Analysis Riemann Integral Suppose a ≤ x1 < x2 < · · · < xn ≤ b. Assume f, g : [a, b] → R are bounded, and f(x)?
Suppose a ≤ x1 < x2 < · · · < xn ≤ b. Assume f, g : [a, b] → R are bounded, and f(x) = g(x) for all x not belonging to {x1, x2, ..., xn}. Show that if f ∈ R[a, b], then g ∈ R[a, b], and the integral from a to b of f = the integral from a to b of g.
Hint: Use Induction
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I dont think induction is a good option here. Observe that, with possible exception of a finite number of points, n points, h = f - g is zero on [a, b]. Let s = |h(x1)| ... + ...|h(xn)|. Given ε > 0, we can always choose a partition P of [a, b] in such a way that each of the points xi is enclosed in an interval of length < ε. Hence, for the upper and lower sum of h over [a, b] with respect to P we have
0 ≤ U(h, P, [a,b]) ≤ (|h(x1|) ε ... + ... |h(xn)| ε) = ε s
and similarly
-ε s ≤ L(h, P, [a,b]) ; 0
Since ε is arbitrary, the infimum of the upper sums and the supremum of the lower sums are both 0. So, h is integrable and its integral is 0. And since f and h are integrable over [a, b] and g = f + h, then so is g. In addition, over [a, b], ∫ g(x) dx = ∫ f(x) dx + &int ;h(x) dx = ∫ f(x) dx + 0 = ∫ f(x) dx.
If we mau use some measure theory, the proof is simpler, because finite sets have Lebesgue measure 0. h is zero almost everywhere on [a, b], hence it is integrable and its integral is 0.
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