This is an edited post of a post I made on sci.math (e.g. to fit MO markup) with
an elementary question on vector measures. Since it is almost a week and I have
received no answers, I am trying here. Below, I will "prove" a theorem that is
false (because it has a simple counter example) but I cannot find where the flaw
in the proof is. This is an elementary question, so I am not sure if it fits MO
standards. If it does not, feel free to close it down -- I will leave that
judgement to the moderators. I am also certain that once someone points out the
hole in my argument I am sure to slap myself on the forehead, cry out what a
complete idiot I am and vow not to show my face in public for the next few
years. But at this point, this problem is driving me bonkers, and I much prefer
my sanity over my reputation (if I have any).


WARNING: long post ahead.


SOME BACKGROUND: All Banach spaces are over the real field. Let $\Omega$ be a
Boolean algebra. In a harmless abuse of notation, the top element of $\Omega$
will be denoted by $\Omega$. By a measure on $\Omega$ I mean a *finitely
additive* map $u$ on $\Omega$ with values in a Banach space. If the codomain $B$
of $u$ is the real field I will call it a scalar measure. Note in particular
that (scalar) measures do *not* take infinite values. Neither
$\sigma$-completeness nor $\sigma$-additivity will be used anywhere (I told you
it was an elementary question).


Recall that if $u$ is a measure, then its *variation* $|u|$ is the (possibly
infinite) quantity,


$$|u|: E\mapsto \sup \{\sum\_{F\in \mathcal{E}}\|u(F)\|\}$$


where the supremum is taken over the set of all finite partitions $\mathcal{E}$
of $E$. Assuming $|u|$ is finite for every $E$, then $|u|$ is a scalar positive
measure (and therefore monotone and bounded).


On the other hand, the *semivariation* $\|u\|$ of $u$ is defined to be the
(possibly infinite) quantity,


$$\|u\|: E\mapsto \sup\{|b^{\ast}u|(E)\}$$


where the supremum is taken over all $b^{\ast}$ in the unit ball of the dual
space $B^{\ast}$. To avoid any possible misunderstandings, $|b^{\ast}u|$ is the
variation of the scalar measure $E\mapsto b^{\ast}(u(E))$.


Assuming $u$ has finite (or bounded) semivariation, that is, $\|u\|(E)$ is
finite for every $E$, then the semivariation is positive, monotone and
subadditive. An application of Hahn-Banach yields that for every $E$:


(A) $\|u(E)\| \leq \|u\|(E)$


Less trivial is the fact that $u$ has finite semivariation iff it is bounded.
This is a consequence of the fundamental inequality for the semivariation. Since
I will not need this inequality, I will just direct you to proposition 11, pg. 4
of the Diestel-Uhl monograph Vector Measures. Another elementary fact is that if
the codomain $B$ is finite-dimensional then the variation and the semivariation
agree (note: this fails in every infinite-dimensional space by
Dvoretzky-Rogers).


EXAMPLE: Let $\Omega$ be a Boolean algebra with an infinite partition of unity
(note: infinite cardinality of $\Omega$ is enough to guarantee this). Consider
the map $\chi:\Omega\to \mathbf{L}^{\infty}(\Omega)$ given by $E \mapsto
\chi(E)$, where $\chi(E)$ is the characteristic function of $E$. Then $\chi$ is
finitely additive and bounded but it is easy to see that its variation is
unbounded.


If you are wondering what is $\mathbf{L}^{\infty}(\Omega)$ for a general Boolean
algebra, suffice to say that such a space can indeed be constructed and with all
the right properties, but to not tarry too long, just take $\Omega$ to be the
power set of $\mathbb{N}$ and replace $\mathbf{L}^{\infty}(\Omega)$ with the
Banach space of bounded real-valued functions on $\mathbb{N}$ with the supremum
norm.


Now, I am going to "prove" that every bounded measure $v$ has bounded variation
in obvious contradiction with the above example. This will be done by
constructing a control measure for $v$ in a very special way. Since what I need
is for someone to tell me where and why I have gone astray, I am going to detail
the argument, even to the point of pedantry.


ARGUMENT: Denote by $\mathbf{BA}(\Omega)$ the space of bounded scalar measures
on $\Omega$. One can introduce a partial order on $\mathbf{BA}(\Omega)$ by
taking the pointwise order:


$$u \leq v \mbox{ iff } u(E) \leq v(E) \mbox{ for all } E \mbox{ in }\Omega$$


It is easy to see that with the pointwise order $\mathbf{BA}(\Omega)$ is a
partially ordered linear space.


Note: For partially ordered linear spaces, Banach lattices, etc. I will take as
my reference chapter 5, volume 3 of Fremlin's 5-volume work on measure theory.
It is available online, so googling will easily get you to it.


We can also put a norm on $\mathbf{BA}(\Omega)$ by taking the total variation,
that is, $\|u\| = |u|(\Omega)$.


Note: the context should make clear when $\|,\|$ denotes the norm of an element
of a Banach space or the semivariation.


Lemma 1: The space $\mathbf{BA}(\Omega)$ with the pointwise order and total
variation norm is a Banach lattice.


