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I was trying to generalize some results from measure theory to functions that are "almost" measures but not additive. Then, I thought it could be interesting to do this in a general fashion, by finding a general way to convert a non-additive set function to an additive one. Here is the general idea.

Suppose $f$ is a a set function. For concreteness, suppose $f$ is defined on the Borel subsets of the real interval $[0,1]$ and satisfies the following properties:

  • Continuity: $f(A)=0$ whenever $A$ has Lebesgue measure $0$ (in particular, $f(\emptyset)=0$).
  • Monotonicity: $f(A)\leq f(B)$ whenever $A\subseteq B$.
  • Non-triviality: $f([0,1])=1$.

I would like to define a function $g$ with the following properties:

  • None: $g(A)=0$ whenever $f(A)=0$
  • Whole: $g(A)=1$ whenever $f(A)=1$
  • Middle:If $C\supseteq B \supseteq A$ and $f(C\setminus B)=f(B\setminus A)$, then $g(B)=[g(C)+g(A)]/2$,

The intuition is that, if $B$ is exactly between $A$ and $C$ (according to the function $f$), then its measure is exactly between their measures (according to $g$).


  • What conditions on $f$ guarantee that such a function $g$ exists?
  • If it exists, is it indeed additive?


Let $f(A)=\text{length}(A)^2$. It is continuous, monotone and non-trivial, but it is not additive. Define $g$ by: $g(A)=\text{length}(A)$. It satisfies the None, Whole and Middle requirements, and it is indeed additive.


Let: $$ a(A) = \text{length}(A \cap [0,1/2]) $$ $$ b(A) = \text{length}(A \cap [1/2,1]) $$ $$ f(A) = a(A) + 2 b(A)^2 $$ So $f$ is continuous, monotone and non-trivial. Suppose that a function $g$ exists. Then by the Middle property: $$ g([0,1/2])=g([1/2,1])=0.5 $$ and: $$ g([0,1/4])=g([1/4,1/2])=g([1/2,3/4])=g([3/4,1])=0.25 $$

If $g$ is additive, then: $$ g([0,1/4]\cup[1/2,3/4])=0.5 $$

Again by the Middle property: $$ g([0,3/16])=g([3/16,1/4]\cup[1/2,3/4])=0.25 $$

But this does not make sense because $g([0,1/4])=g([1/2,3/4])=0.25$.

My conjecture is that an additive $g$ exists only if $f$ is an increasing monotone transformation of an additive function.

  • $\begingroup$ I do not know the answer to you question, but your positive example is an example of a distortion of a measure and treated in Example 2.1 in Denneberg's book "Non-additive measure and integral" $\endgroup$ – Dirk May 6 '19 at 15:11

take a look at https://core.ac.uk/download/pdf/81942549.pdf

Morris, Robert J. T., Optimal constrained selection of a measurable subset, J. Math. Anal. Appl. 70, 546-562 (1979). ZBL0417.49032.

non-additive set functions, how to differentiate them, "convex" set functions, etc

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  • $\begingroup$ Judging by your username, you are probably the author of the article that you link to; it would have been nice if you had mentioned that. Also, link-only answers are discouraged because: 1) links tend to rot (maybe not the case here); 2) their content is not indexable by search engines. $\endgroup$ – Alex M. May 6 '19 at 15:13

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