9.6 A Simple Procedure for Finding Integrals
Suppose f is a continuous function and F is an increasing integrator function. How do you find ∫baf(x)dF? Is there some sort of easy way to do it which will handle lots of simple cases? It turns out there is a way. It is based on Lemma 9.3.22. First of all
$$
F(x+) \equiv \lim {y \rightarrow x+} F(y), F(x-) \equiv \lim {y \rightarrow x-} F(y)
Foranincreasingfunction$F$,thejumpofthefunctionat$x$equals$F(x+)−F(x−)$.Procedure9.6.1Suppose$f$iscontinuouson$[a,b]$and$F$isanincreasingfunctiondefinedon$[a,b]$suchthattherearefinitelymanyintervalsdeterminedbythepartition$a=x0<x1<⋯<xn=b$whichhavethepropertythaton$[xi,xi+1]$,thefollowingfunctionisdifferentiableandhasacontinuousderivative.
G_{i}(x) \equiv\left{F(x) on (xi,xi+1) F(xi+) when x=xi F(xi+1−) when x=xi+1\right.
Alsoassume$F(a)=F(a+),F(b)=F(b−)$.Then \int_{a}^{b} f(x) d F=\sum_{j=0}^{n-1} \int_{x_{j}}^{x_{j+1}} f(x) G_{j}^{\prime}(x) d x+\sum_{i=1}^{n-1} f\left(x_{i}\right)\left(F\left(x_{i}+\right)-F\left(x_{i}-\right)\right) 202CHAPTER9.INTEGRATIONHereiswhythisprocedureworks.Let$δ$beverysmallandconsiderthepartition
a=x0<x1−δ<x1<x1+δ <x2−δ<x2<x2+δ< ⋯xn−1−δ<xn−1<xn−1+δ<xn−δ<xn=b
where$δ$isalsosmallenoughthatwhenever$|x−y|<δ$,itfollows$|f(x)−f(y)|<ε$.Thenfromthepropertiesoftheintegralpresentedabove,
∫x1−δafdF+∫x2−δx1+δfdF+⋯+∫bxn−1+δfdF+n−1∑i=1(f(xi)−ε)(F(xi+δ)−F(xi−δ)) ≤∫bafdF≤∫x1−δafdF+∫x2−δx1+δfdF+⋯+∫bxn−1+δfdF +n−1∑i=1(f(xi)+ε)(F(xi+δ)−F(xi−δ))
ByLemma9.3.22thisimplies
∫x1−δafG′0dx+∫x2−δx1+δfG′1dx+⋯+∫bxn−1+δfG′n−1dx +n−1∑i=1(f(xi)−ε)(F(xi+δ)−F(xi−δ))≤∫bafdF≤
$$
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