Yes,Riemann Integrable function is continous a.e.

But Weierstrass' approximation theorem is:

Every function defined and continuous on the finite interval $\displaystyle [a,b] $can be approximated uniformly on $\displaystyle [a,b]$ by polynomials to any degree of accuracy.

Because Weierstrass' approximation theorem must be use on interval ,so we must make another function $\displaystyle g(x)$
If we denote the set $\displaystyle E$ of all discontinuity point,then define:

$\displaystyle g(x)=\{ \begin{array}{cc}f(x), &\mbox{if x is a continuous point of} f(x) \\ \lim_{t \rightarrow x}f(t) &\mbox{if x is a discontinuity point of} f(x) \end{array}$

then $\displaystyle g(x)$ is a continuous function on$\displaystyle [a,b]$,

right? then use the conclusion above,we get the conclusion $\displaystyle g(x)\equiv0$.
so $\displaystyle f(x)=0$a.e.

Is this proof right?