# A detail about MCT application

I have a indirect question about Monotone Class Theorem (MCT), in its functional form. Here is a version which should be sufficiently general for my purpose.

Functional Monotone Class Theorem :

Let H a collection of real valued and bounded functions such that:
- H is a vector space
- H contains constants
- If $f_n$ is an increasing sequence of real and bounded functions of H, admitting a limit $f$ that is a bounded real valued function, then $f$ belongs to H.

Now let K be another collection of real bounded functions which is stable by multiplication, and such that $K \subset H$.

Conclusion :

Then H contains all real bounded functions that belongs to $\sigma(K)$ (the sigma algebra generated by K).

The statement here is clear but I have a question about applications of it.

Usually the collection K is built by considering a collection P over real valued functions satisfying some propetry and intersecting it with the collections of bounded functions. Then using MCT with a nice space H, you have your conclusion for $\sigma(K)$.

My question is the follwing is there a general way or sufficient conditions to see if $\sigma(K)=\sigma(P)$. There is a trivial inclusion so the question is better stated as under what conditions does $\sigma(P) \subset \sigma(K)$ holds ?

The motivation behind this question comes from the fact that when I see MCT used in proofs of theorems, it is often the case that the theorem's conclusion is stated for the unbounded real valued functions collection P rather than K (P+the bounded property).

Best Regards

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I think that Proposition 8.15 and Theorem 8.16 in these lecture notes answer your question to some extent. See also Nate Eldredge's answer here where I learned about the existence of these notes. – t.b. Aug 1 '11 at 11:35
@Theo Buehler: Hi thank's for the link to those rich and interesting Notes. But regarding the theorems that you quote I don't think they answer the question unfortunately (or I don't see how). – – TheBridge Aug 1 '11 at 12:53
I apologize for promising too much in haste, I didn't think very deeply about your question but it seemed close enough. – t.b. Aug 1 '11 at 13:14
@Theo Buehler: Indeed rather close and please do not apologize. As a matter of fact, the question isn't really directly about MCT, but rather under what conditions 2 sigma fields can be the same. My best guess would be that if for any $f$ in P (bounded or not) there's a sequence of functions in K that ponctually converge to $f$ then we are done. Best regards – TheBridge Aug 1 '11 at 13:40