Are you getting at the measure of the irrationals is 1 over the interval [0,1]. Sure. Are you saying if we do not define, or define it, at the rationals, it doesn't matter which, the measure of the real line between 0 and 1 is the measure of the irrationals ie one. Again sure - just different language to what I said before. But how does that resolve the original query. Under the usual definition of the Riemann Integral it means if it is discontinuous at the rationals it is not Riemann integrable. Thus one could say it cant be discontinuous at the rationals because its Riemann Integrable. - so my argument since the rationals are dense it must be continuous at those points - hence f=g? Have I got your argument correct?It is discontinuous on both the rationals and the irrationals. Every irrational is the limit of a Cauchy sequence of rationals. Every rational is the limit of a Cauchy sequence of irrationals. You do not need to know the measure of the rationals.
My suspicion, and I have never seen a development along those lines, Riemann Integration can be defined in a more general way as detailed previously.