Page 281 - DMTH401_REAL ANALYSIS

Unit 22: Introduction to Riemann-Stieltjes Integration, using Riemann Sums

Proof: If f Î BV[a, b] and a < c < b, let partitions | [a, c] and | [c, b] be given. Then :=  U  is Notes
a partition of [a, b] so V + V = V  V, hence V  V and V  V. Thus f Î BV[a, c] and f Î BV[c, b].
    
Conversely, suppose that a < c < b and that f Î BV[a, c] and f Î BV[c, b]. Let |[a, b]. Then
 : = U {c} is a refinement of . Therefore V  V = V + V , where :=  [a, c] and  is defined
c   c   c
similarly. By hypothesis, V  V = V + V  V(f, [a, c]) + V(f, [c, b]). Thus V(f, [a, b])  V(f, [a, c])
  c  
+ V(f, [c, b]) < ¥. This proves part of the asserted equality. To show the other inequality, now
that we know V < ¥ let partitions | [a, c] and | [c, b] be given. We recall that earlier we had
V + V = V  V, so V + V  V whenever | [a, c] and | [c, b] were arbitrary partitions of [a, c]
   c  
and [c, b], respectively.
Thus sup (V + V ) = V(f, [a, c]) +V  V, and so sup (V(f, [a, c]) + V ) = V(f, [a, c]) + V(f, [c, b])  V.




 |[a,c]  |[c,b]

Note The first inequality holds for an arbitrary |[c, b], making the second one valid.

Example: Prove that the equality in (25) holds for every function f: [a, b] ® , whether f
is a function of bounded variation or not.
Motivated by (25), when f: [a, b] ®  and a  x  b we can define the three functions
V(x) := V(f, [a, x]), P(x) := P(f, [a, x]) and N(x) := N(f, [a, x]).
Each of these is an increasing function of x. Jordan’s Theorem asserts that if f Î BV[a, b] we can
represent f in terms of P(x) and N(x).
Theorem: Jordan
A function f Î BV[a, b] if and only if there exist functions g and h, both increasing on [a, b], such
that f(x) = g(x) – h(x) for a  x  b. If this is the case, then P(x)  g(x) – g(a), N(x)  h(x) – h(a) and
f(x) = f(a) + P(x) – N(x) for a  x  b.

Proof: Suppose first that f(t) = g(t) – h(t), t Î [a, b], where the functions g and h are both increasing
on [a, b]. Let |[a, b] (later, we will apply this when x Î [a, b] and |[a, x]). Then
  g
f = f (x ) – f(x ) = g – h i .
i i i – 1 i i{ - h i
Thus –h  f  g , so |f |  max{g , h }  g +h for 1  i  n . Hence V (f)  V (g) + V (h) =
i i i i i i i i    
g(b) – g(a) + h(b) – h(a) < ¥, so f Î BV[a, b].
Next, we show that f(x) = f(a) + P(x) – N(x) for a  x  b. But we will do this just by showing it for
x = b. Then we can use (25) and let each x Î [a, b] play the role of b. This will show the existence
of the functions g(x)(= f(a) + P(x)) and h(x)(= N(x)). After that is done, we’ll prove the P–g and
N–h inequalities.
By the definitions of P, N and V we know there exist sequences { }, { } and { } such that P ® P,
k k k  k
N ® N and V ® V. Let us define  :=      . As P  P  P. By the Squeeze Principle
 k  k k k k k  k  k
P ® P. Similarly, N ® N and V ® V. By Limit Theorems
 k  k  k
P + N = V and P – N = f(b) – f(a) and the second is the same as f(x) = f(a) + P(x) – N(x)
when x = b. By above we can use any x Î [a, b] in place of b by restricting our attention to f on
[a, x].
Now suppose that f(x) is defined as the difference of two increasing functions on [a, b]: f(t) =
g(t) – h(t). We have the following observation: t ® t is increasing and t ® t is decreasing.
–
+
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