9
$\begingroup$

Let $\Sigma = \sigma(\mathcal C)$ be the $\sigma$-algebra generated by the countable collection of sets $\mathcal C \subset \mathcal{P}(X)$. How can I prove that if $\mu$ is a $\sigma$-finite measure on $(X,\Sigma)$ then $L^p(X)$ is separable for $1 \le p < \infty$?

I know that simple functions are dense in $L^p(X)$, so I would like to find a countable subset of the set of simple functions that is dense in them. Could you help me please?

$\endgroup$
11
$\begingroup$

Before going into a formal proof, here is the idea. The space is $\sigma$-finite, so we can "break" it into countably many spaces of finite measure. Up to some technical considerations, we are reduced to the case $X$ of finite measure. An algebra generated by a countable class is countable and we can approximate elements of finite measure by those of a generating algebra.

Let $(A_n,n\geqslant 1)$ be a partition of $X$ into measurable sets of finite measure.

  1. We show that $(A_n,A_n\cap \Sigma,\mu_{\mid A_n\cap \Sigma})$ is separable. Consider $f\in L^p(A_n)$ and fix $\varepsilon>0$. There is $f'=\sum_{j=1}^J a_j\chi_{B_j}$ simple simple such that $\int_{A_n}|f-f'|^p\mathrm d\mu\lt\varepsilon^p$. Define $\mathcal A_n$ the algebra generated by sets of the form $A_n\cap C,C\in\mathcal C$. Then $\mathcal A_n$ is countable. Approximate $B_j$ by $B'_j$, an element of $\mathcal A_n$, that is, such that $\mu(B'_j\Delta B_j)\lt \frac 1{J(|a_j|^p+1)}\varepsilon$. Defining $f'':=\sum_{j=1}^Ja_j\chi_{B'_j}$, we get $\lVert f-f''\rVert^p\lt 2\varepsilon$.

  2. Define $D_n$ as the set of linear combinations with rational coefficients of characteristic functions of elements of $\mathcal A_n$. Since $\mathcal{A}_n$ is countable, so is $D_n$. Finally, define $$D:=\bigcup_{N\geqslant 1}\left\{\sum_{i=1}^Nd_i,d_i\in D_i\right\}.$$ Then $D$ is countable and dense in $L^p(X)$.

$\endgroup$
  • 1
    $\begingroup$ Thank you! I was playing with the algebra generated by $C$ but I could not find a way to approximate sets with that! This helped me so much :D $\endgroup$ – user67133 Oct 24 '13 at 16:55
  • 1
    $\begingroup$ You are welcome! (in the OP I replaced $C$ by $\mathcal C$, as these letter denote generally a class of sets, while $C$ is used for sets) $\endgroup$ – Davide Giraudo Oct 24 '13 at 17:33
0
$\begingroup$

Begin as follows. You have a countable collection of sets generating the algebra. Now take the finite intersections of all elements of this collection; it is countable. Now take the finite unions of the countable family of finite intersections. This is an algebra of sets; the smallest $\sigma$-algebra generating it is $\Sigma$; let us denote this algebra by $\mathcal{A}$.

The simple functions $$\mathcal{S} = \left\{\sum_{k=1}^n c_k \chi_{A_k}, A_1, A_2, \cdots A_k \in \mathcal{A}, c_1, c_2, .... c_n\in \mathbb{Q}, n\in \mathbb{N}\right\}$$ constitute a countable set.

Choose $E\in \sigma$. For any $\epsilon > 0$ you can choose $A\in \mathcal{A}$ so that $\mu(A\Delta E) < \epsilon$. This says that every characteristic function is in the closure of $\mathcal{S}$. Can you continue and show that $\mathcal{S}$ is dense in $\mathcal{L}^p$?

$\endgroup$
-1
$\begingroup$

How about simple function with rational coefficients?

$\endgroup$
  • $\begingroup$ they are not countable. (every infinite sigma algebra is uncountable, so even characteristic functions are too many) $\endgroup$ – user67133 Oct 23 '13 at 0:47
  • $\begingroup$ only use elements of your countable set. Can you show that these approximate any simple function measurable wrt your $\sigma$-algebra. $\endgroup$ – ncmathsadist Oct 23 '13 at 0:53
  • $\begingroup$ Do you mean the sets in $C$ with finite measure? Otherwise it could be meaningless.. $\endgroup$ – user67133 Oct 23 '13 at 1:07
  • $\begingroup$ Yes. You can localize this then spread it via $\sigma$-finiteness. $\endgroup$ – ncmathsadist Oct 23 '13 at 1:09
  • $\begingroup$ "Only use elements of your countable set": You may also need their complements, and finite intersections. $\endgroup$ – Nate Eldredge Oct 23 '13 at 2:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy