One should probably replace [a,b] by [0,a].

Then this is just because the product of two Riemann-integrable functions is Riemann-integrable (note that a Riemann-integrable function on [0,a] is bounded, contrary to a Lebesgue-integrable function on [0,a]). And if g is Riemann-integrable on [0,a], then so is for any , because it is obtained by symmetry from , which is R-integrable on (remember g is zero on ).