A little question about payoff functions being continuous. In the mixed extension of a finite game $G$, why are the payoff functions of players continuous?
Does it has something to do with being von Neumann and Morgenstern utility functions?
Is there other way to prove them continuous? For instance, by using directly the definition of continuity on these functions?
To be more explicit, suppose there are $n$ players, denoted $N=\{1,2,...,n\}$. 
Define $G=(A,u)$ to be a finite strategy game such that
(i) $A_i$ is the finite strategy set of player $i$, $A=A_1\times A_2\times\cdots\times A_n$ is then the set of strategy profiles;
(ii) Let $u_i$ be the payoff function of player $i$ and $u=u_1\times u_2\times\cdots\times u_n$.
Then the mixed extension of $G$ is defined to be the strategy game $E(G):=(S,p)$ such that
(i) For each $i\in N$, $S_i:=\Delta A_i$(the set of lotteries over $A_i$, or the simplex in $\mathbb{R}^{|A_i|}$) and $S:=S_1\times S_2\times\cdots\times S_n$. For each $s\in S$ and each $a\in A$, let $s(a):=s_1(a_1)s_2(a_2)\cdots s_n(a_n)$, where $a_i\in A_i$ and $a=(a_1,a_2,\cdots,a_n)$;
(ii) For each $s\in S$, $p_i:S\rightarrow\mathbb{R}$ is defined by $p_i(s)=\Sigma_{a\in A}u_i(a)s(a)$ and $p:=p_1\times p_2\times\cdots\times p_n$.
With definitions above, my question is that why $p_i$'s are continuous.
 A: An important way of thinking of $p_{i}(s)$ is that it's the expectation of $u_{i}$ with respect to the measure $s$ on $A$.
Since each $A_{i}$ is finite, the function $p_{i}$ is given by a finite sum, so it suffices to show that each summand is continuous. This follows from the fact that, for each fixed action profile $a$, the map $s \mapsto s(a)$ is continuous (roughly speaking, if $s$ is very close to $s'$, then by definition $s_{i}(a_{i})$ is very close to $s_{i}'(a_{i})$).
A: Okay, this is an awful lot of notation, and I'm also not sure I understand what all things mean --- for example, what is $s_1(a_1)$? is it the probability to choose action $a_1$? --- but here's some intuition. Let first $n=1$ and $A_1=\{A,B\}$. Then,
$p_1(s) = u(A)s(A)+u(B)s(B)=u(A)x+u(B)(1-x)$,
whereby $x=s(A)\in[0,1]$. Now, clearly, $p_1$ is a continuous function of $x$ and this is simply because it is the weighted sum of elementary continuous functions (namely, $x$ and $1-x$). The general case is completely analogous, I think. 
