A measure space which satisfying $\inf\{\mu(E): E\in M \text{ such that } \mu(E)>0\}=0$

Question

I need help with the following questions. Am I correct for the 2. and 3.? How can we solve the 1.?

Problem: We denoted a measure space $(X,M, \mu)$ subatomic if $\inf\{\mu(E): E\in M \text{ such that } \mu(E)>0\}=0$.

1. Show that in a subatomic space there are disjoint measurable sets $E_n$ ($n\geq 1$) with $0<\mu(E_n)<\frac{1}{2^n}$.
2. If $(X,M,\mu)$ is a subatomic space, show that $L^1(X,\mu)\not\subset L^2(X,\mu)$.
3. Conversely, show that if $L^1(X,\mu)\not\subset L^2(X,\mu)$ then $(X,M, \mu)$ is a subatomic space.

Solution

1. Since $\displaystyle{\big( X , M, \mu \big)}$ is subatomic space, that is $\inf \big \{ \mu(E) : E \in M \text{ such that } \mu(E)>0 \big \} = 0$, so \begin{align*} \exists A_{n} \in M \text{ } \text{ , } \text{ } \mu ( A_{n} ) > 0 \text{ } \text{ , } \text{ } \forall n \in \mathbb{N} \text{ } \text{ : } \text{ } \mu ( A_{k} ) \xrightarrow[]{k \to \infty} 0 \Longrightarrow \\ \Longrightarrow \exists n_{k} \in \mathbb{N} \text{ , } n_{1} < n_{2} < \cdots \text{ : } \text{ } \mu ( A_{ n_{k} } ) < \frac{1}{2^{k}} \text{ } \text{ , } \text{ } \forall k \in \mathbb{N}. \end{align*} We define \begin{align*} B_{1} = A_{n_{1}} \\ \\ B_{k} = A_{ n_{k} } \setminus \left( \bigcup\limits_{m=1}^{k-1} A_{n_{ m }} \right), \forall k \geqslant 2 \end{align*} such that $\displaystyle{B_{k}}$ are disjoint measurable sets(as $\displaystyle{A_{n_{k}}}$ are measurable) for all $k \geqslant 1$ and \begin{align*} m ( B_{1} ) = m ( A_{n_{1}} ) < \frac{1}{2^{n_{1}}} < \frac{1}{2} \\ \\ m ( B_{k} ) = m \left( A_{n_{k}} \setminus \left( \bigcup\limits_{m=1}^{n-1} A_{n_{m}} \right) \right) < m ( A_{n_{k}} ) < \frac{1}{2^{k}} , \forall k \geqslant 2. \end{align*}

My question is that: Is it true $m(B_{k})>0$ for all $k\geq2$? OR how can we find that sets $E_{k}$ for all $k\geq1$?

