Let $B$ be a Banach space. Let $\{Y\_{n}\}$ be a sequence of $B$ valued random variables. 


Assume


1. $P(\{Y\_{n}\} \mbox{is bounded}) = 1$,
2. fo every $\epsilon>0$, there exists a finite dimensional subspace $F$ such that $P(\limsup\_{n} q\_{F}(Y\_{n})\leq \epsilon) = 1$.


Then show that $P(\{Y\_{n}\} \mbox{ is relatively compact}) = 1$. Where $q\_{F}(x) = d(x,F)$ is the distance of $x$ from $F$.



Almost sure convergence Banach Space valued Random Variable
For $\omega\in\Omega$, define $S(\omega):=\(Y\_n(\omega),n\in\mathbb N\)$. We have to show that $P(\omega\mid S(\omega)\mbox{ is relatively compact})=1$. 


Considering $\varepsilon=1/j$, we can see that there is $\Omega'$ of probability $1$ such that for each $\omega\in \Omega'$, 


1. $\sup\_n\lVert Y\_n(\omega)\rVert<\infty$, and
2. for each integer $j$, $\limsup\_{n\to +\infty}d(Y\_n(\omega),F\_j)\leqslant \frac 1j$, where $F\_j$ corresponds to the subspace for $\varepsilon=\frac 1j$.


We say that a set $S\subset B$ is *flately concentrated* if for each $
\varepsilon>0$, we can find a finite dimensional subspace $F$ of $B$ such that for all $x\in S$, $d(x,F)<\varepsilon$. 


It's linked to relative compactness (using completeness) by the following:



> 
> A subset $S$ of the Banach space $B$ has a compact closure if and only $S$ is bounded and flately concentrated . 
> 
> 
> 


To conclude here, we have to check that $S(\omega)$ is flately concentrated for each $\omega\in\Omega'$. Argue by contradiction (the problem will come from *infinitely many* $n$). 


