# Hahn Decomposition Theorem and Jordan Decomposition for Finite Signed Pre-measures

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.

• you can construct the measure corresponding to $r$, decompose it and restrict the two parts to the original algebra of sets May 14, 2021 at 9:57
• 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? May 14, 2021 at 14:30
• $\sigma$-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. May 20, 2021 at 4:47

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.

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.

• 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. May 23, 2021 at 5:53
• @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. May 23, 2021 at 12:52
• @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 $f^-$. May 23, 2021 at 13:08
• 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 $p.$ May 23, 2021 at 13:26
• Thanks a lot for the discussion here, I will post the specific problem as a new question quite soon. May 24, 2021 at 0:55