= Solution
Use the riskless <financial asset> as <numéraire>, normalized to have initial value one and terminal value $R>0$. Write $S_0\in\mathbb R^s$ for the initial risky prices and $S_1$ for the finite, measurable terminal price <random vector>. Define the discounted gain vector $X=S_1/R-S_0$. A risky holding $h$ financed by the cash holding $-h\cdot S_0$ has zero initial cost and discounted terminal gain $h\cdot X$. An <arbitrage> is such a <portfolio> with $h\cdot X\geq0$ almost surely and $\mathbb P(h\cdot X>0)>0$. Allowing nonpositive initial cost is equivalent: invest any initial surplus in the riskless <financial asset> to obtain a zero-cost <arbitrage>.
An <equivalent martingale measure> is a <probability measure> $Q$ with exactly the same null events as the physical <probability measure> $P$, under which discounted asset prices are integrable and satisfy $\mathbb E_Q[S_1/R]=S_0$. Equivalently, $\mathbb E_QX=0$. Such a measure rules out <arbitrage>: every zero-cost <portfolio> has mean discounted gain zero, whereas a nonnegative gain positive with positive $P$-probability is also positive with positive $Q$-probability and has strictly positive $Q$-expectation.
For the converse, the <probability space> need not be finite, and the payoffs need not be integrable under $P$. Here is a direct <Gaussian-damped martingale density construction>. Set
$$
N=\{h\in\mathbb R^s:h\cdot X=0\text{ almost surely}\},\qquad L=N^\perp,\qquad F(h)=\mathbb E_P e^{-|X|^2-h\cdot X}\quad(h\in L).
$$
The set $N$ is a <linear subspace>. Choosing a finite <basis> of $N$ shows that $X\in L$ almost surely: intersect the probability-one events on which each basis vector is orthogonal to $X$. If $h\in L\setminus\{0\}$, then $h\cdot X$ is not almost surely zero. Absence of <arbitrage> implies both $P(h\cdot X<0)>0$ and $P(h\cdot X>0)>0$, by applying the definition to $h$ and $-h$.
All the analytical steps in minimizing $F$ can be checked directly. On $|h|\leq H$ the integrand is bounded by $e^{H^2/4}$, since $-|X|^2+H|X|\leq H^2/4$. Its first two directional <derivatives> are bounded uniformly in $X$ by constants depending only on $H$ and the direction, because the functions $r^j e^{-r^2+(H+1)r}$ are bounded for $j=0,1,2$. The mean-value bound therefore proves continuity of $F$, and the <Taylor theorem> remainder bound, integrated over $P$, proves
$$
D_vF(h)=-\mathbb E_P[(v\cdot X)e^{-|X|^2-h\cdot X}],\qquad v\in L.
$$
Thus differentiation under the integral does not require an unproved physical moment assumption.
To prove <coercivity>, suppose $|h_k|\to\infty$ in $L$. A subsequence of $h_k/|h_k|$ converges to a unit vector $v\in L$. This elementary finite-dimensional <compactness> follows by successively taking convergent subsequences of the bounded coordinates. Since $P(v\cdot X<0)>0$, some event $A=\{v\cdot X\leq-\varepsilon,\ |X|\leq K\}$ has positive <probability>: the negative-gain event is the countable union of such events with $\varepsilon=1/j$ and integer $K$. Eventually $(h_k/|h_k|)\cdot X\leq-\varepsilon/2$ on $A$, and hence
$$
F(h_k)\geq P(A)e^{-K^2}e^{|h_k|\varepsilon/2}\longrightarrow\infty.
$$
If <coercivity> failed, an unbounded sequence with bounded $F$ would have just such a contradictory subsequence. Consequently a minimizing sequence for $F$ is bounded and has a convergent subsequence; continuity makes its limit $h_*$ a minimizer. If $L=\{0\}$, simply take $h_*=0$. In every case the directional <derivatives> vanish at $h_*$. The vector $\mathbb E_P[Xe^{-|X|^2-h_*\cdot X}]$ lies in $L$ and is orthogonal to $L$, so it is zero. Therefore
$$
\boxed{\frac{dQ}{dP}=\frac{e^{-|X|^2-h_*\cdot X}}{\mathbb E_Pe^{-|X|^2-h_*\cdot X}}}
$$
is a positive mean-one density defining an <equivalent martingale measure>. Integrability of $X$ under $Q$ follows from the same Gaussian bound with $j=1$. This establishes \b[absence of arbitrage if and only if an equivalent martingale measure exists], without quoting a financial duality theorem or assuming a finite state space.
Now suppose the <probability space> is finite. Discard null states and enumerate its positive-probability atoms by $1,\ldots,m$. Every measurable terminal payoff is a vector in $\mathbb R^m$. Let $A$ be the $m\times(s+1)$ <matrix> whose columns are the discounted terminal payoffs of the traded <financial assets>, with first column $\mathbf1$, and let $p=(1,S_0)$ be their initial price vector. A <portfolio> with holding vector $z$ replicates the discounted payoff $Az$, and an <equivalent martingale measure> is exactly a vector $q$ satisfying
$$
q_j>0,\qquad A^Tq=p.
$$
The first pricing equation includes $\sum_jq_j=1$. A <complete market> means $\operatorname{im}A=\mathbb R^m$. If $q,q'$ both price the <financial assets>, then $(q-q')\cdot Az=0$ for every $z$. Completeness allows $Az$ to be each coordinate vector, so $q=q'$.
Conversely, if the <financial market> is incomplete, there is a nonzero vector $v$ with $A^Tv=0$. To see this without invoking a duality theorem, perform row elimination on the homogeneous system: its number of pivots is $\operatorname{rank}A<m$, leaving a free variable and hence a nonzero solution. The constant payoff column gives $\sum_jv_j=0$. Existence of an <equivalent martingale measure> was already proved, so choose one strictly positive vector $q$. For
$$
0<\varepsilon<\min_{v_j\ne0}\frac{q_j}{|v_j|},
$$
both $q+\varepsilon v$ and $q-\varepsilon v$ are distinct strictly positive probability vectors satisfying $A^T(q\pm\varepsilon v)=p$. Thus uniqueness fails. \b[On a finite arbitrage-free market, completeness is equivalent to uniqueness of the equivalent martingale measure.]
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