The setting is the one period model of a financial market with $d+1$ assets, where the $0$-th asset is a riskless bond:
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. Let $\overline\pi \in \mathbb{R}_+^{d+1}$ be an initial price vector with $\pi^0 := 1$. Let $\overline S:=(S^0,S^1,\dots,S^d)$ be the final price vector, where $S^0 \equiv 1+r, r>-1$ and $S^i, i \ge 1$ are non-negative RVs. At inital time, a portfolio $\overline\xi \in \mathbb{R}^{d+1}$ can be chosen. $\overline\pi \cdot \overline\xi$ is the initial price of the porfolio, $\overline\xi \cdot \overline S$ is the final value of the portfolio, where $x \cdot y$ denotes the standard inner product.
Definition: $\overline\xi \in \mathbb{R}^{d+1}$ is an arbitrage opportunity, if
$\quad$1. $\overline\pi \cdot \overline\xi \le 0$,
$\quad$2. $\mathbb{P}(\overline\xi \cdot \overline S \ge 0 ) = 1$ and
$\quad$3. $\mathbb{P}(\overline\xi \cdot \overline S > 0 ) > 0$.
(a): There exists an arbitrage opportunity $\overline\xi$ such that $\overline\pi \cdot \overline\xi = 0$.
(b): There exists $\overline\eta \in \mathbb{R}^{d+1}$ such that $\overline\pi \cdot \overline\eta <0$ and $\overline\eta \cdot \overline S \ge 0 \ \ [\mathbb{P}]$.
So far, I was able to show that
(a) $\iff$ Existence of an arbitrage opportunity $\Leftarrow$ (b).
Problem. What can be said about the implication "(a) $\implies$ (b)"?
"(a) $\implies$ (b)" is false.
Proof. Suppose "(a) $\implies$ (b)", then "$\neg$(b) $\implies$ $\neg$(a)".
Suppose $\neg$(b). Then $$\forall \ \overline \eta\in \mathbb{R}^{d+1} \text{ holds } \quad \textbf{(I) } \overline\pi \cdot \overline\eta \ge 0 \quad \text{ or } \quad \textbf{(II) } \mathbb{P}(\overline\eta \cdot \overline S < 0)>0.$$
If (II) is satisfied for $\overline\eta$, by 2., $\overline\eta$ cannot be an arbitrage opportunity. By (I) and 1., for any arbitrage opportunity $\overline\xi$ must hold $\overline\pi \cdot \overline\xi = 0.$ This already a contradiction to $\neg$(a), which is $$\forall \text{ arbitrage opportunities } \overline\xi \text{ holds } \overline\pi \cdot \overline\xi <0.$$