2. Assume $(X,M,\mu)$ is a subatomic space. Then, there are disjoint measurable sets $\displaystyle{E_{n} \in \mathcal{M}}$ with $\displaystyle{0 < \mu ( E_{n} ) < \frac{1}{2^{n}}}$ for all $n \geqslant 1$. \ Set $\displaystyle{f = \sum\limits_{n \geqslant 1} c_{n} \chi_{E_{n}}}$ for $\displaystyle{c_{n} = \frac{1}{ ( \sqrt{2} )^{n} \mu ( E_{n} ) }}$ we have that: \begin{align*} || f ||_{ L^{1} } = \int\limits_{X} | f | d \mu = \int\limits_{X} \sum\limits_{n \geqslant 1} c_{n} \chi_{E_{n}} d \mu = \sum\limits_{n \geqslant 1} c_{n} \int\limits_{X} \chi_{E_{n}} d \mu = \sum\limits_{n \geqslant 1} c_{n} \mu ( E_{n} ) = \\ = \sum\limits_{n \geqslant 1} \frac{1}{ ( \sqrt{2} )^{n} \mu ( E_{n} ) } \mu ( E_{n} ) = \sum\limits_{n \geqslant 1} \left( \frac{1}{\sqrt{2}} \right)^{n} = \frac{ \frac{1}{\sqrt{2}} }{ 1 - \frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1 < \infty \end{align*} and we find also \begin{align*} || f ||_{ L^{2} }^{2} = \int\limits_{X} | f |^{2} d \mu = \int\limits_{X} \sum\limits_{n \geqslant 1} c_{n}^{2} \chi_{E_{n}} d \mu = \sum\limits_{n \geqslant 1} c_{n}^{2} \int\limits_{X} \chi_{E_{n}} d \mu = \sum\limits_{n \geqslant 1} c_{n}^{2} \mu ( E_{n} ) = \\ = \sum\limits_{n \geqslant 1} \frac{1}{ 2^{n} ( \mu ( E_{n} ) )^{2} } \mu ( E_{n} ) = \sum\limits_{n \geqslant 1} \frac{1}{ 2^{n} \mu ( E_{n} ) } > \sum\limits_{n \geqslant 1} 1 = \infty. \end{align*} Therefore $\displaystyle{f \in L^{1}}$ and $\displaystyle{f \notin L^{2}}$.
3. We assume that: \begin{align*} \inf \Big \{ \mu ( E ) : E \in \mathcal{M} , \mu ( E ) > 0 \Big \} = \delta > 0. \end{align*} Let $\displaystyle{f \in L^{1}}$, set $\displaystyle{\int\limits_{X} | f | d \mu = c < \infty}$. \ We are proving that: $\displaystyle{f \in L^{2}}$. \ Consider the sets $\displaystyle{A_{n} = \Big \{ x \in X : | f | > n \Big \}}$ for all $n \geqslant 1$. \ We have $\displaystyle{A_{n+1} \subseteq A_{n}}$ for all $n \geqslant 1$. \ Then \begin{align*} \infty > \int\limits_{X} | f | d \mu = c > \int\limits_{A_{n}} | f | d \mu > \int\limits_{A_{n}} n d \mu = n \mu ( A_{n} ) , \forall n \geqslant 1 \Longrightarrow \\ \Longrightarrow 0 < \delta \leqslant \mu ( A_{n} ) < \frac{c}{n} , \forall n \geqslant 1 \Longrightarrow \mu ( A_{n} ) < \infty , \forall n \geqslant 1 \Longrightarrow \\ \Longrightarrow \exists N \in \mathbb{N} : \mu ( A_{n} ) = 0 , \forall n \geqslant N. \end{align*} It is true that: $\displaystyle{|f| \leqslant n+1}$ in $\displaystyle{A_{n} \setminus A_{n+1}}$ for all $\displaystyle{n \geqslant 1}$. \ Therefore \begin{align*} \int\limits_{A_{1}} | f |^{2} d \mu \leqslant 2^{2} \mu ( A_{1} \setminus A_{2} ) + 3^{2} \mu ( A_{2} \setminus A_{3} ) + & \cdots + (N-1)^{2} \mu ( A_{N-1} \setminus A_{N-2} ) < \infty \\ \text{and} \\ \int\limits_{A_{1}^{c}} | f |^{2} d \mu \leqslant \int\limits_{A_{1}^{c}} | f | d \mu < \infty. \end{align*} Therefore $\displaystyle{\int\limits_{X} | f |^{2} d \mu < \infty \Longrightarrow f \in L^{2}}$.

Answer 1

For part (1), start with $A_0\in\mathcal{M}$ so that $0<\mu(A_0)<1$. Once $A_0,\ldots,A_k$ have been selected, choose $A_{k+1}\in\mathcal{M}$ so that $0<\mu(A_{k+1})<\frac12\mu(A_k)$. It is clear that $0<\mu(A_{k+m})<2^{-m}\mu(A_k)$ for all $m\in\mathbb{N}$ and $k\in\mathbb{Z}_+$.

Define $B_k=A_k\setminus \bigcup_{j>k}A_k$. The $B_k$'s are pairwise disjoint and \begin{align} 2^{-k}>\mu(B_k)&=\mu(A_k)-\mu\big(\bigcup^\infty_{j=1}A_k\cap A_{k+j}\big)\geq\mu(A_k)-\sum^\infty_{j=1}\mu(A_k\cap A_{k+j})\\ &\geq\mu(A_k)-\sum^\infty_{j=1}\mu(A_{k+j})>\mu(A_k)-\mu(A_k)\sum^\infty_{j=1}2^{-j}=0 \end{align}

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Parts (2) and (3) seem correct.