# How can one show that the topology of convergence in measure is separable?

Let $X$ be a polish space equiped with the borel sigma-algebra and a probability measure $\mu$. How can one show that the set of all borel measurable functions $f:X\rightarrow R$ ($R$ being the real numbers), where two a.e. equal functions are identified, equiped with the topology of convegence in measure is separable?

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I've given you many references in my last answer to you. Fisher-Witte Morris-Whyte (rem 2.4) give a detailed reference to Wheeden-Zygmund, and it is contained in at least three of the other references I've given to you. Please look around a bit before you ask. You can also see Proposition 7 here for a more precise and general result (there polonais = Polish). – t.b. Aug 19 '11 at 10:49
Thank you Theo, you have helped me greatly. I did look around, but didn't find it. I will look in some of your references... BTW I haven't found it in Wheeden-Zygmund – Arnold Aug 19 '11 at 11:24
Hmmm. Not having one of my most charitable days, I guess... In fact, in the books I mentioned it's in the exercises (e.g. Kechris) and at times not even explicitly mentioned (e.g. in Wheeden-Zygmund), you're right. It's one of those results that people like to leave as an exercise... Sorry about that. Moore's proof of proposition 7 mentioned above suffers a bit from its generality. BTW: I flagged your account for merging with the other one. You could consider registering your account, then it is easier for the software to recognize you. – t.b. Aug 19 '11 at 12:12
@Arnold: I have merged your other account, A_0, into your current account. If you have trouble logging in, or if you accidentally create duplicate accounts, simply flag one of your own questions for moderator attention, and we will help out. – Zev Chonoles Aug 19 '11 at 14:39
Thanks (at least 15 characters...) – Arnold Aug 19 '11 at 14:53

Here's an outline of an argument, and it should be easy to fill in the details:

1. Note that $L_0(X)$ is a metric space e.g. with respect to the metric $\displaystyle d(f,g) = \int \frac{|f-g|}{1+|f-g|}$.
2. Choose a countable base $\{A_n\}_{n \in \mathbb{N}}$ for the topology on $X$.

• Every open set is equal to the union of elements in $\{A_n\}$.
• For every measurable set $E$ there is a $G_\delta$-set $G$ such that $\mu(E \triangle G) = 0$, that is $[E] = [G]$ in $L_{0}(X)$.
3. Show that a non-negative measurable function $f$ is a pointwise monotone limit of simple functions.
Hint: Put $B_{k,n} = \{x\in X : 2^{-n} k \leq f(x) \lt 2^{-n}(k+1)\}$ and consider $f_n = 2^{-n} \sum\limits_{k=0}^{2^{2n}}k \cdot[B_{k,n}]$.
4. Split a general measurable function into positive and negative parts.

Use these observations to build a countable dense set of $L_{0}(X)$.

For completeness and further properties of $L_0(X)$, I recommend Driver's notes on probability Section 12, especially Theorem 12.8 on page 179. (Thanks to Nate Eldredge from whom I learned about these notes).

Edit: In view of Byron's answer, note that Driver's notes contain various forms of the functional monotone class theorem in Part II, Section 8 on pages 111ff. Of course, the main point in both our answers is that there is a countable generating and separating set for the $\sigma$-algebra. The assumption that $X$ be Polish ensures that.

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@Didier: Yes, the formula is a bit nicer this way; but doesn't it obscure the idea somewhat? – t.b. Aug 20 '11 at 11:02
Sorry, mainly I meant to add the x in X in the definition of your set B. The other modification is a (possibly not so nice after all) nicety. Theo, really, proceed at your convenience, this is your post... – Did Aug 20 '11 at 11:08

An alternative is to use the Functional Monotone Class Theorem. Let $\cal A$ be a countable collection of sets that generates ${\cal B}(X)$, and put $${\cal K}=\{1_A: A\mbox{ is a finite intersection of }{\cal A}\mbox{ sets }\}.$$

Let ${\cal K}^\prime$ be the (countable!) $\mathbb{Q}$-vector space generated by $\cal K$, and set $${\cal M}=\{h: k^\prime_n\to h \mbox{ in probability for some }k^\prime_n\in{\cal K}^\prime\}.$$

Then $\cal K$ and $\cal M$ satisfy the conditions of the FMCT, and hence $\cal M$ includes all bounded ${\cal B}(X)$-measurable functions. A truncation argument now shows that any ${\cal B}(X)$-measurable function can be approximated in probability by a sequence in ${\cal K}^\prime$.

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I just wanted to point out that various useful variants of the FMCT are contained in Driver's notes I linked to, see the edit to my post. The dense set $\mathcal{K}'$ is of course the one I had in mind in my answer. – t.b. Aug 21 '11 at 8:01