$L_p(X)$ separable if $(X,\mu)$ is separable measure space.

Question

A measure space $(X,\mu)$ is separable if there is a countable family of measurable subsets $\{E_k \}_{k=1}^\infty $ so that if $E$ is any measurable set of finite measure , then $$\mu(E \triangle E_{n_k}) \to 0 \,\,\,\,\,\,\,as \,k\to0$$ for an appropriate subsequence $\{n_k \}$ which depends on $E$ .

Prove that if the measure space $X$ is separable, then $L_{p}$ is separable when $1 ≤ p < ∞$.

I try to prove it like this:

We have for every measurable set $E$ with finite measure associated measurable set $E_{nk}$ s.t. $\mu(E \triangle E_{n_k})<\epsilon$.

CLAIM: The collection$$ F:= \{\sum_{i=1}^Nr\chi_{E_{n_{i}} }\}\,\,\,\,\,\,\, r\in Q $$ is countable dense in $L^p$.

Since simple functions are dense in $L^p$ given $f \in L^p$, Let $\epsilon >0$ and choose $\phi$ such that $\|\phi-f\|_{L^p} < \frac{\epsilon}{2}$. Now, let $$\phi = \sum_{i=1}^{N} c_i \chi_{E_i}$$ with $E_i$ pairwise disjoint meaurable with finite measure. Let $\psi \in F$ with $$\psi = \sum_{i=1}^{N} r_i \chi_{E_{n_i}}$$ be such that $\mu(E_n \triangle E_{n_i}) < {\epsilon^p}$ with the $E_{n_i}$ pairwise disjoint. Then,

\begin{eqnarray*} \left( \int_\mathbb{X} |\phi-\psi|^p \, d\mu \right)^\frac{1}{p} &\leq& \left( \int_\mathbb{X} \left(\sum_{i=1}^{N}|c_i\chi_{E_i}-r_i\chi_{E_{n_i}}| \right)^p \, d\mu \right)^\frac{1}{p} \\ & \stackrel{Minkowski}{\leq}& \sum_{i=1}^{N} \left( \int_\mathbb{X} |c_i\chi_{E_i}-r_i\chi_{E_{n_i}}|^p \, d\mu \right)^\frac{1}{p} \\ \end{eqnarray*}

I got stuck..!! I try to connect the last above inequality with symmetric difference between $E_n$ and $E_{n_i}$ but I have difficulties in that because of existences of $r_i$ and $c_i$.

I need help to complete the proof.

Thanks.

Answer 1

We claim that $\mathcal{A} := \{\sum_{i=1}^N r_i \chi_{E_{k_i}} : r_i \in \mathbb{Q}, k_1,\dots,k_N \ge 1\}$ is dense in $L^p$. Since simple functions are dense, it suffices to show simple functions are in the closure of $\mathcal{A}$. By linearity and the density of the rationals in the reals, it suffices to show that $\chi_E$ is in the closure of $\mathcal{A}$ for any measurable set $E$. So fix a measurable set $E$, and take $\epsilon > 0$. By assumption, there is some $E_k$ so that $\mu(E\Delta E_k) < \epsilon$. Then $||\chi_E - \chi_{E_k}||_p \le \epsilon^{1/p}$.