= Solution
Normalize the riskless <financial asset> to $B_0=1$, $B_1=b>0$, and write $S_0,S_1$ for the $s$ risky asset prices. Their discounted gain <random vector> is $Y=S_1/b-S_0$. A zero-cost <portfolio> holding $h\in\mathbb R^s$ in the risky assets and $-h\cdot S_0$ in the riskless asset has terminal value $b\,h\cdot Y$. Thus an <arbitrage> means
$$
h\cdot Y\geq0\quad\text{almost surely},\qquad \mathbb P(h\cdot Y>0)>0.
$$
An <equivalent martingale measure> is a <probability measure> $Q$ with the same <null sets> as $P$, under which $Y$ is <integrable> and $\mathbb E_QY=0$. The discounted risky prices are then <martingales> over the single trading period.
If such $Q$ exists, a nonnegative terminal gain has <expectation> zero under $Q$. A nonnegative <random variable> with zero <expectation> vanishes <almost surely>: on $\{X\geq1/n\}$ its <expectation> is at least $\mathbb P(X\geq1/n)/n$, and the union of these events is $\{X>0\}$. Equivalence transfers this conclusion to $P$, excluding <arbitrage>.
For the converse we prove the needed <positive-density separation proof of the one-period asset-pricing theorem>. First make the gain <integrable> without changing its <null sets>. Define an <equivalent probability measure> $P_0$ by
$$
\frac{dP_0}{dP}=\frac{k}{1+\|Y\|},\qquad
k^{-1}=\mathbb E_P\frac1{1+\|Y\|}.
$$
Let $N=\{h:h\cdot Y=0\text{ almost surely}\}$ and let $L=N^\perp$ be its <orthogonal complement>. Taking a finite <basis> of $N$ shows $Y\in L$ <almost surely>. If $L=\{0\}$, the gain is zero and $P_0$ already gives the required measure. Otherwise consider the <positive-weight expectation cone>
$$
C=\{\mathbb E_{P_0}[ZY]: Z\text{ bounded and }Z>0\text{ almost surely}\}\subset L.
$$
Positive combinations of weights show that $C$ is a <convex cone>. It is open relative to $L$. Indeed, for any admissible weight $Z$, perturb it to
$$
Z_\varepsilon=Z\left(1+\frac{\varepsilon\cdot Y}{1+\|Y\|}\right),\qquad \varepsilon\in L,\quad \|\varepsilon\|<1.
$$
These weights remain bounded and strictly positive, while
$$
\mathbb E_{P_0}[Z_\varepsilon Y]=\mathbb E_{P_0}[ZY]+A_Z\varepsilon,
\qquad A_Z=\mathbb E_{P_0}\frac{ZYY^T}{1+\|Y\|}.
$$
For every nonzero $v\in L$, $v\cdot Y$ is nonzero on an event of positive <probability>; otherwise $v\in N\cap L$. Consequently
$$
v^TA_Zv=\mathbb E_{P_0}\frac{Z(v\cdot Y)^2}{1+\|Y\|}>0.
$$
The <matrix> is finite because $\|Y\|^2/(1+\|Y\|)\leq\|Y\|$. The <kernel> of $A_Z$ on $L$ is zero, so it defines an <invertible matrix> on the <finite-dimensional vector space> $L$. For any sufficiently small $\delta>0$, every $y$ with $\|y\|<\delta/\|A_Z^{-1}\|$ is the image of a vector in the radius-$\delta$ ball, since $\|A_Z^{-1}y\|<\delta$. Hence the perturbations give a <neighborhood> of zero, proving relative openness of $C$.
If $0\notin C$, the <hyperplane separation theorem> gives a nonzero $h\in L$ with $h\cdot c\geq0$ for every $c\in C$. The separating constant can be taken to be zero because $C$ is a cone. Put $A=\{h\cdot Y<0\}$ and use the strictly positive weights $Z=\mathbf1_A+\varepsilon$. Taking the <limit> as $\varepsilon\downarrow0$ yields $\mathbb E_{P_0}[(h\cdot Y)\mathbf1_A]\geq0$, forcing $P_0(A)=0$. But $h\notin N$, so $P_0(h\cdot Y>0)>0$. This is an <arbitrage>, a contradiction. Hence $0\in C$: some bounded strictly positive $Z$ satisfies $\mathbb E_{P_0}[ZY]=0$. Normalize it by $dQ/dP_0=Z/\mathbb E_{P_0}Z$. This gives an <equivalent martingale measure> with <integrable> gains, establishing the <fundamental theorem of asset pricing> here without any further existence theorem.
Now retain only the $m$ positive-probability states of the finite <probability space>. Let $V\subseteq\mathbb R^m$ be the <linear span> of the discounted terminal asset payoffs; it contains the constant vector $\mathbf1$. A <complete market> means $V=\mathbb R^m$. An <equivalent martingale measure> is a vector $q$ of strictly positive state probabilities whose pairing with each asset payoff gives its initial price. By linearity it prices every attainable payoff. If $V=\mathbb R^m$, the <indicator functions> of individual states are attainable, so these prices determine every $q_j$ uniquely.
If $V$ is proper, choose $0\ne v\in V^\perp$. Since $\mathbf1\in V$, $\sum_jv_j=0$. For sufficiently small $\varepsilon>0$, both $q+\varepsilon v$ and $q-\varepsilon v$ have strictly positive entries, sum to one, and give the same pairings with every asset payoff. They are distinct <equivalent martingale measures>. Therefore <market completeness> is equivalent to uniqueness of the <equivalent martingale measure>.
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