As Richard already said, the difference between the ordinary integral(s) and the Choquet integral is that the latter allows you to integrate against set functions which are not additive (i.e. not measures or probability measures). If you calculate the Choquet integral against an additive measure it will give you the same result as the ordinary integrals.

Let me give a concrete example which may be useful. Imagine I am trying to hire an assistant for my research work. I am judging the candidates giving them a score in each of the following three skills:

(e) Their ability to understand and write in English,

(t) Their ability to type papers using math software such as TeX,

(m) Their fluency with maths.

One thing I could do is give each skill a weight W so that they all sum up to one, and aggregate the scores S(e), S(t), S(m) using those weights. The result

S(e)W(e)+S(t)W(t)+S(m)W(m)

would be the integral of the function S against the unique discrete probability measure which gives each skill a probability equal to its weight.

However, that is problematic, because a probability measure assigns the set {e,t} a probability W(e)+W(t).If we look closely at what I need, a person who can both write in English and type maths can release me from typing papers, but a person with only one of those skills cannot type a paper. The combination of both skills is needed. Therefore the relevance of having both skills {e,t} is larger than the sum of each skill {e} and {t} separately.

It is in that sort of situation in which one might define a non-additive measure with something like

W(e)=0.1, W(t)=0.1 and W({e,t})=0.7

and use the Choquet integral.

I know of Choquet integrals in the context of fuzzy measure theory. It is a generalization of the Lebesgue integral for monotone functions that allows non-additive measures to be used. Unless you are working with imprecise probabilities or Choquet capacities (k-additive measures), you're probably best off using Lebesgue integration (I'm assuming that a Riemann-Stiltjes integral is not going to cut it in your case). In any case, the Choquet integral is defined as a special Lebesgue integral that reduces to the Lebesgue integral if your measure is additive. If you have a non-additive measure, once you construct the appropriate integrand, the computation proceeds as a normal Lebesgue integral.

For a good paper on integrals and the Choquet integral, see

D. Denneberg, Non-additive measure and integral, basic concepts and their role for applications. In M. Grabsch, T. Murofushi, M. Sugeno, Eds., Fuzzy Measures and Integrals. Theory and Applications, Studies in Fuzziness and Soft Computing 40,

Physica-Verlag, Berlin, 2000, 42-69.

Dear Kerboui

Enclosed an article that can help you:

Choquet Integral and Its Applications: A Survey

Zengwu Wang and Jia-an Yan

Academy of Mathematics and Systems Science, CAS, China

August 2006

Regards

As Richard already said, the difference between the ordinary integral(s) and the Choquet integral is that the latter allows you to integrate against set functions which are not additive (i.e. not measures or probability measures). If you calculate the Choquet integral against an additive measure it will give you the same result as the ordinary integrals.

Let me give a concrete example which may be useful. Imagine I am trying to hire an assistant for my research work. I am judging the candidates giving them a score in each of the following three skills:

(e) Their ability to understand and write in English,

(t) Their ability to type papers using math software such as TeX,

(m) Their fluency with maths.

One thing I could do is give each skill a weight W so that they all sum up to one, and aggregate the scores S(e), S(t), S(m) using those weights. The result

S(e)W(e)+S(t)W(t)+S(m)W(m)

would be the integral of the function S against the unique discrete probability measure which gives each skill a probability equal to its weight.

However, that is problematic, because a probability measure assigns the set {e,t} a probability W(e)+W(t).If we look closely at what I need, a person who can both write in English and type maths can release me from typing papers, but a person with only one of those skills cannot type a paper. The combination of both skills is needed. Therefore the relevance of having both skills {e,t} is larger than the sum of each skill {e} and {t} separately.

It is in that sort of situation in which one might define a non-additive measure with something like

W(e)=0.1, W(t)=0.1 and W({e,t})=0.7

and use the Choquet integral.