Gaussian quadrature is exact through degree 2m−1 and has positive weights

By [F4], write $f=p_{m}s+t$, where $t=0$ or $\mathrm{deg}t<m$. If $s\mathrm{\ne }0$, the leading term of $p_{m}s$ has degree $m+\mathrm{deg}s$ and cannot cancel against $t$, so $m+\mathrm{deg}s=\mathrm{deg}f\le 2m-1$ and $\mathrm{deg}s<m$. The moments in these polynomials are finite by [F1]. If $s=0$, the integral of $p_{m}s=0$ is zero; otherwise orthogonality [F2] gives $\int p_{m}sd\mu =0$. Thus $\int fd\mu =\int td\mu$ by [F8].

At every node $x_i$, $p_m(x_i)=0$, so $f(x_i)=t(x_i)$ and [F5] gives $Q_m(f)=Q_m(t)$.

The cardinal interpolation identity [F6] gives $t(x)={\sum }_{i}t(x_i){\mathrm{\ell }}_i(x)$. Integrating this finite sum and using [F5] and [F8] yields $\int td\mu ={\sum }_{i}t(x_i)\int {\mathrm{\ell }}_{i}d\mu ={\sum }_i{w}_{i}t(x_i)=Q_m(t)\mathrm{.}$

Steps 1.1–1.3 give $\int fd\mu =\int td\mu =Q_m(t)=Q_m(f)$, proving exactness for every such $f$.

For a fixed $i$, ${\mathrm{\ell }}_i^2$ has degree at most $2m-2$, so step 2.1 applied to this polynomial gives $\int {\mathrm{\ell }}_i^{2}d\mu =Q_m({\mathrm{\ell }}_i^2)={\sum }_j{w}_j{\mathrm{\ell }}_i(x_j{)}^2=w_i$ by the cardinal values in [F6]. The cardinal polynomial is nonzero because ${\mathrm{\ell }}_i(x_i)=1$; [F7] therefore gives $w_i>0$.

Applying exactness to the constant polynomial $1$ and then [F9] gives ${\sum }_i{w}_i=\int 1d\mu =\mu ([a,b])$.