The following question refers to Ehud Lehrer's and Roee Teper's paper The Concave Integral over Large Spaces, which was published August 2008 in Fuzzy Sets and Systems 159(16):2130-2144.
The above paper defines a capacity as follows. (Definition 1 under section 2, The Concave Integral for Capacities, p. 4.)
Let $X$ be some set, $\mathcal{F}$ be a $\sigma$-algebra over $X$. Consider the extended set function (or simply, set function) $\nu:\mathcal{F}\rightarrow[0,\infty]$ such that $\nu(\emptyset)=0$. $\nu$ is monotone iff $\nu(A)\leq\nu(B)$ for all $A\subseteq B$ where $A,B\in\mathcal{F}$. [...] $\nu$ is a capacity iff it is monotone and $\nu(X) = 1$.
The above paper defines the Choquet integral as follows. (Section 3, The Concave Integral and the Choquet Integral, p. 6.)
Given a capacity $\nu$ and a non-negative measurable function $f$, the Choquet integral of $f$ w.r.t. $\nu$ over $A\in\mathcal{F}$ is defined by $$ \int_A^{\text{Cho}} f\ d\nu := \int_0^{\infty} \nu\big(\{s\in A:\ f(s)\geq t\}\big)\ dt, $$ where the latter integral is an extended Riemann integral.
The above paper makes the following claim. (Immediately following the definition of Choquet integral.)
By the definition of Riemann integral $$ \int_A^{\text{Cho}} fd\nu = \sup\bigg\{\sum_{i=1}^N\lambda_i\nu(A_i):\sum_{i=1}^N\lambda_i\mathbb{1}_{A_i}\leq f\mathbb{1}_A,\{A_i\}_{i=1}^N\subseteq\mathcal{F}\text{ is decreasing},\lambda_i\geq0, N\in\mathbb{N}\bigg\}, $$ where by decreasing we mean that $A_{i+1}\subseteq A_i$ for every $1\leq i\leq N-1$.
No further derivation or explanation of this claim is given in the paper. I'd appreciate it if someone derives the claim in greater detail.
Fix some non-empty set $X$, some $\sigma$-algebra $\mathcal{F}$ on $X$, some capacity $\nu$ on $\mathcal{F}$, some $\mathcal{F}$-measurable function $f:X\rightarrow[0,\infty]$, and some $A \in \mathcal{F}$.
Define \begin{align*} g &= t_{\in[0,\infty)}\mapsto\nu\big(\{s\in A\ |\!:\ f(s)\geq t\}\big),\\ \mathcal{S} &= \{(N,\lambda,B)\ :\!|\ N\in\{0,1,2,\dots\},\ \lambda\in[0,\infty)^N,\ B\in\mathcal{F}^N\},\\ \mathcal{T} &= \{(N,\lambda,B) \in \mathcal{S}\ |\!:\ B_1\supseteq B_2\supseteq\cdots\supseteq B_N,\ \sum_{i=1}^N\lambda_i\mathbb{1}_{B_i} \leq f\mathbb{1}_A\},\\ F &= (N,\lambda,B)_{\in\mathcal{S}}\mapsto\sum_{i=1}^N\lambda_i\nu(B_i). \end{align*}
Note that if $(N,\lambda,B)\in\mathcal{S}$ is such that $N=0$, then $\lambda$ and $B$ are empty sequences, and we have $\sum_{i=1}^N\lambda_i\mathbb{1}_{B_i}=0\mathbb{1}_X$, and $F(N,\lambda,B) = 0$.
