= Solution
We prove the contrapositive. Let
$$
N=\{h\in\mathbb R^n:h\cdot X=0\text{ almost surely}\}.
$$
The function $F$ is constant along $N$, so minimize it on $N^\perp$. Suppose there is no $H$ with $H\cdot X\geq0$ almost surely and strict inequality with positive probability. If a sequence $h_k\in N^\perp$ satisfies $\|h_k\|\to\infty$, pass to a subsequence with
$$
\frac{h_k}{\|h_k\|}\longrightarrow H\in N^\perp,
\qquad\|H\|=1.
$$
Because $H\notin N$ and there is no arbitrage direction, $\mathbb P(H\cdot X<0)>0$. On that event, $e^{-h_k\cdot X}\zeta\to\infty$, and <Fatou lemma> gives $\liminf_kF(h_k)=\infty$. Hence every finite sublevel set of $F$ in $N^\perp$ is bounded. It is also closed, so $F$ attains its infimum there and has a bounded minimizing sequence. This contradicts the assumption. Therefore there is a unit vector $H$ satisfying
$$
\boxed{H\cdot X\geq0\text{ almost surely},
\qquad\mathbb P(H\cdot X>0)>0.}
$$
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