[proofplan] We prove uniqueness by comparing two finitely-supported laws with the same expectation. Suppose $\nu = \sum_{j=1}^m q_j \delta_{g_j}$ is a second law on distinct atoms with $\sum p_i h_i = \sum q_j g_j$ in $T((V))$. Pooling the atoms $\{h_i\} \cup \{g_j\}$ into a single indexed family of distinct signatures and rewriting both expectations against this pool yields a single linear relation among distinct signatures with real coefficients. The [linear independence of distinct signatures](/theorems/???) forces every coefficient to vanish, which translates back into $\mu = \nu$ as measures. [/proofplan]

[step:Set up a competing law and reduce to a single linear relation among distinct signatures]Suppose $G$ is another $\mathcal{S}_p$-valued random variable on $(\Omega, \mathcal{F}, \mathbb{P})$ (or any probability space — we are arguing about laws, not realisations) with finitely-supported law \begin{align*} \nu : \mathcal{B}(\mathcal{S}_p) &\to [0,1], \\ \nu &= \sum_{j=1}^m q_j \delta_{g_j}, \end{align*} where $g_1, \ldots, g_m \in \mathcal{S}_p$ are distinct, $q_j > 0$ for all $j$, and $\sum_{j=1}^m q_j = 1$. Assume $\mathbb{E}[H] = \mathbb{E}[G]$ in $T((V))$, i.e. \begin{align*} \sum_{i=1}^n p_i h_i = \sum_{j=1}^m q_j g_j. \end{align*} Our goal is to deduce that the multisets $\{(h_i, p_i)\}$ and $\{(g_j, q_j)\}$ coincide, which is exactly $\mu = \nu$. Form the union of atoms as a set: \begin{align*} S := \{h_1, \ldots, h_n\} \cup \{g_1, \ldots, g_m\} \subset \mathcal{S}_p. \end{align*} Enumerate $S = \{s_1, \ldots, s_N\}$ with the $s_k$ pairwise distinct. Define coefficients \begin{align*} \alpha_k &:= \begin{cases} p_i & \text{if } s_k = h_i \text{ for some } i, \\ 0 & \text{if } s_k \notin \{h_1, \ldots, h_n\}, \end{cases} \\ \beta_k &:= \begin{cases} q_j & \text{if } s_k = g_j \text{ for some } j, \\ 0 & \text{if } s_k \notin \{g_1, \ldots, g_m\}. \end{cases} \end{align*} The cases are mutually exclusive because the $h_i$ are distinct (so each $s_k = h_i$ for at most one $i$) and similarly for the $g_j$. Then \begin{align*} \sum_{k=1}^N \alpha_k s_k = \sum_{i=1}^n p_i h_i = \sum_{j=1}^m q_j g_j = \sum_{k=1}^N \beta_k s_k, \end{align*} so subtracting yields \begin{align*} \sum_{k=1}^N (\alpha_k - \beta_k) s_k = 0 \quad \text{in } T((V)). \end{align*}[/step]

[guided]The crux of the argument is that two formal expectations $\sum p_i h_i$ and $\sum q_j g_j$ may use different atoms, so we cannot equate coefficients term by term unless we first put both sums over a common set of atoms. The cleanest way is to take the union $S$ of all atoms appearing in either law, list it as $\{s_1, \ldots, s_N\}$, and re-express each expectation as a linear combination indexed by $S$ — putting a zero coefficient wherever an atom does not appear in that particular law. Concretely, we define $\alpha_k = p_i$ if $s_k$ happens to equal some $h_i$ (and $0$ otherwise) and $\beta_k = q_j$ if $s_k$ equals some $g_j$ (and $0$ otherwise). The case split is well-defined: distinctness of $\{h_1, \ldots, h_n\}$ guarantees $s_k$ matches at most one $h_i$, and similarly for the $g_j$. With these coefficients, \begin{align*} \sum_{k=1}^N \alpha_k s_k = \sum_{i=1}^n p_i h_i, \qquad \sum_{k=1}^N \beta_k s_k = \sum_{j=1}^m q_j g_j, \end{align*} because the $\alpha_k = 0$ summands and $\beta_k = 0$ summands contribute nothing. The hypothesis $\mathbb{E}[H] = \mathbb{E}[G]$ now reads \begin{align*} \sum_{k=1}^N (\alpha_k - \beta_k) s_k = 0 \end{align*} in $T((V))$. We have a single linear relation among the distinct signatures $s_1, \ldots, s_N$ — exactly the form to which a linear-independence theorem applies.[/guided]

[step:Apply linear independence of distinct signatures to force all coefficients to vanish]The signatures $s_1, \ldots, s_N \in \mathcal{S}_p$ are pairwise distinct by construction. By the [linear independence of distinct signatures](/theorems/???), the only real linear combination of distinct signatures that equals $0$ in $T((V))$ is the elementary one. Applied to the relation \begin{align*} \sum_{k=1}^N (\alpha_k - \beta_k) s_k = 0, \end{align*} this forces $\alpha_k = \beta_k$ for every $k \in \{1, \ldots, N\}$.[/step]

