
“real: chapter_07” — 2011/5/22 — 23:09 — page 11 — #11
Functions of Bounded Variation 7-11
Thus V
1
0
f ≤ 2a/(1 − a), which implies that V
1
0
f = 2a/(1 − a).
Hence f is of bounded variation on [0, 1]. However given any
subinterval of (0, 1), say [c, d] with c < d we can always find
points r of the form r
k
and points s, t which are not of the form
r
k
such that c < t < r < s < d. Then f (r)>0, f (s) = 0 and
f (t) = 0 and so f (t)<f (r) and f (s)<f (r). This shows that f
is neither decreasing nor increasing in the interval [c, d].
2. Here is an example of a function that is of bounded variation on
[,1] for each >0 but is not of bounded variation on [0, 1].