Choquet integral

Remarks

If $f$ takes the values $0 = a_0 < a_1 < \cdots < a_k$, then $(C)\!\int f \, d\nu = \sum_{i=1}^{k} (a_i - a_{i-1}) \, \nu(\set{f \ge a_i})$. If $\nu$ is a measure, the Choquet integral equals the Lebesgue integral, by the layer cake representation. Example: let $X = \set{1, 2}$ with all subsets, $\nu(\set{1}) = \nu(\set{2}) = 0$ and $\nu(X) = 1$. Then $\nu$ is monotone with $\nu(\emptyset) = 0$, and the indicators $1_{\set{1}}$ and $1_{\set{2}}$ have Choquet integral $0$. Their sum $1_X$ has Choquet integral $\int_0^1 \nu(X) \, dt = 1$, so the integral is not additive. Non-example: if $f$ is not measurable, some set $\set{f \ge t}$ lies outside $\mathcal{F}$, so $\nu$ is not defined on it and the formula makes no sense. Some authors use $\set{f > t}$; this changes the integrand at only countably many $t$, so the integral is the same.