[guided]The hypothesis we need is that $s_1, \ldots, s_N$ are distinct elements of $\mathcal{S}_p$. We verify this directly: $S$ was defined as the union $\{h_i\} \cup \{g_j\}$ of two finite subsets of $\mathcal{S}_p$, listed without repetition as $\{s_1, \ldots, s_N\}$, so the $s_k$ are pairwise distinct by construction. The linear-independence theorem says: distinct elements of $\mathcal{S}_p$ are linearly independent in $T((V))$ over $\mathbb{R}$. Plugging this into our relation \begin{align*} \sum_{k=1}^N (\alpha_k - \beta_k) s_k = 0, \end{align*} the only solution is $\alpha_k - \beta_k = 0$ for every $k$, i.e. $\alpha_k = \beta_k$ for all $k = 1, \ldots, N$. It is essential that we built the case split so that $\alpha_k$ and $\beta_k$ are real numbers (not measures or formal symbols) — only then does the conclusion of the linear-independence theorem give us a usable scalar equality.[/guided]

[step:Translate the equality of coefficients back into equality of measures]We show $\mu = \nu$ as Borel measures on $\mathcal{S}_p$. Since both measures are concentrated on the finite set $S = \{s_1, \ldots, s_N\}$, it suffices to show $\mu(\{s_k\}) = \nu(\{s_k\})$ for every $k$. Fix $k \in \{1, \ldots, N\}$. From the construction of $\alpha_k$, \begin{align*} \mu(\{s_k\}) = \sum_{i=1}^n p_i \, \delta_{h_i}(\{s_k\}) = \begin{cases} p_i & \text{if } s_k = h_i \text{ for some } i, \\ 0 & \text{otherwise} \end{cases} = \alpha_k, \end{align*} where again the case split is unambiguous because the $h_i$ are distinct. By the same argument, $\nu(\{s_k\}) = \beta_k$. The previous step gave $\alpha_k = \beta_k$, hence $\mu(\{s_k\}) = \nu(\{s_k\})$. For $s \in \mathcal{S}_p \setminus S$, we have $\mu(\{s\}) = 0 = \nu(\{s\})$ since $s$ is not among the $h_i$ or $g_j$. Both measures are finitely supported on $S$, so by countable additivity they agree on every Borel set: for $B \in \mathcal{B}(\mathcal{S}_p)$, \begin{align*} \mu(B) = \sum_{s_k \in B} \mu(\{s_k\}) = \sum_{s_k \in B} \nu(\{s_k\}) = \nu(B). \end{align*} Therefore $\mu = \nu$, which proves that any other discrete law on $\mathcal{S}_p$ with the same expectation as $\mu$ must equal $\mu$. The expectation determines $\mu$ uniquely.[/step]

[guided]We have established the scalar equality $\alpha_k = \beta_k$ for every $k$ and must now translate it into the measure equality $\mu = \nu$. The measure-theoretic content here is that two discrete (finitely-supported) Borel probability measures coincide iff they put the same mass on each atom — the abstract Borel-set agreement follows automatically by countable additivity. So we just need to compute $\mu(\{s_k\})$ and $\nu(\{s_k\})$ in terms of the coefficients. **Computing $\mu(\{s_k\})$.** By definition $\mu = \sum_i p_i \delta_{h_i}$ and $\delta_{h_i}(\{s_k\}) = \mathbb{1}_{\{h_i = s_k\}}$. Therefore \begin{align*} \mu(\{s_k\}) = \sum_{i=1}^n p_i \, \delta_{h_i}(\{s_k\}) = \sum_{i : h_i = s_k} p_i. \end{align*} Distinctness of the $h_i$ implies that the sum on the right has at most one term: either $s_k = h_i$ for exactly one $i$ (in which case the value is $p_i$), or $s_k$ is not among the $h_i$ (in which case the sum is empty, value $0$). Comparing with the definition of $\alpha_k$ from Step 1, this is exactly $\mu(\{s_k\}) = \alpha_k$. The same calculation gives $\nu(\{s_k\}) = \beta_k$. Step 2 produced $\alpha_k = \beta_k$, so $\mu(\{s_k\}) = \nu(\{s_k\})$ for every $k$. **Atoms outside $S$.** For $s \in \mathcal{S}_p \setminus S$, $s$ is not equal to any $h_i$ (else $s \in S$) and not equal to any $g_j$, so both sums above are empty and $\mu(\{s\}) = 0 = \nu(\{s\})$. **From atom equality to measure equality.** $\mu$ and $\nu$ are concentrated on the finite (hence countable) set $S$, meaning $\mu(\mathcal{S}_p \setminus S) = 0 = \nu(\mathcal{S}_p \setminus S)$. By countable additivity, every Borel set $B \in \mathcal{B}(\mathcal{S}_p)$ satisfies \begin{align*} \mu(B) = \mu(B \cap S) + \mu(B \setminus S) = \sum_{s_k \in B} \mu(\{s_k\}) = \sum_{s_k \in B} \nu(\{s_k\}) = \nu(B). \end{align*} This is the reduction "discrete measures agree on Borel sets iff they agree on atoms". Hence $\mu = \nu$ as Borel measures on $\mathcal{S}_p$. This establishes uniqueness: any other discrete law on $\mathcal{S}_p$ with $\mathbb{E}[G] = \sum_i p_i h_i$ must equal $\mu$, so the expectation in $T((V))$ determines the underlying discrete law on signatures.[/guided]