= Solution
Write $W=\int_{-1}^1w(x)dx$, finite and positive. Exactness on the constant <polynomial> gives $\sum_i a_i^{(n)}=W$. Positivity therefore implies the uniform operator bound $|I_n(h)|\leq W\|h\|_\infty$, and the <integral> has the same bound.
For a continuous $f$ and $\epsilon>0$, the <Weierstrass approximation theorem> gives a fixed <polynomial> $Q$ of degree $m$ with $\|f-Q\|_\infty<\epsilon$. For every $n\geq m$, <polynomial> exactness gives $I_n(Q)=I(Q)$. Thus
$$
\begin{aligned}
|I(f)-I_n(f)|&\leq |I(f-Q)|+|I_n(f-Q)|\\&\leq2W\|f-Q\|_\infty<2W\epsilon.
\end{aligned}
$$
Since $\epsilon$ is arbitrary, $\boxed{I_n(f)\to I(f)}$ for every continuous $f$. This is <convergence of positive quadrature on continuous functions>. The argument requires nodes in $[-1,1]$, implicit in a rule defined on all continuous functions there; otherwise their function values would not even be specified by the stated domain. Exactness alone would not provide the norm bound if arbitrarily signed weights were allowed.
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