Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 41 / 1 / b

Suppose $A_0\in \mathrm{ri}K$. We first justify the finite-dimensional fact $\mathrm{ri}K\subseteq D$, since $D$ itself need not be closed. Around any relative-interior point $z$, choose a small simplex in $L$ whose vertices lie in $K$ and which contains $z$ strictly inside. Approximate its vertices by points of $D$, possible because $\overline{D}=K$. For close enough approximations, the barycentric coordinates of $z$ remain strictly positive; hence $z$ is a convex combination of points of $D$ and belongs to $D$. If $L=\{0\}$, the assertion is immediate.

Let $m=\mathbb{E}[w{A}_1]\in L$. Relative openness lets us choose $\epsilon >0$ such that $A_0-\epsilon m\in \mathrm{ri}K\subseteq D$. There is therefore a bounded nonnegative measurable $f$ satisfying $A_0-\epsilon m=\mathbb{E}[wf{A}_1]$. SetThen $\mathbb{E}[\nu A_1]=A_0$ and $\mathbb{E}\Vert \nu A_1\Vert \le \Vert f{\Vert }_{\infty }+\epsilon <\infty$. This proves existence with strict positivity even when the unweighted asset vector is not integrable.

It also proves the strictly positive barycentre cone lemma. Every strictly positive weighted barycentre must lie in $\mathrm{ri}K$: otherwise the supporting vector constructed for the relative boundary would have zero value at that barycentre while its strictly positive weighted payoff had positive expectation. Thus the relative interior is exactly the set of such barycentres.