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Is there a Hahn Decomposition Theorem and Jordan Decomposition for Finite Signed Pre-measures?

Let $A$ be an algebra of sets. A set function $r \colon A \to [0,\infty)$ is called a pre-measure if $r(\emptyset)=0$ and $r(\bigcup_{i=1}^{\infty} B_i) = \sum _{i=1}^{\infty} r(B_i)$ whenever $B_i \in A,$ disjoint with $\bigcup_{i=1}^{\infty} B_i \in A.$

A set function $r \colon A \to (-\infty,\infty)$ is called a signed pre-measure if $s(\emptyset)=0$ and $s(\bigcup_{i=1}^{\infty} B_i) = \sum _{i=1}^{\infty} s(B_i)$ whenever $B_i \in A,$ disjoint with $\bigcup_{i=1}^{\infty} B_i \in A.$

Is it possible to have positive and negative parts and more importantly, given a signed pre-measure $s$, can we find two pre-measures $r$ and $t$ such that $$s =r-t$$

The answer to the above question seems negative based on Ramiro's answer. If we add another, sort of regulatory, assumption that

There exists a probability measure $p$ on $\sigma(A)$ such that $s \ll p $ on $A$ in the sence that for any $B \in A,$ $p(A)=0$ implies $s(A)=0.$

then would the answer still be negative?

Because in my particular problem, $s$ is defined as an iterated, not double, integral of a function which is measurable componentwise, yet not jointly. My goal is to extend $s$ to a signed finite measure on $\sigma(A).$ Traditional approaches that are based on taking infimums over countable covers seem to fail since infimum will most likely result in $-\infty .$ That is why I am interested in decomposing my signed finite pre-measure into two finite pre-measures and extending them by conventional means and considering their difference as a way of extending $s$ to a finite signed measure.

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  • $\begingroup$ you can construct the measure corresponding to $r$, decompose it and restrict the two parts to the original algebra of sets $\endgroup$
    – Federico
    May 14, 2021 at 9:57
  • $\begingroup$ I don't see a straightforward way of constructing a measure out of $r$ since the extension (that I know of) from pre-measure to measure is based on taking infimum over countable covers of a set. When the values are negative, infimum might not be the thing to consider. Could you please provide with some details on how to get a measure? $\endgroup$ May 14, 2021 at 14:30
  • $\begingroup$ $\sigma$-continuity is required in order to define variation and then positive and negative sets. The details are rather ling, but you can check Bichteler's integration: a functional approach where a full discussion of variation and lattices of measures is presented; the Aliprantis Hichthiker's guide is also a good reference as it also contains a treatment of charges of finite variation. $\endgroup$
    – Mittens
    May 20, 2021 at 4:47

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If I understood correctly, you are defining:

Let $\mathcal{A}$ be an algebra of sets. A set function $s \colon \mathcal{A} \to (-\infty,\infty)$ is called a signed pre-measure if $s(\emptyset)=0$ and $s(\bigcup_{i=1}^{\infty} B_i) = \sum _{i=1}^{\infty} s(B_i)$ whenever $B_i \in \mathcal{A}$, disjoint with $\bigcup_{i=1}^{\infty} B_i \in \mathcal{A}$.

This definition of signed pre-measure, although it looks like just as a generalisation of the pre-measure definition, allows several "pathological" cases. Moreover, a Hahn decompostion or a Jordan decomposition may not exist and it may not be possible to extend a signed pre-measure defined in $\mathcal{A}$ to $\sigma(\mathcal{A})$.

Here is a simple example.

Let $X=[0,1]$ and $\mathcal{A}=\{E\subseteq [0,1] : E \text{ or } E^c \text{ is finite} \}$. It is easy to check that $\mathcal{A}$ is in fact an algebra.

Let $s \colon \mathcal{A} \to (-\infty,\infty)$ be defined as $s(E) = \# E $ if $E$ finite and $s(E) = -\# E^c $ if $E^c$ is finite. (where $\#$ denotes the counting measure).

Now,

  1. Since it is not possible that $E$ and $E^c$ be both finite, we have that $s$ is well defined;

  2. $s(\emptyset) = \# \emptyset = 0$ (it is also true that $s(X)=0$);

  3. Suppose $E_i \in \mathcal{A}$, pairwise disjoint with $\bigcup_{i=1}^{\infty} E_i =E\in \mathcal{A}$. Then,

3.a If $E$ is finite, then there is at most a finite number of $i$ such that $E_i \ne \emptyset$.

3.b If $E^c$ is finite, then $E$ is uncountable. Since a countable union of finite set can not be uncountable, there is at least one $i_0$ such that $E_{i_0}^c$ is finite. Since the sequence $E_i$ is pairwise disjoint, there is at most one $i_0$ such that $E_{i_0}^c$ is finite. So, we conclude that there is one and only one $i_0$ such that $E_{i_0}^c$ is finite.

Since the sequence $E_i$ is pairwise disjoint, it follows that $\bigcup_{i \ne i_0} E_i \subseteq E_{i_0}^c$. So, $\bigcup_{i \ne i_0} E_i $ is finite. So, we have again that there is at most a finite number of $i$ such that $E_i \ne \emptyset$.

From 3.a and 3 .b, it is easy to prove that $s(\bigcup_{i=1}^{\infty} E_i) = \sum _{i=1}^{\infty} s(E_i)$. So $s$ is a signed pre-measure.

