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Suppose $Q_1,Q_2$ are equivalent martingale measures. Fix a finite maturity $T$ and an event $E\in {\mathcal{F}}_T$. Because the market is a complete market, replicate the claim $H=B_T{1}_E$, whose discounted payoff is bounded. For a self-financing strategy, the discounted value satisfiesUnder either pricing measure its conditional mean increment is zero, since $h_t^S$ is known at time $t$ and the discounted stock is a martingale. In the standard admissible discrete-time pricing class these finite-horizon values are integrable; a constant lower bound on discounted wealth also gives integrability successively from the conditional increment formula. Therefore replication and taking expectations giveThe replicating portfolio and its initial cost are the same for both measures, so $Q_1(E)=Q_2(E)$. Since $E$ was arbitrary, the measures agree on ${\mathcal{F}}_T$, for every finite $T$, and hence on the sigma-algebra generated by the market filtration. Thus $\boxed{\text{the equivalent martingale measure is unique}}$. This is the measure version of complete markets have unique state-price densities.

A one-period self-financing strategy is a deterministic holdings vector $h=(h^0,\dots ,h^d)$, with initial cost $h\cdot S_0$ and terminal wealth $h\cdot S_1$. In the zero-initial-cost convention, an arbitrage satisfiesThe convention allowing a nonpositive initial cost is equivalent here: invest any initial proceeds in the strictly positive numéraire to obtain a zero-cost portfolio with nonnegative terminal wealth.

Suppose a strictly positive state-price density $\rho$ satisfies the asset-pricing equations, with the indicated products integrable. By linearity,For an arbitrage, the right-hand side is zero while the left-hand side is strictly positive, since the integrand is nonnegative and is positive with positive probability. This contradiction proves absence of arbitrage.

An attainable European contingent claim $X$ is a terminal random payoff for which some holdings vector $h$ satisfies $h\cdot S_1=X$ almost surely; its replication cost is $h\cdot S_0$. To prove uniqueness in a complete market, suppose ${\rho }^{\prime }$ is another positive state-price density. For every event $A$, attainability gives a portfolio replicating $S_1^{(0)}{1}_A$. Its initial cost has both representations, soThe signed random variable here is integrable, because $\mathbb{E}[\rho S_1^{(0)}]=\mathbb{E}[{\rho }^{\prime }{S}_1^{(0)}]=S_0^{(0)}<\infty$. Take $A$ to be the event where it is positive and then where it is negative. Both its positive and negative parts have expectation zero, and hence vanish almost surely. Since $S_1^{(0)}>0$,This proves that complete markets have unique state-price densities without requiring an extra lower bound on the random terminal numéraire.

For the finite-state example, order the states from top to bottom in the diagram and write the density values as ${\rho }_1,{\rho }_2,{\rho }_3$. The two pricing equations areAdding gives ${\rho }_1+{\rho }_2+{\rho }_3=3$. Put $t={\rho }_3$; substitution then givesThese are all the strictly positive solutions, and each satisfies both equations. The parameter is a state-price-density value, not a probability.

For a concrete unattainable contingent claim, take the payoff vector $(0,1,0)$. A replicating portfolio $(a,b)$ would have to satisfy $3a+6b=0$ and $6a+3b=0$ in the first and third states. These imply $a=b=0$, contradicting the required payoff one in the middle state. Thus the market is an incomplete market, despite having strictly positive state-price densities.