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\vspace*{.3in}
\noindent{\bf Rudin, page 165/22:}

\proof
Let $\epsilon>0$ be given.  By Exercise 12 from Chapter 6, there is a 
continuous function $h:[a,b]\to\R$ such that 
$\norm[2]{h-f}< \sqrt{\epsilon/4}$.
In other words,
$$
\int_a^b |h(x) - f(x)|^2\,dx < \epsilon/4.
$$
By Weiestrass' Approximation Theorem, there is a polynomial $P$ such that
$$
|P(x) - h(x)| < \sqrt{\frac{\epsilon}{4(b-a)}}
$$
for all $x\in[a,b]$.  Thus
$$
\int_a^b |P(x) - h(x)|^2\,dx < (b-a)\frac{\epsilon}{4(b-a)} = \epsilon/4.
$$
Finally, 
$$
|P(x) - f(x)|^2 \leq (|P(x) - h(x)| + |h(x)-f(x)|)^2 
\leq 2(|P(x) - h(x)|^2 + |h(x)-f(x)|^2),
$$
so
$$
\int_a^b |P(x) - f(x)|^2\,dx < 2(\epsilon/4 + \epsilon/4) = \epsilon.
$$
If I now choose a sequence $\epsilon_n > 0$ tending to $0$ and let
$P_n$ be the corresponding polynomials, then it follows that
$$
\lim_{n\to\infty} \int_a^b |P_n(x) - f(x)|^2\,dx =0
$$
\qed

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