With these notations we can restate what is to prove as follows: $$ \int^{\text{Cho}}_A\ f\ d\nu = \sup_{(N,\lambda,B)\in\mathcal{T}}F(N,\lambda,B) $$
We leave it to the reader to verify that $$ \sup_{(N,\lambda,B)\in\mathcal{T}}F(N,\lambda,B) = \sup_{(N,\lambda,B)\in\mathcal{T}'}F(N,\lambda,B), $$ where $\mathcal{T}'$ is like $\mathcal{T}$, except the $\lambda_i$'s are strictly positive, i.e. \begin{align*} \mathcal{S}' &= \{(N,\lambda,B)\ :\!|\ N\in\{0,1,2,\dots\}, \lambda\in(0,\infty)^N, B\in\mathcal{F}^N\},\\ \mathcal{T}' &= \{(N,\lambda,B)\in\mathcal{S}'\ |\!:\ B_1\supseteq B_2\supseteq \cdots\supseteq B_N,\ \sum_{i=1}^N\lambda_i\mathbb{1}_{B_i}\leq f\mathbb{1}_A\}. \end{align*}
(Hint: If, for some $k \in \{1,\dots,N\}$, $\lambda_k=0$, we can "drop" the summand $\lambda_k\mathbb{1}_{B_k}$ from $\sum_{i=1}^N\lambda_i\mathbb{1}_{B_i}$ without affecting the sum.)
Case 1: $f$ is bounded from above
We will firstly assume that $f$ is bounded from above. Set $M=\sup f\mathbb{1}_A$. Then $M\in[0,\infty)$. We leave it to the reader to verify that the conclusion holds if $M=0$, and we assume from now on that $M>0$.
Observe that, for every $t\in(M,\infty)$, we have $\{s\in A\ |\!:\ f(s)\geq t\}=\emptyset$. Therefore, by the definition of capacity, for every $t\in(M,\infty)$, $g(t)=0$. Hence, by the definition of Choquet integral, $\int_A^{\text{Cho}} f\ d\nu = \int_0^M g(t)\ dt$, where the integral on the right hand side is Riemann.
Since the Riemann integral of an integrable function equals the Darboux integral of this function, which, in turn, equals the lower Darboux sum of the function, we have $$ \int_0^M g(t)\ dt = \underline{\int_0^M} g(t)\ dt, $$ where the expression on the right hand side is the lower Darboux sum, namely \begin{align*} \underline{\int_0^M} g(t)\ dt = \sup\bigg\{&\sum_{i=1}^N(t_i-t_{i-1})m_i\ :\!|\\ &N\in\{1,2,\dots\};\\ &(t_0,t_1,\dots,t_N) \subseteq [0,M]^{N+1};\\ &0=t_0<t_1<\cdots<t_N=M;\\ &m_i=\inf_{s \in [t_{i-1},t_i]}g(s),\ \ \ i \in \{1, 2, \dots, N\}\bigg\}, \end{align*} or, equivalently, since $g$ is non-increasing, \begin{align*} \underline{\int_0^M} g(t)\ dt =\sup \bigg\{&\sum_{i=1}^N(t_i-t_{i-1})g(t_i) :\!|\\ &N\in\{1,2,\dots\};\\ &(t_0, t_1, \dots, t_N) \subseteq [0,M]^{N+1};\\ &0=t_0<t_1<\cdots<t_N=M\bigg\}. \end{align*} This can be restated as follows: \begin{align*} \underline{\int_0^M} g(t)\ dt &=\sup_{(N,t) \in \mathcal{P}} F(N,\lambda^t,B^t), \end{align*} where $$ \mathcal{P} = \{(N,t)|\!:N\in\{1,2,\dots\};t = (t_0,t_1,\dots,t_N)\in [0,M]^{N+1};0=t_0<t_1<\cdots<t_N=M\}, $$ and for every $(N,t) \in \mathcal{P}$ we define $\lambda^t = (\lambda^t_1,\dots,\lambda^t_N)\in(0,\infty)^N$ and $B^t=(B^t_1,\dots,B^t_N)\in\mathcal{F}^N$ as follows: \begin{align*} \lambda^t_i &= t_i-t_{i-1},\quad\forall i \in \{1,2,\dots,N\},\\ B^t_i &= \{s \in A\ |\!:\ f(s) \geq t_i\},\quad\forall i \in \{1,2,\dots,N\}. \end{align*}
