$L^1$ weak topology Let $T=[0,1]$, $$P=\{x=(x^1,x^2,…,x^n)\in\mathbb{R}^n \vert \; 
x^{i}\geq 0,i=1,2,…,n; x^1+x^2+…+x^n=1\}.$$ 
A $T$-strategy is a measurable function $y$ from $T$ to $P$. Hence $y$ is Lebesgue-integrable and we write $\int y$ for $(\int y^1(t)d\lambda,\dots,\int y^n(t)d\lambda)$, where the integration is on $T$. That is, a $T$-strategy $y\in L_1(T\times\{1,2,\dots,n\})$. Let $S$ denote the set of all $T$-strategies endowed with $L_1$ weak topology.  Claim: The set $S$ is a compact space.
Questions:


*

*Does $L_1$ weak topology mean that if a sequence $\{y_{k}\}\subset L_1(T\times\{1,2,\dots,n\})$ converges weakly to $ y\in L_1(T\times\{1,2,\dots,n\})$ if $\int y_{k}^{i}(t)v(t)d\lambda\to \int y^{i}(t)v(t)d\lambda$,$\forall v\in L_{\infty}(T)$, $\forall i=1,2,…,n$?

*How to show the set $S$ is compact?


1. Yes, weak convergence  means convergence of integrals against every $L^\infty$ function. 
2. First, the set is closed. It's easy to see that it's closed in norm (use the relation of norm convergence and a.e. convergence). And since it's also convex, it is weakly closed as well. 
It is awkward to work with the weak topology on $L^1$. Instead, use the fact that your functions are bounded by $1$. Therefore, they form a bounded set in $L^p$ for every $p$. For $1<p<\infty$, closed bounded sets in $L^p$ are weakly compact. Recall that the weak topology on $L^p$ is determined by integration against $L^q$ with $1/p+1/q=1$. Since $L^\infty$ is smaller than $L^q$ for finite $q$, it follows that the weak $L^1$ topology is weaker than the weak $L^p$ topology for $p>1$. Having fewer open sets implies having more compact sets. Thus, the set is  weakly compact in $L^1$. 
