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 $\inf\{\mu(E): E\in M \text{ such that } \mu(E)>0\}=0$.
- Show that in a subatomic space there are disjoint measurable sets $E_n$ ($n\geq 1$) with $0<\mu(E_n)<\frac{1}{2^n}$.
- If $(X,M,\mu)$ is a subatomic space, show that $L^1(X,\mu)\not\subset L^2(X,\mu)$.
- Conversely, show that if $L^1(X,\mu)\not\subset L^2(X,\mu)$ then $(X,M, \mu)$ is a subatomic space.
Solution
- 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 $k\geq1$?
- 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 $\displaystyle{f \notin L^{2}}$. - 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 $\displaystyle{\int\limits_{X} | f |^{2} d \mu < \infty \Longrightarrow f \in L^{2}}$.
Best Answer
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}
Parts (2) and (3) seem correct.