= Solution
We prove exhaustiveness, including existence of the positive density when neither form of <arbitrage> occurs. Use the subspace $L$ from part (a)(ii), and consider
$$
\mathcal K=\{\mathbb E[\nu A_1]:\nu>0\text{ almost surely},\ \mathbb E[\nu\|A_1\|]<\infty\}\subseteq L.
$$
This is a nonempty <convex cone> closed under positive scaling: $\nu_0=(1+\|A_1\|)^{-1}$ supplies one element and guarantees weighted <integrability> even when $A_1$ itself has no first moment. Crucially, $\mathcal K$ is relatively open in $L$; it need not be closed.
To prove openness at a point $z=\mathbb E[\nu A_1]$, put $Z=\nu A_1$. The vectors $\mathbb E[hZ]$, for bounded real <measurable> $h$, span $L$. Indeed, a vector orthogonal to all of them would satisfy $\mathbb E[h\,u^\top Z]=0$ for every such $h$, forcing $u^\top Z=0$ <almost surely>. Since $\nu>0$, this says $u\in L^\perp$. Choose finitely many bounded $h_j$ for which $\mathbb E[h_jZ]$ form a basis of $L$. For coefficients $t_j$ sufficiently small,
$$
\nu\left(1+\sum_jt_jh_j\right)>0,\qquad z+\sum_jt_j\mathbb E[h_jZ]\in\mathcal K.
$$
This gives a relative neighbourhood of $z$. It also proves the openness step behind the <strictly positive barycentre cone lemma> without assuming a closed moment cone.
If $A_0\notin L$, the second <arbitrage> form was already constructed. If $A_0\in L\setminus\mathcal K$, the <separation of a point and an open convex set>, applied inside the finite-dimensional space $L$, gives a nonzero $x\in L$ with $x^\top A_0\leq\inf_{z\in\mathcal K}x^\top z$. Positive scaling in the cone forces
$$
x^\top A_0\leq0,\qquad x^\top z\geq0\quad(z\in\mathcal K).
$$
We claim $x^\top A_1\geq0$ <almost surely>. Otherwise $E=\{x^\top A_1<0\}$ has positive <probability>, and the admissible positive weights $\nu_m=\nu_0(1+m\mathbf1_E)$ give
$$
x^\top\mathbb E[\nu_m A_1]=\mathbb E[\nu_0x^\top A_1]+m\mathbb E[\nu_0x^\top A_1\mathbf1_E]\longrightarrow-\infty,
$$
contradicting the separation inequality. Since $x\in L$ is nonzero, $x^\top A_1$ cannot vanish <almost surely>; thus it is strictly positive on an <event> of positive <probability>. This is the first <arbitrage> form.
Consequently absence of both forms forces $A_0\in\mathcal K$, which supplies the required $\nu$. Conversely either form excludes every such density by part (a). We have proved \b[exactly one of arbitrage or a strictly positive pricing density holds]. If $L=\{0\}$, the same argument reduces to the elementary cases $A_0=0$ and $A_0\ne0$, so no nondegeneracy assumption has been hidden.
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