If $\mu$ is a finite measure on a measure space $(X, \Sigma,\mu)$, then how do we show that $L^p(d \mu)$ being separable implies $\Sigma$ is separable in the metric $d(E_1, E_2) = \mu(E_1 \Delta E_2)$? I know how to show the other direction.
Separable $L^p$ with a finite measure
2
$\begingroup$
real-analysis
measure-theory
1 Answers
3
Consider a countable dense subset $\{f_n\} \subset L^p(\mu)$.
Then define sets $E_n: = f_n^{-1}((1/2,3/2))$. I claim that the $E_n$ form a countable dense subset of $\Sigma$ with respect to the metric $d$.
To prove this, take any set $E \in \Sigma$. By density we can find a subsequence $f_{n_k}$ such that $\|f_{n_k}-1_E\|_p \to 0$ as $k \to \infty$. Since convergence in $L^p$ implies convergence a.e. along a subsequence, we can find a further subsequence $f_{n_{k_j}}=:g_j$ such that $g_j \to 1_E$ a.e.
Define a function $h:\Bbb R \to \{0,1\}$ by $h(x):=1_{(1/2, 3/2)}(x)$. That is, $h(x)=1$ iff $|x-1| < 1/2$. Since $h$ is continuous at $0$ and $1$, and since $g_j \to 1_E$ which only takes values in $\{0,1\}$, it follows that $h \circ g_j \to h \circ 1_E = 1_E$ a.e. (as $j \to \infty$).
But $h \circ g_j$ is the indicator function of the set $E_{n_{k_j}} = :F_j \in \mathcal E$. So we see that $|1_{F_j}-1_E| \to 0$ a.e. as $j \to \infty$. By the dominated convergence theorem, it follows that $$d(F_j,E)=\mu(F_j \Delta E)=\int_X |1_{F_j}-1_E| d\mu\; \stackrel{j \to \infty}{\longrightarrow}\;0$$ which shows that the $E_n$ are dense in $\Sigma$.