Let $\mathcal F$ be a convex set of probability measures or distribution functions and $F, G$ be two elements in $\mathcal F$. Let $T$ be a functional on $\mathcal F$ defined as follows. Note that $h$ is symmetric. $$T(F):= \int \cdots \int h(x_1, \dots, x_c) dF(x_1) \cdots dF(x_n).$$ Then how to derive the following result, please? Thank you!
$$T[F+\lambda(G-F)] = \sum_{j=0}^c {c \choose j} \lambda^{c-j} \int \cdots \int h(x_1, \dots, x_c) \prod_{i=1}^j dF(x_i) \prod_{i=j+1}^c d[G(x_i) - F(x_i)].$$
Moreover, what is the convention for the following formula, please? Thank you!
$$\prod_{i=1}^0dF(x_i) := ?$$
When it comes down to it, this is really just a multiplication problem. First, \begin{equation*} \prod_{i=1}^{n}dF(x_i)=dF(x_1)\dots dF(x_n) \end{equation*} Now, \begin{align*} T(F+\lambda(G-F))&=\int\!\dots\!\int h(x_1,\dots,x_c)\prod_{i=1}^{c}d(F(x_i)+\lambda(G(x_i)-F(x_i)))\\ &=\int\!\dots\!\int h(x_1,\dots,x_c)\prod_{i=1}^{c}(dF(x_i)+\lambda(dG(x_i)-dF(x_i))) \end{align*} Now you just want to rewrite the last part. To make typing easier, I'm going to use $f_i=dF(x_i)$, and $g_i=dG(x_i)$. So, \begin{align*} \prod_{i=0}^{c}(f_i+\lambda(g_i-f_i))&=(f_1+\lambda(g_1-f_1))\dots(f_c+\lambda(g_c-f_c))\\ &=\sum_{j=0}^{c}\binom{c}{j}f_1\dots f_j (\lambda g_{j+1}-\lambda f_{j+1})\dots(\lambda g_{c}-\lambda f_{c})\\ &=\sum_{j=0}^{c}\binom{c}{j}f_1\dots f_j \lambda^{c-j}(g_{j+1}-f_{j+1})\dots(g_{c}-f_{c}) \end{align*} since you're "choosing" how many of the "first" ($a$ in $(a+b)$) elements you're multiplying and how many of the "second" ($b$ in $(a+b)$), but each term in the summation must be the product of $c$ terms in total. The last line is what you're looking for. Think cubic or quartic if that's easier.
P.S. I suspect that I haven't actually done anything!
Sorry, I misread the last part of your question: \begin{equation*} \prod_{i=1}^{0}dF(x_i)=1, \end{equation*} for all intents and purposes. We just skip it--it means only the $\lambda(g_i-f_i)$ terms are being multiplied. It's the same with \begin{equation*} \prod_{i=c+1}^{c}(dG(x_i)-dF(x_i))=1 \end{equation*}