I believe the answer is no. Have a look at my counter-example.

The answer is basically no: take $1\leq s

Here is a follow up question. I call a subset S of L_{p,\infty} "controlled" if there is a g in S so that D_f

The distribution function is $D_f(s) = measure(\{ x : |f(x)| > s\})$, isn't it?

Yes. I also think that combining "controlled" with "weak compactness"(p>1) may imply some interesting properties of a set of functions in weak Lp. For example, I THINK these 2 conditions imply that the set is closed in measure, if a sequence in such a set converges in measure then its limit is in the set and the sequence converges in weak Lp too. I may be wrong as I have not though about this enough yet.

I believe that the ball of weak Lp is controlled by a multiple of the function 1/t^(1/p). This in turn says that "controlled"= bounded in weak Lp.

That is true. However, does every bounded set contain a control function? The answer is no. Therefore, "controlled" is stronger than "bounded". Given a bounded subset of weak Lp which is not controlled, can you always identify a function outside the set that controls the set. In other words, can every bounded set be made into a controlled set by simply adding a single function? I'm not sure this is possible for general domains but I may be wrong. In other words, if (A,m) is a non-atomic probability space must weak-Lp(A,m) contain a function whose distribution is 1/t^p? If so, then all bounded sets can be made "controlled" by adding a single function.

I should update the "controlled" condition for a set of measurable functions F to be there is a g in F so that D_f(t)0 and all f in F. With this better condition, I believe I have a proof that any compact subset of a separable closed subspace of L_0 has a control function g. I posted a related question which asks for which Banach spaces of measurable functions does weak-convergence imply convergence in measure. In such spaces, my proof goes through(as long as it is correct) with weak-comapactness in place of compactness. Some or all of these things may be known but I have not found anything yet.

Oh, I missed the part about the control being within the set.

For your other questions, I think that it should follow from Maharam's classification of measure spaces that all nonatomic probability spaces contain a sub sigma algebra measure isomorphic to [0,1], or more precisely 2^N. So there is a meas function with dist 1/t^(1/p).

But without using this big hammer, we can observe that we don't need precisely 1/t^(1/p), within a multiple will do (for control). This is easy to achieve since we can construct disjoint meas sets so that the n th one has measure 1/2^n.

In order to assume that the domain is [0,1] and the measure is Lebesgue measure, I thought one needed the domain to be a complete separable metric space. It is not clear to me if the above mentioned isomorphism of a sub-sigma algebra will preserve a condition like "controlled" for the entire sigma-algebra. I would have to take some serious time to think about this.

The real issue with the existence of a control function is that it is a limit point of the set with respect to the topology of the space of measurable functions in which the set resides.

Could it be that you are mixing things up with the Sobolev embedding theorem. There are some people that denote by L^p_q the space of function which such that all derivatives up to the q'th one in L_p (i.e. what many people denote by W^p_q) . If D \subset R^n, is compact with smooth boundary, then

L^p_q(D) \subset L^r_s(D) if q > s (i.e. if you relax on differentiability) and q - n/p >= s - n/r (i.e. you can trade differentiabiliy for higher integrability) and here 1

No. I am thinking of Lorentz spaces.

Geoff,

Maybe I am confused, but consider the Rademacher sequence in L^p (1\le p

Thanks Anton and good to hear from you again. Actually, the real issue is whether there are meaningful spaces in which weak convergence implies convergence in measure. I talked with Chris Lennard about this and he explained that this would be a weakened Schur property but it is unclear if there are such spaces which don't already have the schur property.

With regard to your comment on compactness, I am interested in which classical spaces do and do not compactly embed into L_0. Moreover, are there any known reasons for why such embeddings would or would not exists?

Dear Geoff Diestel

I am not used to play with weird measured spaces, but on the part of the set where the measure has no atoms, one may use a theorem of Liapounov (not the famous one who worked on ODE) for which Zvi Artstein has given a quite simple proof in the mid 70s (Look for the Extreme Points), and construct a sequence of characteristic functions converging weakly * in L^{\infty} to 1/2 (hence a sequence like the Rademacher one mentioned by Anton Schep).

Luc TARTAR, mathematician