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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

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Edited to make some TeX display right. –  Mark Meckes May 18 '10 at 15:55

2 Answers 2

up vote 2 down vote accepted

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.

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Thanks. I just started studying Banach lattices and lemma 2 did indeed looked fishy, but I was unable to see why. Now, excuse me, I am going to slap myself in the forehead. Repeatedly. –  G. Rodrigues May 18 '10 at 16:21

"Since $A$ is not order-bounded there is $u_n$ in $A$ such that $u_n>v_n$ for every $n$".

Really? Either your definition of "order boundedness" is different from mine, or we can only conclude that there is $u_n$ in $A$ such that it is false that $u_n\le v_n$ (which is a far weaker statement).

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