Let $c_1, \dots, c_\nu$ be the zeros of $p_\nu$, and define weights $b_k = \int_a^b w(x)\,\ell_k(x)\,dx$ where $\ell_k$ are the Lagrange cardinal polynomials for $\{c_j\}$.

**Exactness for $\deg f \leq 2\nu - 1$.** Let $f \in P_{2\nu-1}[x]$. Divide $f$ by $p_\nu$: $f(x) = p_\nu(x)\,q(x) + r(x)$ where $q, r \in P_{\nu-1}[x]$ (since $p_\nu$ is monic of degree $\nu$ and $\deg f \leq 2\nu - 1$). Then \begin{align*} \int_a^b w(x)\,f(x)\,dx = \int_a^b w(x)\,p_\nu(x)\,q(x)\,dx + \int_a^b w(x)\,r(x)\,dx. \end{align*} The first integral vanishes: $q \in P_{\nu-1}[x]$, so $\langle p_\nu, q \rangle = 0$ by orthogonality. For the quadrature sum: $f(c_k) = p_\nu(c_k)\,q(c_k) + r(c_k) = r(c_k)$ (since $p_\nu(c_k) = 0$), so \begin{align*} \sum_{k=1}^\nu b_k f(c_k) = \sum_{k=1}^\nu b_k r(c_k). \end{align*} Since $r \in P_{\nu-1}[x]$, the Lagrange interpolant of $r$ at $c_1, \dots, c_\nu$ is $r$ itself (by [Existence and Uniqueness](/theorems/473), as $\nu$ points determine a polynomial of degree $\leq \nu - 1$). Therefore $r(x) = \sum_{k=1}^\nu r(c_k)\,\ell_k(x)$ and \begin{align*} \int_a^b w(x)\,r(x)\,dx = \sum_{k=1}^\nu r(c_k) \int_a^b w(x)\,\ell_k(x)\,dx = \sum_{k=1}^\nu b_k r(c_k). \end{align*} Combining: $\int_a^b w\,f\,dx = \int_a^b w\,r\,dx = \sum b_k r(c_k) = \sum b_k f(c_k)$.

**Positivity of weights.** For each $j$, define $g_j(x) = \ell_j(x)^2 \in P_{2\nu-2}[x] \subseteq P_{2\nu-1}[x]$. By exactness, $\int_a^b w\,\ell_j^2\,dx = \sum_{k=1}^\nu b_k\,\ell_j(c_k)^2 = b_j$ (since $\ell_j(c_k) = \delta_{jk}$). The left-hand side is positive ($w > 0$ and $\ell_j \not\equiv 0$), so $b_j > 0$.

**Optimality.** The monic polynomial $f(x) = \prod_{k=1}^\nu (x - c_k)^2 \in P_{2\nu}[x]$ satisfies $f(c_k) = 0$ for all $k$, so $\sum b_k f(c_k) = 0$. But $\int_a^b w\,f\,dx > 0$ (since $w > 0$ and $f \geq 0$ with $f \not\equiv 0$). Therefore no $\nu$-point rule can be exact for all of $P_{2\nu}[x]$. $\blacksquare$