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I'm trying prove this proposition, i got a part.

Proposition: Let $(X,M, \mu)$ a finite measure space and define a metric space $M'$ as follows: for $A,B \in M$, define: $d(A,B) = \mu(A\triangle B)$.

The space $M'$ is defined as all sets in $M$ where sets $A$ and $B$ are identified if $\mu(A\triangle B) = 0$.

(a) Prove that $(M', d)$ is a complete metric space

(b) Prove that $(M', d)$ is separable iff $L^p(X, M', \mu)$ is, $1\leq p < \infty$.

My attempt: (a) I will prove that $(M', d)$ is complete. Let $\displaystyle(E_i)_{i\in \mathbb{N}}$ a Cauchy sequence relative to the metric $d$. Given $\epsilon > 0$ there exists $k_0 \in \mathbb{N}$ such that, for $m, n \geq k_0$, we have $d(E_m, E_n) < \epsilon$. Fix $E_{k_{0}}$.

Consider the set $\bigcup \bigcap E_{j}$; we have $\displaystyle\left(\bigcup \bigcap E_{j}\right)^{c} = \left(\bigcap \bigcup E_{j}^c\right)$. By the definitions of $\limsup, \liminf$, there exists $m > k_0$ such that $\bigcup \bigcap E_{j} \subset E_m$ and $\bigcap\bigcup E_{j}^c \subset E_{m}^c$, so $\displaystyle\left(\bigcup \bigcap E_{j}\right)\cap E_{k_0} \bigcup E_{m}\cap\left(\bigcup \bigcap E_{j}\right)^c \displaystyle\subset (E_{m}\cap E_{k_0}^c) \cup (E_{m}^c \cap E_{k_0})$.

Therefore $\displaystyle d \left(\bigcup \bigcap E_{j}, E_{k_0}\right) \leq d(E_{m}, E_{k_0}) < \epsilon$. Since $\bigcup \bigcap E_{j} \in M'$, we conclude that $(M', d)$ is a complete metric space.

(b)($\Rightarrow$) Let $S \subset M$ countable and dense and $\displaystyle \{A_i\}_{i \in \mathbb{N}}$ a finite partition of $X$. For each $A_i$, there exists a sequence $(B_{j}^i)_{j}$ of elements of $S$ such that $B_{j}^i \rightarrow A_i$ in the sense that $d(B_{j}^i, A_i)\rightarrow 0$. Let $f\in L^p$ a simple function. We can approximate $f$ by simple functions defined on $S$. Then we can aproximate any $f \in L^p$ this way,. As $S$ is countable, the same holds for the simple functions obtained from $S$, moreover this set is dense in $L^p$.

Lack the converse. Any error? How proceed to the converse?

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    Since you have a finite measure space, the characteristics functions of element of $M$ are in $L^p$. That might be useful, idk. Just two comments: 1) you consider a finite partition of $X$, yet it is indexed on $i\in\mathbf{N}$. 2) Also in (b), you wrote $L^p(X,M',\mu)$. Shouldn't it be $M$? – 2017-02-10
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    Ok; but as the soluction, seems ok? – 2017-02-10

1 Answers 1

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($\Leftarrow$) let $B$ the countable and dense subset of $L^p(\mu)$. Let $A\in M$ and consider the characteristic function $\chi_{A}$ (note that $\chi_{A} \in L^p(\mu)$), there is a sequence $(f_n)$ of functions of $B$ such that $||f_n - \chi_{A}||_p = 0$, then we can take $0 < \eta < 1$ such that $\eta < f_n$ a.e. for $n$ big. Consider the sequence $(D_n)$ where $D_n$ is domain of $f_n$, if $\mu(A\triangle D_n)\nrightarrow 0$ then there is $\epsilon > 0$ such that for a infinity $n's$ $\mu(A - D_n) > \epsilon$ or $\mu(D_n - A) > \epsilon$. In any case $\displaystyle\mu(A - D_n) = \int_{A - D_n} |f_n - \chi_{A}|^p d\mu \leq \int |f_n - \chi_{A}|^p d\mu$ or $\displaystyle\eta\mu(D_n - A) \leq \int_{C_n - A}|f_n - \chi_{A}|^pd\mu \leq \int|f_n - \chi_{A}^p|d\mu$, contradiction.

Therefore $d(A, D_n) \rightarrow 0$; take the class of all domains of functions of $B$, this class is enumerable and dense in $M$.