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 $h$.



Do 1-additive maps admit tensor products?
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.


