\int_{\Gamma} \frac{z^{\nu} p_{n}(z) \, \mbox{d} z}{\displaystyle  \prod_{i=1}^{\ell}\left(z-t_i\right)^{2d_i-1}q(z)} .$$ Since $\displaystyle \frac{z^{\nu} p_{n}(z)}{\displaystyle  \prod_{i=1}^{\ell}\left(z-t_i\right)^{2d_i-1}q(z)}  \in \mathcal{H} \left(\mbox{Int}(\Gamma)\right)$ the second term vanishes. From (\[pnexpand\]), using the Cauchy integral formula, we obtain: $$0=\sum_{j=1}^{n-\ell} y_j^{\nu}\widetilde{p}(y_j)\frac{\widetilde{\lambda}_{j}}{\displaystyle  \prod_{i=1}^{\ell}\left(y^j-t_i\right)^{2(d_i-1)}q(y_j)}=0, \quad \nu=0,\ldots, n-\ell-1.$$ Taking into account that for each $j=1, \ldots, n-\ell$, $\displaystyle \frac{\widetilde{\lambda}_{j}}{\displaystyle  \prod_{i=1}^{\ell}\left(y^j-t_i\right)^{2(d_i-1)}q(y_j)}>0$, we conclude that the above orthogonality relations imply that $\widetilde{p}$ must change sign at least $n-\ell$ times, hence $\deg \widetilde{p} \geq n-\ell$. Since $\deg \widetilde{p}= \deg p_{n-1}-\ell\leq n-\ell-1$ we arrive at a contradiction which completes the proof.

Consider a monic polynomial $\displaystyle P_n(x)=\prod_{j=1}^n(x-x_{j})$ with degree $n\in \mathbb{N}$ which is $\mu$ admissible. We say that a weight function $\rho_n$ on $\mbox{supp}(\mu)$ is orthogonal with respect to $P_n(x)$ and $\mu$ if $P_n \equiv q_{\mu_n,n}$, where $\mbox{d} \mu_n(x)=\rho_n(x) \, \mbox{d} \mu(x)$, $x\in \mbox{supp}(\mu)$. We also say that $\rho_n$ is orthogonal with respect to ${\bf x}_n=(x_{1}, \ldots, x_{n})$ and $\mu$. A sequence of weight functions $\left\{\rho_n\right\}_{n\in \mathbb{N}}$ is a family of orthogonal weight functions with respect to the sequence of polynomials $\left\{P_n\right\}_{n\in \mathbb{N}}$, if for each $n\in \mathbb{N}$, $P_n\equiv q_{\mu_n,n}$.

Let $q_{m(n)}$ be an arbitrary polynomial with degree $\deg q_{m(n)}(x)=2n-m(n)-1$ being positive on $[-1,\,1]$. Let $\mu_n$ denote the measure with differential form $\mbox{d} \mu_n(x) = q_{m(n)}^{-1}(x) \mbox{d} 