With this notation we need to prove that $$ \sup_{(N,\lambda,B)\in\mathcal{T}'}F(N,\lambda,B) = \sup_{(N,\lambda,B)\in\{(N,\lambda^t,B^t)\ :\!|\ (N,t) \in \mathcal{P}\}} F(N,\lambda,B). $$
Equivalently, we can prove:
To see that $\{(N,\lambda^t,B^t)\ :\!|\ (N,t) \in \mathcal{P}\} \subseteq \mathcal{T}'$, let $(N,t) \in \mathcal{P}$. We leave it to the reader to verify that $(N,\lambda^t,B^t)\in\mathcal{S}'$ and that $B^t_1\supseteq B^t_2\supseteq \cdots\supseteq B^t_N$. We will show that $\sum_{i=1}^N\lambda^t_i\mathbb{1}_{B^t_i} \leq f\mathbb{1}_A$. Let $s \in X$. We need to show that $\sum_{i=1}^N\lambda^t_i\mathbb{1}_{B^t_i}(s) \leq f(s)\mathbb{1}_A(s)$. If $s \notin B^t_i$, $i \in \{1,2,\dots,N\}$, then $\sum_{i=1}^N\lambda^t_i\mathbb{1}_{B^t_i}(s) = 0$, and the inequality holds. So assume that $s \in B^t_i$ for some $i \in \{1,2,\dots,N\}$. Since $B^t_1 \supseteq B^t_2 \supseteq \cdots \supseteq B^t_N$, there exists a unique $k \in \{1,2,\dots,N\}$ such that $s \in B^t_1,\dots,B^t_k$ and $s\notin B^t_{k+1},\dots,B^t_N$. Therefore $$ \sum_{i=1}^N\lambda^t_i\mathbb{1}_{B^t_i}(s) = \sum_{i=1}^k\lambda^t_i\mathbb{1}_{B^t_i}(s) = \sum_{i=1}^k\lambda^t_i = \sum_{i=1}^k(t_i-t_{i-1}) = t_k-t_0 = t_k-0=t_k, $$ and, since $B^t_1,\dots,B^t_N\subseteq A$, we have $s \in A$, and therefore $$ f(s)\mathbb{1}_A(s) = f(s)\cdot1 = f(s). $$ Thus, we need to show that $f(s)\geq t_k$, i.e. that $s \in B^t_k$. This holds by the definition of $k$.
We show that, for every $(N,\lambda,B)\in\mathcal{T}'$, there is some $(N',\lambda',B')\in \{(N,\lambda^t,B^t)\ :\!|\ (N,t) \in \mathcal{P}\}$ such that $F(N',\lambda',B')\geq F(N,\lambda,B)$. In other words, we show that, for every $(N,\lambda,B)\in\mathcal{T}'$, there is some $(N',t')\in\mathcal{P}$ such that $F(N',\lambda^{t'},B^{t'})\geq F(N,\lambda,B)$.
Let $(N,\lambda,B)\in\mathcal{T}'$. If $B_i=\emptyset$ for every $i\in\{1,2,\dots,N\}$ (this includes the degenerate case $N=0$), then $F(N,\lambda,B) = 0$, and therefore any selection of $(N',t')\in\mathcal{P}$ will do (since $F\geq0$).
So assume that $N>0$ and that $B_i\neq\emptyset$ for some $k\in\{1,2,\dots,N\}$. Define $$ K = \max\{k\in\{1,2,\dots,N\} |\!:\ B_k\neq\emptyset\}. $$ Then for every $k \in \{1,2,\dots,K\}$, $B_k\neq\emptyset$, and for every $k \in \{K+1,\dots,N\}$, $B_k=\emptyset$.
