7-6

Show that if \alpha>0, then f(x)=x^\alphais absolutely continuous on every bounded subinterval of [0,\infty).
Proof:
Let g(x)=x^{\alpha-1}, then f(x)=\alpha\int_a^x g\,\,\,+a^\alpha.
We want to show \int_a^x g is absolutely continuous. \int_a^x g is absolutely continuous if g\in L([a,b]), where [a,b]\in [0,\infty). We see that \int_a^b g is bounded for \alpha\ge1. For \alpha<1 and a=0, The improper Riemann integral \int_0^b g(t) dt exists, so the Lebesgue integral \int_0^b g exists and is equal to Riemann integral. These show us that g\in L([a,b]). Since f(x) is combination of multiple of absolutely continuous function and constant, therefore it is absolutely continuous.