Page 21 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Measure Theory and Functional Analysis

Notes
n 1 n 1
Now V f x – f x f x – f x [using (i)]
r 1 r r 1 r
r 0 r 0
V f x n f x 0 f b f a .

Now f is monotonic f b and f a are finite quantities.

V a finite quantity independent of the mode subdivision. Hence f is of bounded variation.

Notes

If f is a monotonic function on [a,b], then
b
T f f b f a
a
Theorem 2: Let V, P, N denote total, positive and negative variations of a bounded function f on
[a,b]; then prove that
V = P+N, and P–N= f(b)–f(a).
Proof: Let the interval [a,b] be divided by means of points

a x x x ... x b.
0 1 2 n
n 1
v f x – f x
r 1 r
r 0
If P denotes the sum of those differences f x – f x which are +n for positive and –n for
r 1 r
negative, then obviously,
v = p + n, f(b) – f(a) = p – n ...(i)
Let P supp,V supv,N supn, ...(ii)
where suprema are taken over all subdivisions of [a,b]. From (i), we have

v = 2p + f(a) – f(b), ...(iii)
v = 2n + f(b) – f(a). ...(iv)
Taking supremum in (iii) and (iv) and using (ii), we get

V = 2P + f(a) – f(b), ...(v)
V = 2N + f(b) – f(a). ...(vi)
By adding and subtracting, (v) and (vi) give
V = P+N and f(b) – f(a) = P–N.

Theorem 3: If f and f are non-decreasing functions on [a,b], then f –f is of bounded variation on
1 2 1 2
[a,b].
Proof: Let f = f – f defined on [a,b].
1 2
Then for any partition P a x ,x ,...,x n b , we have
1
0

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