\tau_n}{\mbox{d} \mu}= \frac{\rho_n}{q_{m(n)}}$. Thus $\mbox{sign}\, w_{j,n}= \mbox{sign}\, S_{1/\rho_n, \tau_n,n}(x_{j})=\mbox{sign}\, S_{1/\rho_n, \tau_n,n+1}(x_{j}),$ $j=1,\ldots, n$.

Take an orthogonal weight $\rho_n$ with respect to the system of $n$ nodes $(x_{1},\ldots,x_{n})$ and the measure with differential form $\mbox{d} \mu(x)/q_{m(n)}(x)$. According to (\[lambda\]) and taking into account that $P_n \equiv q_{\mu_n,n}$ where the measure $\tau_n$ has the differential form $\displaystyle \mbox{d} \tau_n(x) =\rho_n(x) \mbox{d} \mu (x)/q_{m(n)}(x)$, we have the following $$w_{j,n}=q_{m(n)}(x_{j})\int \frac{q_{\tau_n,n}(x)}{q_{\tau_n,n}^{\prime}(x_{j})(x-x_{j})} \, \frac{d\mu(x)}{q_{m(n)}(x)}, \quad j=1,\ldots,n.$$ Arranging the above formula and using the identity (\[CDidentity\]) we obtain that $$w_{j,n}=\frac{q_{m(n)}(x_{j})\left|\left|q_{\tau_n,n}\right|\right|_{2,\tau_n}^2}{q_{\tau_n,n+1}(x_{j})q_{\tau_n,n}^{\prime}(x_{j})} \int \frac{q_{\tau_n,n+1}(x_{j})q_{\tau_n,n}(x)}{\left|\left|q_{\tau_n,n}\right|\right|_{2,\tau_n}^2(x-x_{j})} \frac{1}{\rho_n(x)} \frac{\rho_n(x)\, d\mu(x)}{q_{m(n)}(x)}$$ $$=-\frac{q_{m(n)}(x_{j})\left|\left|q_{\tau_n,n}\right|\right|_{2,\tau_n}^2}{q_{\tau_n,n+1}(x_{j})q_{\tau_n,n}^{\prime}(x_{j})} S_{1/\rho_n, \tau_n,n}(x_{j}),$$ which proves the second identity in (\[identityknown\]). Since $q_{\tau_n,n+1}(x_{j})q_{\tau_n,n}^{\prime}(x_{j})<0, \quad j=1,\ldots,n,$ then $\mbox{sign}\,  w_{j,n}= \mbox{sign}\, S_{1/\rho_n, \tau_n,n+1}(x_{j})$. Following the above steps we can prove the first equality in (\[identityknown\]) and $\mbox{sign}\,  w_{j,n}= \mbox{sign}\, S_{1/\rho_n, \tau_n,n}(x_{j})$.

The following two results are consequences of the above Lemma \[mainsec\]

\[theokey1\] An admissible scheme of nodes $\displaystyle {\bf x}=\left\{{\bf x}_n=(x_{1,n},\ldots,x_{n,n})\right\}_{n\in \mathbb{N}}$ is convergent if there exists a family of orthogonal weights $\displaystyle \left\{\rho_n\right\}_{n\in \mathbb{N}}$ with respect to ${\bf x}$ and the sequence of measures $\displaystyle \left\{\mbox{d} \tau_n(x) =\mbox{d} \mu (x) / q_{m(n)}(x)\right\}_{n\in \mathbb{N}}$ satisfying $$\label{keyeq}
    \lim_{n \to \infty} \left|\left|1-\rho_n(x)S_{1/\rho_n, \tau_n,n}(x)  \right|\right|_{[-1,\,1],\infty}=0,$$ where $\displaystyle \left|\left|\cdot\right|\right|_{[-1,1],\infty}$ denotes the supremum norm on $[-1,\,1]$.

Assuming the 