The Real And Complex Number Systems

number
f T supfx fy : x, y T
is called the oscillation (or span) of f on T. Ifx S, the oscillation of f at x is defined to
be the number
f x lim
h0
fBx; h S.
Prove that this limit always exists and that f x 0if,andonlyif,f is continuous at x.
Proof: 1. Note that since f is bounded, say |fx| M for all x, wehave
|fx fy| 2M for all x, y S. So, f T, the oscillation of f on any subset T of S,
exists. In addition, we define gh f Bx; h S. Note that if T 1 T 2 S, wehave
f T 1 f T 2 . Hence, the oscillation of f at x, f x lim h0
gh g0 since g is
an increasing function. That is, the limit of f Bx; h S always exists as h 0 .
2. Suppose that f x 0, then given 0, there exists a 0 such that as
h 0, , wehave
|gh| gh f Bx; h S /2.
That is, as h 0, , wehave
supft fs : t, s Bx; h S /2
which implies that
/2 ft fx /2 as t x , x S.
So, as t x , x S. wehave
|ft fx| .
That is, f is continuous at x.
Suppose that f is continous at x, thengiven 0, there exists a 0 such that as
t x , x S, wehave
|ft fx| /3.
So, as t, s x , x S, wehave
|ft fs| |ft fx| |fx fs| /3 /3 2/3
which implies that
supt fs : t, s x , x S 2/3 .
So, as h 0, , wehave
f Bx; h S supt fs : t, s x , x S .
Hence, the oscillation of f at x, f x 0.
Remark: 1. The compactness of S is not used here, we will see the advantage of the
oscillation of f in text book, Theorem 7.48, in page 171. (On Lebesgue’s Criterion for
Riemann-Integrability.)
2. One of advantage of the oscillation of f is to show the statement: Let f be defined on
a, b Prove that a bounded f does NOT have the properties:
f is continuous on Q a, b, and discontinuous on R Q a, b.
Proof: Since f x 0if,andonlyif,f is continuous at x, we know that f r 0for
r R Q a, b. DefineJ 1/n r : f r 1/n, then by hypothesis, we know that
n1 J 1/n R Q a, b. It is easy to show that J 1/n is closed in a, b. Hence,
intclJ 1/n intJ 1/n for all n N. It means that J 1/n is nowhere dense for all
n N. Hence,