Bounded operators and axiom of choice In the article below, it is shown that the proposition "Every linear operator defined on a whole Hilbert space is bounded" is consistent with the axioms of ZF + a weakened version of the axiom of choice (called DC). 
So, if I want to prove that an operator A defined on a Hilbert space H is bounded, is it enough to just check that the axiom of choice was not used to define it?
And a related question: To show that a set is measurable, is it enough to check that the definition of this set didn't use the axiom of choice? (As similarly, the statement "all subsets of R" are Lebesgue measurable is consitent with ZF without the axiom of choice.)
~ Link: http://www.ams.org/journals/bull/1973-79-06/S0002-9904-1973-13399-3/S0002-9904-1973-13399-3.pdf
 A: In the case of measurable sets, something like your
statement is true, but your particular criteria is not literally correct. It is an
over-simplification, since once one rises to a certain
level of complexity, the issue isn't whether the definition
of your set "uses the axiom of choice", but rather the
issue is a matter of the properties of the ambient set
theoretic universe in which you are defining your set.
The basic fact is that yes, sets of reals defined by
particularly simple definitions are indeed automatically
measurable. (One should assume at least a small amount of
choice, say the DC principle, in order to have a
satisfactory theory of Lebesgue measure.)
To illustrate the lowest level of this, you probably know that every Borel set is
Lebesgue measurable. This is an instance of the
definability claim, because the Borel sets are precisely
those sets that have complexity $\Delta^1_1$ in the descriptive set-theoretic hierarchy, which means
that they can be defined by a property involving
quantification over finite objects plus a single universal
quantifier over the reals, and equivalently by a definition
with a single existential quantifier over the reals.
More generally, the $\Sigma^1_1$ sets are the projections
of Borel sets, and these are also measurable. These are the
sets that can be defined with a single existential real
quantifier, followed by any number of quantifiers over a
fixed countable realm. If one has Martin's axiom plus
$\neg$CH, then this rises to the $\Sigma^1_2$ sets.
Above this, one has the very interesting phenomenon that
the assertions that classes of sets are Lebesgue measurable
begin to have large cardinal consistency strength. For
example, Solovay proved that the assertion that every
$\Sigma^1_3$ set is measurable--these are the sets
definable with the quantifier structure $\exists x\forall
y\exists z\ \varphi(\cdot,x,y,z)$, where $\varphi$ uses
only arithmetic quantifiers---is equiconsistent with the
existence of an inaccessible cardinal.
The existence of much stronger large cardinals have
outright consequences for the measurability of projective
sets. For example, the principle known as Projective
Determinacy, which is equiconsistent with and implied by
the existence of infinitely many Woodin cardinals, implies
that every projective set is determined. Thus, under PD,
any set of reals that can be defined by quantifying only
over the reals and over the natural numbers will
automatically be Lebesgue measurable.
Some of these issues were considered in other questions
here on mathoverflow, such as ZFC plus every analytical
set is
measurable?.
Finally, to show that particular criterion you mention is not correct, observe that by forcing we can make any particular set of reals be the reals that are coded into a block of the GCH pattern on the cardinals $\aleph_\alpha$. Thus, I can define a set of reals by saying "the set of reals whose binary expansion occurs as a part of the GCH pattern". This definition does not use the axiom of choice, but it can be used to define any given set, which may or may not be measurable. There are numerous other ways to see a similar effect.
A similar observation applies the bounded operator case---in principle any set can be defined by reference to the GCH pattern, and these definitions do not in principle refer to the axiom of choice.
A: Joel has explained very well the situation in ZF+DC and the effect of large cardinals.  Let me add that a consistency result due to Solovay comes very close to achieving what the question hoped for, but (because it's a consistency result) only in certain models of set theory.  Solovay's result says that the following theory is consistent (relative to ZFC plus an inaccessible cardinal): ZFC plus Lebesgue measurability of all sets of reals that are definable (in the language of set theory) with a countable sequence of ordinals as a parameter.  (Note that the permitted parameters include reals and, via coding, countable sequences of reals.) I believe that this sort of definability is extensive enough to cover anything likely to arise in analysis (when no set theorists are involved).  
I also believe that the analogous result holds, in the same model, for the part of the question about boundedness of Hilbert space operators, but I have not thought enough about this to make any guarantees.
A: if I want to prove that an operator A defined on a Hilbert space H is bounded, is it enough to just check that the axiom of choice was not used to define it? 
In a certain sense, the answer is "yes", taking into account the peculiarities that Joel notes.  But if you actually have explicitly written down the definition of your operator on an explicit Banach space: in every known case it will be easy to simply apply the closed graph theorem to conclude continuity.  Far easier than doing the meticulous checking that the Axiom of Choice has been avoided.
