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for P a partition of [a,b], show that L(g,P)<=L(f,P)

I don't know whether I should show by cases since I don't know the monotonicity of the both functions f and g. It seems like that the graphs of both functions have to behave the same way from the given condition g(x)<=f(x), isn't it? Or else how would I compare g(Xi) with f(Xi-1) or g(Xi-1) with f(x) though?

Case 1) suppose both f, g are monotone increasing

Case 2) suppose both f, g are monotone decreasing

Case 3) suppose f is monotone increasing and g is monotone decreasing

Case 4) suppose f is monotone decreasing and g is monotone increasing

Proof:

Since f, g are bounded, therefore there exists upper and lower integrals.

L(g,P)=sum g(Xi-1)(change of Xi) or L(g,P)=sum g(Xi)(change of Xi)

L(f,P)=sum f(Xi-1)(change of Xi) or L(f,P)=sum f(Xi)(change of Xi)

Since change of Xi are the same, we only need to compare sum g(Xi-1) with sum f(Xi-1) or sum g(Xi) with sum f(Xi).

Since g(x)<=f(x), it follows that sum g(Xi-1) <= sum f(Xi-1) or sum g(Xi) <= sum f(Xi)

Hence L(g,P) <= L(f,P)

Does it make sense? Any comments?