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

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.

• (1) in first centered thing, $k \to 0$ should be $k \to \infty$. (2) in the CLAIM, you sum over $i$ but there's no $i$. Commented May 8, 2020 at 22:42
• From Stein Analysis book k goes to 0...second one typo error
– ks1
Commented May 8, 2020 at 22:45
• i don't see how $k$ could possibly be going to $0$. that makes absolutely no sense Commented May 8, 2020 at 22:49
• also, the collection you defined in CLAIM is for a particular $E$? that seems wrong Commented May 8, 2020 at 22:50
• my family contains all associated $E_{n_k}$ for each measurable set $E$
– ks1
Commented May 8, 2020 at 22:52

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 $$||\chi_E - \chi_{E_k}||_p \le \epsilon^{1/p}$$.

• I tried to give a detailed answer for the question...So it's enough to show that $χ_E$ is in the closure of $A$
– ks1
Commented May 8, 2020 at 23:55
• what did you mean about the closure...and isn't my family $F$ like the family $A$?
– ks1
Commented May 8, 2020 at 23:56
• what do u think "dense" means? Commented May 8, 2020 at 23:57
• Any element can be written as a linear combination of the dense set elements
– ks1
Commented May 8, 2020 at 23:58
• @ks1 finite union of balls with rational radius Commented May 9, 2020 at 13:36