= Solution
Fix a <continuous function> $f$ and $\varepsilon>0$. By the <Weierstrass approximation theorem>, choose a polynomial $p\in\mathcal P_m$ such that
$$
\lVert f-p\rVert_\infty<\varepsilon.
$$
For every $n\geq m$, exactness gives $I_n(p)=I(p)$. Positivity of the quadrature weights and exactness on the constant polynomial give
$$
\sum_{i=0}^na_i^{(n)}=I_n(1)=I(1)=\int_a^bw(x)\,dx.
$$
Consequently
$$
\begin{aligned}
|I_n(f)-I(f)|
&\leq |I_n(f-p)|+|I(f-p)|\\
&\leq \varepsilon\sum_{i=0}^na_i^{(n)}
+\varepsilon\int_a^bw(x)\,dx\\
&=2\varepsilon\int_a^bw(x)\,dx.
\end{aligned}
$$
Since $\varepsilon$ is arbitrary, $I_n(f)\to I(f)$. This is the <convergence of positive quadrature rules>.
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