Lemma: Let and the indicator function on . Then, on and .
Proof: Note that evidently for every partition of . Thus, it suffices to show that for every there exists some partition of such that . To do this choose such that and . It's clear then that if then from where integrability follows. The integral then follows by recalling that . The conclusion follows.
From this note then that if on and is such that where then evidently is continuous since by our lemma we have that is integrable and thus so is and