty$, as desired. (2) For $(f_n estriction_{X/E})(x) = (A_n\cdot 1_{E_n} estriction_{X/E})(x) = (A_n\cdot 1_{E_n/E})(x)$ we have $\emptyset = \bigcap_{N=1}^{\infty}\bigcup_{n\geq N} (E_n/E) = \left(\bigcap_{N=1}^{\infty}\bigcup_{n\geq N} E_n ight)/E$, or in other words, taking unions from both sides we obtain $\bigcap_{N=1}^{\infty}E_n^* \subseteq E$, for the null set $E$, as desired. ewline Now suppose conversely one of the opposite statements. In case of $\left(A ight)_{n\in\mathbb{N}} o 0$, $\left(f ight)_{n\in\mathbb{N}}$ converges to 0 pointwise by the previous part. Now suppose that $\bigcap_{N=1}^{\infty}E_n^*$ is a null set. Then pretty straigthforwardly it follows that $\forall x \in X / \bigcap_{N=1}^{\infty}E_n^*: 1_{E_n}(x) \leq 1_{E_n^*}(x) o 0$, which is almost everywhere. \item Suppose that $\left(f ight)_{n\in\mathbb{N}}$ converges to 0 in measure, i.e., for any arbitrary $\delta>0$ we have $$\lim_{n o\infty}\mu\left(\left\{x\in X: A_n1_{E_n}(x)\geq \delta ight\} ight) = 0$$ Suppose for the sake of contradiction that both $\left(A ight)_{n\in\mathbb{N}}$ is bounded away from zero, and that $\lim_{n o\infty}\mu(E_n) eq 0$. Set $\delta := \liminf_{n o\infty}A_n$. We then have $$\mu\left(\left\{x\in X: 1_{E_n}(x)\geq \frac{\delta}{A_n} ight\} ight) = \mu\left(\left\{x\in X: 1_{E_n}(x)=1 ight\} ight) = \mu(E_n)$$ The last term does not converge to 0 as $n o\infty$ - a contradiction. ewline Fix an arbitrary $\delta>0$ in $\mu\left(\left\{x\in X: A_n1_{E_n}(x)\geq \delta ight\} ight)$ and suppose conversely that $\left(A ight)_{n\in\mathbb{N}} o 0$. Then, for our $\delta>0$ we can find a $N\in\mathbb{N}$ such that for all $n\geq N$ (and for all $x\in X$) we have $A_n\cdot 1_{E_n}(x) \leq A_n < \delta$, making the set $\left\{x\in X: A_n1_{E_n}(x)\geq \delta ight\}$ eventually empty. \ Now suppose that $\mu(E_n) o 0$ as $n o\infty$. Notice that by positivity of $\delta$ for any $x\in X$ we have $A_n1_{E_n}(x) \geq \delta \implies 1_{E_n}(x) = 1$, alas $$\mu\left(\left\{x\in X: A_n1_{E_n}(x) \geq \delta ight\} ight) \leq \mu\left(\left\{x\in X: 1_{E_n}(x)= 1 ight\} ight)=\mu(E_n) o 0.$$ \item Suppose that $\left(f ight)_{n\in\mathbb{N}}$ converges to $0$ in $L^1$-norm. By \href{./exercise_1-5-2}{Exercise 1.5.2} this implies that $\left(f ight)_{n\in\mathbb{N}}$ converges to $0$ in measure, and thus both $\left(A ight)_{n\in\mathbb{N}}$ and $\mu(E_n)$ converge to $0$, meaning their product converges to $0$ as well. ewline Conversely, suppose that $A_n\mu(E_n)$ converges to $0$. But $$A_n\cdot \mu(E_n) = \intx{A_n \cdot 1_{E_n}} = \intx{f_n} = \intx{|f_n - 0|}$$ and so $\left(f ight)_{n\in\mathbb{N}}$ converges in $L^1$ norm to $0$. \end{enumerate}

Exercise 1.5.3 (Convergence for step functions)

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Exercise 1.5.3 (Convergence for step functions)

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