MATH1115 Lecture Notes - Lecture 5: Riemann Sum, Riemann Integral, Bounded Function
MATH1115: Test for Integrability of Riemann Sum proofs
● Test for Integrability:
○ Suppose that a,b ER, with a<b and that f: [a,b] -> R is a bounded function.
Then f is Riemann integrable if and only if for every E>0 there is a partition
P of [a,b] such that U (f,P) - L(f,P) < E
■ Proof: <- Start by the right hand statement first
● Suppose there is such a partition P for each E>0. (Referring
to a previous line).
● Then we have, for such P, L (f,P) [lower sum is part of the
set that we take the sup from in order to calculate the lower
integral] < or equal to the L integral from b to a (f) < or equal
to the U integral from b to a (f) < or equal to U(f,P)
● So 0 < or equal to the U integral b to a (f) - L integral b to a <
or equal to U(f,P) - L(f,P) < E
● Hence for all E>0 the upper integral of f and the lower
integral of f < E
● Thus the upper integral (f) - lower integral (f) = 0
● I.e. L integral = Upper integral, so F is integrable by definition
■ Proof: -> OTHER DIRECTION
● Suppose that f is integrable. Let E > 0
● Since the lower integral of f is = to the sup {L (f,P):....} about
to find the member of the set that lies below L integral. Tiny
element is not E but E/2.
● By the prop. There exists partition P of [a,b] such that L (f,P)
> lower integral (f) - E/2 = definite integral of f - E/2 and there
exists partition Q of [a,b] such that upper sum (f,Q) < upper
integral + E/2 = integral of f from b to a + E/2.
● Choose E/2 to make sure the difference between the lower
integral and upper integral to be E but they are part of
different partitions hence take different refinements.
● But then PUQ (p union q) is a refinement of both P and Q.
So integral of f - E/2 < L (f,P) < or equal to (f, PUQ) < or
equal to U (f,PUQ) < or equal to U (f,Q) < integral of f + E/2.
● Therefore the U (f,PUQ) - L(f, PUQ) < E
■ Application:
● Let f: [0,3] -> R be given by f(x) = { 0 - 0 if 0 < x < or equal to
1, if 1 < x < or equal to 2 and if 2 < x < or equal to 3} ->
discontinuous graph/diagram
● This is integrable by the test we’ve just proven.
● Let E>0 be given and assume E<2 without loss of generality.
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Document Summary
Math1115: test for integrability of riemann sum proofs. Suppose that a,b er, with a r is a bounded function. Then f is riemann integrable if and only if for every e>0 there is a partition. P of [a,b] such that u (f,p) - l(f,p) < e. Proof: 0. (referring to a previous line). So 0 < or equal to the u integral b to a (f) - l integral b to a < or equal to u(f,p) - l(f,p) < e. Hence for all e>0 the upper integral of f and the lower integral of f < e. Thus the upper integral (f) - lower integral (f) = 0. L integral = upper integral, so f is integrable by definition.
