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Chapter

# PPP

DOI link for PPP

PPP book

# PPP

DOI link for PPP

PPP book

## ABSTRACT

PROOF We will establish the equality for upper integrals. To this end, let P be a partition of [a, b]. Then P ∪ {c} = P1 ∪P2 where P1 is a partition of [a, c], and P2 is a partition of [c, b]. Now, since

SP (f ) ≥ SP∪{c}(f ) = SP1 (f ) + SP2 (f ) by Exercise 7.2 on page 340,

≥

∫ c a f (x) dx +

∫ b c f (x) dx,

it follows that ∫ b a f (x) dx ≥

∫ c a f (x) dx +

∫ b c f (x) dx. (7.7)

We now establish the reverse inequality. To this end, for any ! > 0 there exists a partition P1 of [a, c], and a partition P2 of [c, b] such that

SP1 (f ) < ∫ c a f (x) dx + (2 , and SP2 (f ) <

∫ b c f (x) dx + (2 . (Why?)

Now, since P1 ∪ P2 is a partition of [a, b], we have∫ b a f (x) dx ≤ SP1∪P2 (f )

= SP1 (f ) + SP2 (f ) by Exercise 7.2 on page 340,

<

∫ c a f (x) dx +

∫ b c f (x) dx + !.