Yes,Riemann Integrable function is continous a.e.

But Weierstrass' approximation theorem is:

Every function defined and continuous on the finite interval

can be approximated uniformly on

by polynomials to any degree of accuracy.

Because Weierstrass' approximation theorem must be use on interval ,so we must make another function
If we denote the set

of all discontinuity point,then define:

then

is a continuous function on

,

right? then use the conclusion above,we get the conclusion .
so

a.e.

Is this proof right?