= Solution
Work in discounted units with a riskless asset of value one, trivial initial information and finitely many risky assets. Let $Y=S_1-S_0\in L^2(P;\mathbb R^d)$ be their discounted gains. A terminal <attainable claim> has the form $v+\theta^TY$, whose initial cost is $v$. Their span
$$
\mathcal H=\operatorname{span}\{1,Y_1,\ldots,Y_d\}\subseteq L^2(P)
$$
is a finite-dimensional closed subspace of a <Hilbert space>. For $H\in L^2(P)$, <one-period least-squares hedging> is its <orthogonal projection> onto $\mathcal H$.
Let $m=\mathbb EY$, $\Sigma=\operatorname{Cov}(Y)$ and $b=\operatorname{Cov}(Y,H)$. Remove redundant risky directions so that $\Sigma$ is invertible. Under absence of <arbitrage>, a gain direction with zero <variance> is a constant and must be zero, so this reduction loses no genuine trading opportunity. For freely chosen capital and holdings, minimizing
$$
\mathbb E(H-v-\theta^TY)^2
$$
first gives $v=\mathbb EH-\theta^Tm$. The remaining centered error is
$$
\operatorname{Var}H-2\theta^Tb+\theta^T\Sigma\theta.
$$
Completing the square yields
$$
\boxed{\theta^*=\Sigma^{-1}b,\quad v^*=\mathbb EH-m^T\Sigma^{-1}b,\quad
\min\mathbb E(H-v-\theta^TY)^2=\operatorname{Var}H-b^T\Sigma^{-1}b.}
$$
The residual $L=H-v^*-(\theta^*)^TY$ satisfies $\mathbb EL=0$ and $\mathbb E[LY]=0$, the <least-squares normal equations>. Exact <claim replication> is possible precisely when this residual is zero. If capital is fixed at $v$, its normal equation instead gives the <fixed-capital quadratic hedge in a one-period market>
$$
\theta(v)=\bigl(\mathbb E[YY^T]\bigr)^{-1}\mathbb E[Y(H-v)].
$$
Thus fixed-capital and freely optimized hedging are distinct problems.
A <dominated martingale measure> is a <probability measure> $Q\ll P$ with integrable gains and $\mathbb E_QY=0$. Its density $Z$ obeys $Z\ge0$, $\mathbb EZ=1$, $\mathbb E[ZY]=0$. An <equivalent martingale measure> additionally has $Z>0$ almost surely. Every such measure prices an attainable discounted claim $v+\theta^TY$ at $v$, but different measures may give different <expectations> to an unattainable claim.
The <minimal martingale measure in a one-period market> is defined by preserving the mean-zero <martingale> directions orthogonal to the gain innovation $Y-m$. Among square-integrable densities it is
$$
\boxed{Z_*=1-m^T\Sigma^{-1}(Y-m).}
$$
Indeed $\mathbb EZ_*=1$, $\mathbb E[Z_*Y]=m-\Sigma\Sigma^{-1}m=0$, and for $\mathbb EL=0$, $\mathbb E[L(Y-m)]=0$ we have $\mathbb E[Z_*L]=0$. Conversely preservation of all these orthogonal directions puts $Z$ in $\operatorname{span}\{1,Y-m\}$; normalization and zero gain <expectations> then determine $Z_*$. Any other square-integrable signed pricing density differs from $Z_*$ by a vector in $\mathcal H^\perp$, so $Z_*$ also has the smallest $L^2$ norm. Its evaluation of a claim is
$$
\mathbb E[Z_*H]=\mathbb EH-m^T\Sigma^{-1}b=v^*.
$$
This is the capital of the optimal quadratic hedge, not automatically an arbitrage-free price. The formula may define only a <signed martingale measure>: $Z_*$ is a genuine dominated measure only if it is nonnegative, and equivalent only if strictly positive. For example, gains $(-1,1,2)$ with <probabilities> $(1/10,4/5,1/10)$ have $m=9/10$, $\Sigma=49/100$ and $Z_*(2)=-50/49$. Yet <probabilities> $(11/20,7/20,1/10)$ are all positive and have zero gain mean. Thus absence of arbitrage does not force a positive minimal density.
For the <market completeness> assertion, assume absence of arbitrage and hence existence of an <equivalent martingale measure> $Q_0$. In finite states this is the usual <fundamental theorem of asset pricing>; the one-period version also holds with finitely many assets on a general <probability> space. <Market completeness> means that every bounded claim is attainable, equivalently $\mathcal H=L^2(P)$ in this square-integrable setting. We now prove <completeness and uniqueness of dominated martingale measures>.
If the model is complete, replicate every indicator $\mathbf1_A$. Two <dominated martingale measures> both evaluate it at its unique <claim replication> cost, and therefore assign every event $A$ the same <probability>. They are identical.
Conversely let $k=\dim\mathcal H$ and suppose the model is incomplete. There exist $k+1$ disjoint events $A_j$ of positive <probability>. Otherwise the <probability> space would have at most $k$ atoms: any non-atomic positive event could be split to increase the number. Its whole payoff space would then have dimension at most $k$, forcing equality with $\mathcal H$ and <market completeness>. Choose a basis $h_1,\ldots,h_k$ of $\mathcal H$. Its members are integrable under $Q_0$, because they are combinations of the constant and the gains. The $k+1$ vectors
$$
\bigl(\mathbb E_{Q_0}[h_i\mathbf1_{A_j}]\bigr)_{i=1}^k\in\mathbb R^k
$$
are linearly dependent. Hence a nonzero bounded $g=\sum_j a_j\mathbf1_{A_j}$ satisfies $\mathbb E_{Q_0}[g h]=0$ for every $h\in\mathcal H$, in particular $h=1,Y_i$. For $0<\varepsilon<1/\|g\|_\infty$, define
$$
\frac{dQ_\pm}{dQ_0}=1\pm\varepsilon g.
$$
These strictly positive densities integrate to one, preserve every gain <expectation>, and give distinct <equivalent martingale measures>. They are also dominated by $P$. Thus uniqueness is impossible in an <incomplete market>, proving the equivalence under the stated no-arbitrage hypothesis.
That hypothesis is essential. On three positive-probability states, the gain $(0,1,2)$ has the unique <dominated martingale measure>, concentrated on the first state, but its attainable span has dimension only two. The market is incomplete and admits arbitrage. This explains why uniqueness among merely dominated <probabilities> must not be used without an equivalent pricing measure or a no-arbitrage assumption.
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