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Let $\mathcal{F}$ be a set algebra (or a Boolean algebra). Following Kalton, let me call a function $f\colon \mathcal{F}\to \mathbb R$ $\delta$-additive ($\delta \geqslant 0$), whenever $f(\varnothing) = 0$ and

$$| f(A) + f(B) - f(A\cup B) | \leqslant \delta$$

as long as $A\cap B=\varnothing$ for $A,B\in \mathcal{F}$. Surely 0-additive functions are nothing but finitely additive signed measures.

I am interested in the notion of a tensor product that would be analogous to a product measure but actually only in a very simplistic setting.

Let $X$ and $Y$ be finite sets and suppose that $f\colon \wp(X)\to \mathbb R, g\colon \wp(Y)\to\mathbb R$ are 1-additive functions. Is there a function $h\colon \wp(X\times Y)\to \mathbb{R}$ such that

  • $h$ is 1-additive,

  • $h(A\times B) = f(A)\cdot g(B)\quad (A\subset X, B\subset Y)$.

The problem is, I think, non-trivial as we need some sort of a canonical decomposition of any given set into a union of rectangles, which is highly non-unique. On the other hand, working only with singletons (trivial rectangles) is not good enough to retrieve the tensorial property of $h$.

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  • $\begingroup$ I don't know the subject, but from an algebraic point of view, the number of "canonical admissible decompositions" of a given set should depend on the symmetries thereof. I think one should study the group of permutations $\sigma$ of $1$-additive "building blocks" that commute to $f$, $g$ and $h$. $\endgroup$ Commented Dec 22, 2019 at 20:28
  • $\begingroup$ Hence the requirement is $\sigma(h(A\times B)=h(\sigma(A\times B))=h(\sigma(A)\times\sigma(B))=\sigma(f(A).g(B))=f(\sigma(A)).g(\sigma(B))$. $\endgroup$ Commented Dec 22, 2019 at 20:41
  • $\begingroup$ Have you tried with some simple cases like X,Y having two elements? Or maybe with arbitrary number of elements but every subset having measure 1? $\endgroup$ Commented Dec 23, 2019 at 0:37
  • $\begingroup$ I found something that can be useful. The space of such functions is a subset of $\mathbb{R}^{N}$ for some $N$, and I claim that it is compact and convex. Compactness come from repeated use of the inequality with singleton, that have assigned measure because they are rectangles. Furthermore, a convex combination preserve both the conditions. Thus, you can add the further condition on h having minimum modulo; such function will exist and will be unique provided that there exist a 'tensor' product. This allows to 'glue' solutions by uniqueness on intersection. $\endgroup$ Commented Dec 23, 2019 at 8:46
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    $\begingroup$ Maybe a reasonable question is: for what $\epsilon$ does there exist the product of two given $\delta$-additive measures? $\endgroup$ Commented Dec 23, 2019 at 12:50

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As you guessed in the comments, it does not exist. The idea is the following: you are searching for some values that have small distance from some given values (the one on rectangles). If for example a new value $v$ must be close to some given values $v_1,v_2$, then necessarily $v_1, v_2$ must be close.

Conditions of Delta additivity in each variable is not enough to guarantee that $v_1$ and $v_2$, will be always close. If you want to find a condition, I guess you should do something like you do in elimination theory with equations, but in this "distance" fashion.

An interesting question so would be, mimicking elimination style: if such inequalities are satisfied, does there exist a solution?

Without such conditions there is a counterexample. Take both sets to have 2 elements that we call $x,y$.

In the first one:

  • $x$ and $y$ have measure 0;

  • $\{x,y\}$ has measure 1.

In the second one:

  • $x$ and $y$ have measure $r$;

  • $\{x,y\}$ has measure $2r$.

Emptyset has zero measure in both. Note that singletons in the product has measure zero.

Now take the L-shaped set $L=\{(x,x), (x,y), (y,x) \}$ and suppose it has measure $A$. If we add the last brick to get the rectangle, we have $$ | A + 0- 2r| \le 1$$

If we take out the the brick $(y,x)$ the rectangle we are left with has projection $x$ on the first set, thus it has measure zero. On balance we get $$ |A -0-0| \le 1$$

In contradiction with the previous one for big $r$. Also, note that this yield that in general it does not exist a delta additive tensor product for any fixed $\delta$, and that even if one of them is a measure the tensor could not exist.

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  • $\begingroup$ @Tomek Kania: if you are interested in, I am finding sufficient (maybe also necessary, not sure about it), for the existence of tensor product. The idea basically is the following: you can split a delta additive function in a sum of a measure and a special linear combination of "characteristic" delta functions, which has measure delta on a prescribed subset and zero anywhere else. The tensor product is bilinear, so it is enough to study the tensor product of two characteristic functions (easy), and a measure with a char. I found out that msr * char of a singleton is a measure, and ... $\endgroup$ Commented Dec 23, 2019 at 13:32
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    $\begingroup$ msr * char of a non singleton subset exist only if the measure is (in modulo) <= 2 (disregarding delta!!) on any set. I am trying to find a formula that works in this case given the condition on the measure. $\endgroup$ Commented Dec 23, 2019 at 14:05
  • $\begingroup$ Dear Tomek, I haven't received the mail; actually I don't know which addres I have put on my profile. Could you send it to [email protected]? Thanks! $\endgroup$ Commented Dec 23, 2019 at 17:03
  • $\begingroup$ Thanks, I've sent you an email. $\endgroup$ Commented Dec 23, 2019 at 18:04

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