Proof: This can also be found in the books. One proof proceeds by noting that
$\mathbf{BA}(\Omega)$ is the dual of $\mathbf{L}^{\infty}(\Omega)$, which is a
Banach lattice. A (sketch of a) more direct proof goes as follows. Completeness
of $\mathbf{BA}(\Omega)$ is straightforward because only finite additivity is
involved. By elementary results on partially ordered linear spaces, to prove the
existence of arbitrary binary suprema and infima it suffices to prove the
existence of the supremum $\sup\{u, 0\}$ for every $u$ (the *positive part*
$u^{+}$ of $u$). This is given by $u^{+}: E \mapsto \sup\{u(F): F\leq E\}$. It
can also be seen that the *absolute value* $|u|$ of u defined by $\sup\{u, -u\}$
is just the variation of u, so that first, there is no ambiguity in my notation,
and second, the Banach lattice condition


(B) if $|u| \leq |v|$ then $\|u\| \leq \|v\|$


is trivially satisfied. Q. E. D.


The cone of positive elements of $\mathbf{BA}(\Omega)$ will be denoted simply by
$P$. Note that if $u$ is positive then it coincides with its variation.


Next, a lemma relating boundedness in the norm with boundedness for the
pointwise order.


Lemma 2: Let $A$ be a subset of $P$. Then $A$ is order-bounded iff it is
norm-bounded.


Proof: The direct implication follows from the Banach lattice condition (B).
For the converse implication, we prove the contrapositive. So suppose $A$ is
not order-bounded. Pick a non-zero positive $v$ and consider the sequence of
positive measures $v\_n = (n v)/\|v\|$. Since $A$ is not order-bounded there is
$u\_n$ in $A$ such that $u\_n \geq v\_n$ for every $n$. By the Banach lattice
condition (B) it follows that $\|u\_n\| \geq \|v\_n\| = n$, so that $A$ is not
norm-bounded. Q. E. D.


The next two lemmas amount to a proof that norm-bounded subsets of the positive
cone have a supremum. The structure of the proof is fairly standard and is
patterned after proofs of similar facts in other contexts (e.g. the fact that a
category has all coproducts if it has finite coproducts and filtered colimits).


Lemma 3: Let $(u\_i)$ be a norm-bounded, monotone net of positive measures. Then
the pointwise limit


(C) $E \mapsto \lim\_i u\_i(E) = \sup\{u\_i(E)\}$


exists and defines a map $u$ that is bounded, finitely additive and the supremum
of $(u\_i)$.


Proof: Since each $u\_i$ is positive (and therefore monotone) and $(u\_i)$ is
norm-bounded, by a constant $C$ say, then, $u\_i(E) \leq u\_i(\Omega) \leq C$ and
the supremum $\sup \{u\_i(E)\}$ exists. Since the net is monotone, this
supremum is just $\lim\_i u\_i(E)$ from which it follows that $u(E) = \lim\_i
u\_i(E)$ is finitely additive. Since it is positive, it is monotone and therefore
bounded by $u(\Omega) = \lim\_i u\_i(\Omega) \leq C$. Since the order is the
pointwise order, $u$ is the supremum of $(u\_i)$. Q. E. D.


Lemma 4: If $A$ is a norm-bounded subset of $P$ then it has a supremum.


Proof: By lemma 2, $A$ is order-bounded, by $v$ say. Consider the net $J
\mapsto \sup J$ where $J$ runs over the filtered partial order of the finite
subsets of $A$. The existence of $\sup J$ for every finite $J$ is guaranteed by
lemma 1. Since $v$ bounds $A$, it follows that $v$ bounds $\sup J$ for every
$J$. Since the net is monotone and norm-bounded by $\|v\|$, lemma 3 gives us its
supremum. A simple check proves that this supremum is precisely the supremum of
$A$. Q. E. D.


Now, let $v:\Omega\to B$ be a bounded measure. Since it is bounded, it has
finite semivariation. This implies that the set of positive measures
$\{|b^{\ast}v|\}$, where $b^{\ast}$ ranges over the unit ball of $B^{\ast}$,
is norm-bounded. By lemma 4, it has a supremum $u$. By the very definition of
the order structure, we have for every $E$, $|b^{\ast}v|(E) \leq u(E)$ and thus
by definition of the semivariation:


$$\|v\|(E) \leq u(E)$$


But by inequality (A) we have,


$$\|v(E)\| \leq u(E)$$


and this implies that $v$ has bounded variation!!!!!! For if $\{E\_n\}$ is a
finite partition of $E$ then


$$\sum\_n \|v(E\_n)\| \leq \sum\_n u(E\_n) = u(E) \leq u(\Omega)$$


Can anyone help me out here and point out where and why is my argument screwed?


Regards and thanks in advance,
G. Rodrigues



Elementary vector measure question: what am I doing wrong?
Lemma 2 is false. Consider, e.g., a sequence of probability measures with mutually disjoint supports.


The problem with your argument is that just because $A$ is not ordered bounded you do not get elements of $A$ that are larger than a given measure. After all, $P$ is only partially ordered; not linearly ordered.