Now, note:

A. $s$ does not extend to $\sigma(\mathcal{A})$. In fact, such extention would have value $+\infty$ on countable infinite set and value $-\infty$ on set whose complement is countable infinite (see Remark below for details).

B. For the Hahn decomposition, we must find a positive set $P$ and a negative set $N$ such that $P \cap N =\emptyset$ and $P\cup N =[0,1]$. Remember that $N$ is a negative set if and only if, for any $C \subseteq N$, $s(C)\leq 0$. It is easy to see that there is no negative set for $s$, because for any single-point set $\{x\}$, $s(\{x\})=1$.

C. It is also possible to show that there is not a Jordan decomposition for $s$. In fact, we can prove a stronger result: There are no pre-measures $r$ and $t$ defined in $\mathcal{A}$ such that $s = r-t$.

Proof: Suppose there are pre-measures $r$ and $t$ defined in $\mathcal{A}$ such that $s = r-t$. Then, we have for all $E \in \mathcal{A}$, $$ s(E) = r(E) - t(E)$$ Since $r(E)\geq 0$ and $t(E) \geq 0$, we can easily see that $r(E) \geq s(E)$ and $t(E) \geq -s(E)$. In particular, we have

  1. $r(E) \geq \# E $ if $E$ is finite, and
  2. $t(E) \geq \# E^c $ if $E^c$ is finite.

Since $r$ and $t$ are pre-measures, they are monotone set functions, so, we have that $r([0,1]) = +\infty$ and $t([0,1]) = +\infty$. But then, we can not have $s([0,1]) = r([0,1]) - t([0,1]) $ because $+\infty - (+\infty)$ is not defined. So we can not have $s = r-t$. $\square$

Remark: Details of item A above.

In the example above $\mathcal{A}=\{E\subseteq [0,1] : E \text{ or } E^c \text{ is finite} \}$. So $\sigma(\mathcal{A}) =\{E\subseteq [0,1] : E \text{ or } E^c \text{ is countable} \}$. We have that $s$ does not extend to $\sigma(\mathcal{A})$.

In fact, any extention of $s$ to $\sigma(\mathcal{A})$ would be a signed measure such the $s(E)=+\infty$ if $E$ is countably infinite and $s(F)=-\infty$ if $F^c$ is countably infinite.

Take $E= \{ \frac{1}{n} : n\in \Bbb N, n>0\} $ and $F= [0,1] \setminus \Bbb Q$. Then $E \cap F= \emptyset$, $s(E)=+\infty$ and $s(F)=-\infty$. Then we should have $s(E \cup F)=+\infty+(-\infty)$ which is not defined.

Comment: Regarding the additional information added to the question.

Since in your particular problem, $s$ is defined as an iterated, not double, integral of a function $f$ which is measurable componentwise, yet not jointly and your goal is to extend $s$ to a signed finite measure on $\sigma(A)$, you could try to proceed as follows:

  1. Decompose $f$ in $f^+$ and $f^-$
  2. Define the pre-measure $r$ using $f^+$ and the pre-measure $t$ using $f^-$
  3. Extend $r$ and $t$ to be, respectively, measures $R$ and $T$ defined in $\sigma(A)$.
  4. If $R$ or $T$ is a finite measure, then you can define $S= R-T$, and $S$ will be a signed measure extending $s$.

Of course, without knowning exactly how you $s$ is defined, we can not be more specific. I hope this can help you.

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  • $\begingroup$ C. I do not see how one can decompose $s$ in this example, I do not necessarily ask for a Jordan-type decomposition, any decomposition is fine for me. Note that if we have two probability measures $r$ and $t.$ Letting $s = r-t,$ we obtain a signed measure, however, $r-t$ is not the Jordan decomposition of $s.$ For signed finite measures, we always have Jordan decomposition. Hence having Jordan-type decomposition is equivalent to having some decomposition. Yet for signed finite pre-measures, despite lacking Jordan-type decomposition we might still have some decomposition as $r-t$ with premeas. $\endgroup$ May 23, 2021 at 5:53
  • $\begingroup$ @vekinpirna , For item A, I added a Remark at the end my answer to present the details. For item C, I update it to include a proof that there are no pre-measures $r$ and $t$ defined in $\mathcal{A}$ such that $s = r-t$. Please, take a look and let me know if you have any further question. $\endgroup$
    – Ramiro
    May 23, 2021 at 12:52
  • $\begingroup$ @vekinpirna , What is the definition of "almost Radon-Nikodym derivative" you are using? If you mean that there is a probability measure $p$ defined on $\sigma(\mathcal{A})$ and a $\sigma(\mathcal{A})$-measurable function $f$ define on $X$, such that, for all $E \in \mathcal{A}$, $ s(E) =\int_E f dp$. Then $s$ can be easily decomposed by decomposing $f$ into $f^+$ and $f^-$. $\endgroup$
    – Ramiro
    May 23, 2021 at 13:08
  • $\begingroup$ My additional assumption is that there is a probability measure $p$ on $\sigma(A)$ such that for $B \in A,$ whenever $p(B)=0,$ we have $s(B)=0.$ So, if $s$ was a signed measure, $s$ would have a Radon Nikodym derivative with respect to $p.$ $\endgroup$ May 23, 2021 at 13:26
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    $\begingroup$ Thanks a lot for the discussion here, I will post the specific problem as a new question quite soon. $\endgroup$ May 24, 2021 at 0:55

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