FOM: RVM vs V=L

[JoeShipman@aol.com]

FOMers may find the following e-mail exchange I had with Harvey yesterday of 
interest:

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Shipman:

I am looking at ways to augment ZFC that have strong consequences in ordinary
mathematics.

There are a number of well-known independent statements (GCH, Suslin Problem,
Whitehead Problem, etc) that are settled by V=L; but V=L doesn't have anything
new to tell us about absolute statements.

I prefer to assume that c has a real-valued measure (or, slightly stronger,
that c is a real-valued measurable cardinal).  This has lots of consequences 
in
both analysis and arithmetic (including all your independent finite statements
and my strong Fubini theorems).

My question is: are any reasonable statements of ordinary mathematics or
descriptive set theory known to be independent of "c is a real-valued
measurable cardinal"?  The best I can think of is Projective Determinacy which
is equiconsistent with cardinals well above measurables; but is there any
reasonable statement about analytic sets or below which is known not to be
settled by "c is RVM"?

Friedman:

I don't know of any reasonable statements about analytic sets or below that
does not follow already from the existence of sharps, which is of course not
really mathematical, but is a weak consequence of RVM. The exceptions are some
statements I formulated about functions on and sets of rationals on the FOM a
while ago. They go up to very very large cardinals.

Any two analytic but not Borel sets of reals are Borel isomorphic is 
equivalent
to sharps. You could use that as an axiom?!

Shipman:

That's a very nice statement -- but I'm not ONLY interested in statements 
about
analytic sets or below, and RVM (which can be given some intuitive motivation)
is much stronger (in particular settling CH and much more).  In this sense
ZFC+RVM is much "more complete" than ZFC+V=L.  All it takes to intuitively
motivate ZFC+RVM is to give up the geometric intuition of the uniformity and
flatness of space, which modern physics makes dubious anyway.  (The only
problem with RVM is that the measure can't be translation-invariant.)

Friedman:

I have discussed in talks at Princeton Phil. Dept such merits of RVM, and the
fact that set theorists do not discuss it very much in terms of being an
axiom. Obviously, they don't accept it since if they did, they wouldn't
regard the continuum hypothesis as an open question. Yet I don't hear much
about why it is not a good axiom. If I recall, it contradicts Martin's
Axiom, however, and so the set theorists may be bothered with that. I think
that RVM actually is a good example of where set theoretic realists and 
Platonists are somewhat challenged. If they can't get a philosophical hold on 
the mere *status* of such a thing as RVM as an axiom - well, maybe their 
subject doesn't make as much well defined sense as they represent.

Shipman:

> I have discussed in talks at Princeton Phil. Dept such merits of RVM, and
> the fact that set theorists do not discuss it very much in terms of being
> an axiom. Obviously, they don't accept it since if they did, they wouldn't
> regard the continuum hypothesis as an open question. Yet I don't hear much
> about why it is not a good axiom. If I recall, it contradicts Martin's 
Axiom, 
> however, 

Of course it does -- MA implies that all sets of reals with cardinality < c
have Lebesgue measure 0 so you can have a subset of the plane that's measure 0
on vertical lines and co-measure 0 on horizontal lines, while RVM allows you 
to
prove my strong Fubini theorems.

> and so the set theorists may be bothered with that.

Why? Do they think MA is more plausible or do they find MA easier to work 
with?

>I think that RVM
>actually is a good example of where set theoretic realists and Platonists are
>somewhat challenged. If they can't get a philosophical hold on the mere
>*status* of such a thing as RVM as an axiom - well, maybe their subject
>doesn't make as much well defined sense as they represent.

I don't understand your point.  It's not set theoretic realism and Platonism
that get into trouble, the problem is simply that most of the set theorists 
out
there right now don't like RVM for (in my opinion) bad reasons (it's easier 
to 
prove things with MA, or RVM is unfashionable because it goes all the way 
back 
to Ulam, or it can't be proved consistent [this didn't used to be a bad 
reason, 
but by now they all believe measurables are consistent, and it's wimpy to 
only 
consider new axioms like MA that have no additional consistency strength]).  
"The current crop of set theorists, who are realists and Platonists, neglect 
RVM" doesn't imply the subject isn't well-defined.

A *good* reason to reject RVM would be an intuitive argument that it is false,
but that the real-valued measure couldn't be translation-invariant is not such
an argument.  It's almost a case of sour grapes -- since our beautiful
well-behaved Lebesgue measure can't be extended nicely, it can't be extended 
at
all.  The set theorists may have an intuition that sets smaller than c in
cardinality ought to have Lebesgue measure 0, which contradicts RVM, but I
dealt with this in my thesis when I said

"It is interesting to contrast Theorem 2, which implies that if Lebesgue
measure can be extended to {\it all} sets of reals then a strong Fubini 
theorem
holds, with Sierpinski's example which shows that if Lebesgue measure is
already defined for all {\it small} sets of reals then the strong Fubini
theorem {\it fails}. In other words, if Lebesgue measure is already defined on
too many sets, then it cannot be extended to all sets.

This is not as counterintuitive as it might appear. Vitali's famous
construction of a nonmeasurable set used the translation-invariance of 
Lebesgue
measure; any measure on {\it all} sets of reals will not be invariant, so
having too many sets Lebesgue measurable might well preclude a noninvariant
extension."

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By the way, the FOM posts Harvey refers to are his "positive postings"  17
and 23 from 4/26/98 and 11/6/98.  The "Strong Fubini theorems" I
refer to say that iterated integrals of nonnegative functions are equal
whenever they exist (the classical Fubini theorem says that for measurable
functions on product spaces, iterated integrals exist and are equal no matter
what order the integrations are done in; my strong versions apply to arbitrary
functions and I showed them to be consistent and independent and to follow 
from
RVM in my thesis, "Cardinal Conditions for Strong Fubini Theorems",
Transactions of the AMS October 1990).

-- Joe Shipman