= Solution
Suppose $A_0\in\operatorname{ri}K$. We first justify the finite-dimensional fact $\operatorname{ri}K\subseteq D$, 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 $m=\mathbb E[wA_1]\in L$. Relative openness lets us choose $\varepsilon>0$ such that $A_0-\varepsilon m\in\operatorname{ri}K\subseteq D$. There is therefore a bounded nonnegative measurable $f$ satisfying $A_0-\varepsilon m=\mathbb E[w fA_1]$. Set
$$
\boxed{\nu=w(f+\varepsilon)>0.}
$$
Then $\mathbb E[\nu A_1]=A_0$ and $\mathbb E\|\nu A_1\|\leq\|f\|_\infty+\varepsilon<\infty$. 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 $\operatorname{ri}K$: 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.
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