For every $k \in \{0,1,\dots,K\}$, define $t_k = \sum_{i=1}^k\lambda_i$. (In particular, $t_0 = 0$.) Since $(N,\lambda,B)\in\mathcal{T}'$, we have, in particular, $\lambda\in(0,\infty)^N$, and therefore $t_0<t_1<\cdots<t_K$.
Firstly we show that $t_K \leq M$. Since $(N,\lambda,B)\in\mathcal{T}'$, we have $B_1\supseteq\cdots\supseteq B_K\neq\emptyset$. We may therefore choose some $s\in B_1,\dots,B_K$, and so $$ t_K = \sum_{i=1}^K\lambda_i = \sum_{i=1}^K\lambda_i\mathbb{1}_{B_i}(s). $$ Since $(N,\lambda,B)\in\mathcal{T}'$, we have $\sum_{i=1}^K\lambda_i\mathbb{1}_{B_i}(s)\leq\sum_{i=1}^N\lambda_i\mathbb{1}_{B_i}(s)\leq f(s)\mathbb{1}_A(s) \leq M$.
Now define \begin{align*} N' &= \begin{cases} K,&t_K=M\\ K+1,&\text{otherwise} \end{cases}, \end{align*} and define $t'=(t'_0,t'_1,\dots,t'_{N'})$ as follows: for every $i\in\{0,1,2,\dots,N'\}$ set \begin{align*} t'_i &= \begin{cases} t_i,&i\leq K\\ M-t_K,&\text{otherwise} \end{cases}. \end{align*} Then $(N',t')\in\mathcal{P}$, and it remains to show that $F(N',\lambda^{t'},B^{t'})\geq F(N,\lambda,B)$. Firstly observe that, for every $i \in \{1,2,\dots,K\}$,
Now, \begin{align*} F(N',\lambda^{t'},B^{t'}) &= \sum_{i=1}^{N'}\lambda^{t'}_i\nu\big(B^{t'}_i\big)\\ &\geq \sum_{i=1}^K\lambda^{t'}_i\nu\big(B^{t'}_i\big)\\ &= \sum_{i=1}^K\lambda^t_i\nu\big(B^t_i\big)\\ &\geq\sum_{i=1}^K\lambda_i\nu\big(B_i\big)\\ &=\sum_{i=1}^N\lambda_i\nu\big(B_i\big)\\ &= F(N,\lambda,B), \end{align*} where the penultimate equality is due to the fact that, for every $i \in \{K+1,\dots,N\}$, $B_i=\emptyset$, and therefore $\nu(B_i)=0$.
Case 2: $f$ is not bounded from above
We leave it to the reader to verify the following claim.
In light of this claim, it suffices that we define sets $\mathcal{T}_0,\mathcal{T}_1,\dots$, such that
For every $n \in \{0,1,\dots\}$ define \begin{align*} f_n &= \min(f,n),\\ \mathcal{T}_n &= \{(N,\lambda,B)\in\mathcal{S}\ |\!:\ B_1\supseteq B_2\supseteq\cdots\supseteq B_N,\ \sum_{i=1}^N\lambda_i\mathbb{1}_{B_i}\leq f_n\mathbb{1}_A\}. \end{align*}
We show that $\mathcal{T}_n\uparrow\mathcal{T}$. We leave it to the reader to show that $\mathcal{T}_0\subseteq\mathcal{T}_1\subseteq\cdots\subseteq \mathcal{T}$. This implies that $\cup_{n=0}^\infty\mathcal{T}_n\subseteq\mathcal{T}$. We will show that $\mathcal{T}\subseteq\cup_{n=0}^{\infty}\mathcal{T}_n$.
