Assume $f\in \mathscr{R}(\alpha)$ on $[a,b]$, and prove that there are polynomials $P_n$ such that $$\lim \limits_{n\to \infty}\int_{a}^{b}|f-P_n|^2d\alpha=0.$$
Proof: Let $\varepsilon>0$ be given then for $f\in \mathscr{R}(\alpha)$ on $[a,b]$ exists continuous function $g$ on $[a,b]$ such that $\lVert f-g \rVert_2<\frac{\sqrt{\varepsilon}}{2}$ (was proven here). By Weierstrass approximation theorem exists $N_{\varepsilon}$ such that for $n\ge N_{\varepsilon}$ we have $|g(x)-P_n(x)|<\sqrt{\dfrac{\varepsilon}{4(\alpha(b)-\alpha(a))}}$ for any $x\in [a,b]$. Then it's easy to show that $\lVert g-P_n \rVert_2<\frac{\sqrt{\varepsilon}}{2}$.
Then by Minkowski inequality $\lVert f-P_n \rVert_2\leqslant \lVert f-g \rVert_2+\lVert g-P_n \rVert_2<\sqrt{\varepsilon}$. Hence $\int_{a}^{b}|f-P_n|^2d\alpha<\varepsilon$. Hence $$\lim \limits_{n\to \infty}\int_{a}^{b}|f-P_n|^2d\alpha=0.$$ Q.E.D.
Dear users! What did you think about my proof please? Is it correct?