Let $(N,\lambda,B)\in\mathcal{T}$. Then \begin{align*} (N,\lambda,B)&\in\mathcal{S},\\ B_1\supseteq B_2&\supseteq\cdots\supseteq B_N,\\ \sum_{i=0}^N\lambda_i\mathbb{1}_{B_i}&\leq f\mathbb{1}_A. \end{align*}
Set $n' = \lceil\sum_{i=1}^N\lambda_i\rceil$. Then $\sum_{i=1}^N\lambda_i\mathbb{1}_{B_i}\leq\min\big(f\mathbb{1}_A,n'\big)=\min(f,n')\mathbb{1}_A=f_{n'}\mathbb{1}_A$. Then $(N,\lambda,B)\in\mathcal{T}_{n'}\subseteq\cup_{n=0}^\infty\mathcal{T}_n$.
We show that $\int^{\text{Cho}}_A\ f\ d\nu = \lim_{n\rightarrow\infty}\sup_{(N,\lambda,B)\in\mathcal{T}_n}F(N,\lambda,B)$. $$ \int^{\text{Cho}}_A\ f\ d\nu = \int_0^\infty g(t)\ dt = \lim_{s\rightarrow\infty}\int_0^s g(t)\ dt = \lim_{n\rightarrow\infty}\int_0^n g(t)\ dt = \lim_{n\rightarrow\infty}\int_0^{\infty}g(t)\mathbb{1}_{[0,n]}(t)\ dt. $$ It therefore suffices to show that, for every $n\in\{0,1,\dots\}$, $$ \int_0^{\infty}g(t)\mathbb{1}_{[0,n]}(t)\ dt = \sup_{(N,\lambda,B)\in\mathcal{T}_n}F(N,\lambda,B). $$
Let $n \in \{0,1,\dots\}$. Since $f_n$ is bounded from above, we know from case 1 that $$ \sup_{(N,\lambda,B)\in\mathcal{T}_n}F(N,\lambda,B) = \int^{\text{Cho}}_A f_n\ d\nu = \int_0^\infty\nu\big(\{s\in A\ |\!:\ f_n(s)\geq t\}\big)\ dt. $$ It therefore suffices to show that, for every $t\in[0,\infty)$, $$ \nu\big(\{s\in A\ |\!:\ f_n(s)\geq t\}\big) = g(t)\mathbb{1}_{[0,n]}(t) = \nu\big(\{s\in A\ |\!:\ f(s)\geq t\}\big)\mathbb{1}_{[0,n]}(t). $$
Let $t\in[0,\infty)$.
Addendum
It is possible to show that, if $f$ is bounded (in addition to being non-negative and $\mathcal{F}$-measurable), then $\int_A^{\text{Cho}} fd\nu$ can also be represented as follows: $$ \int_A^{\text{Cho}} fd\nu = \inf\bigg\{\sum_{i=1}^N\lambda_i\nu(A_i):\sum_{i=1}^N\lambda_i\mathbb{1}_{A_i}\geq f\mathbb{1}_A,\{A_i\}_{i=1}^N\subseteq\mathcal{F}\text{ is decreasing},\lambda_i\geq0, N\in\mathbb{N}\bigg\}. $$
However, in order that the proof proceed analogously to case #1 above, it is first required to observe that $$ \int^{\text{Cho}}_A\ f\ d\nu = \int_0^\infty \hat{g}(t)\ dt, $$ with $$ \hat{g} = t_{\in[0,\infty)}\mapsto\nu\big(\{s\in A\ |\!:\ f(s)>t\}\big)\ dt. $$ (This is true because, by Lebesgue's criterion for Riemann integrability, the set of discontinuities of $g$ has Lebesgue-measure zero, and at every point of continuity $t$ of $g$, we obtain from the Sandwich theorem, $g(t) = \hat{g}(t)$, since $\nu$ is monotone.)
Now replace all occurrences of $g$ in the proof above by $\hat{g}$, and also change the definition of $B^t$ to $$ B^t_i = \{s\in A\ |\!:\ f(s) > t_{i-1}\},\quad i\in\{1,2,\dots,N\}. $$