? As in the case of integrals of functions of one variable,
this is an important question. It turns out the Riemannn integrable functions are
characterized by being continuous except on a very small set. This has to do with Jordan
content.
Definition C.4.1A bounded set E, has Jordan content 0 or content 0if forevery ε > 0 there exists a grid G such that
∑
v(Q ) < ε.
Q∩E ⁄=∅
This symbol says to sum the volumes of all boxes from G which have nonemptyintersection with E.
Next it is necessary to define the oscillation of a function.
Definition C.4.2Let f be a function defined on ℝ^{n}and let
ωf,r(x) ≡ sup {|f (z) − f (y)| : z,y ∈ B (x,r)}.
This is called the oscillation of f on B
(x,r)
. Note that this function of r is decreasing inr. Define the oscillation of f as
ωf (x) ≡ lim ωf,r(x).
r→0+
Note that as r decreases, the function ω_{f,r}
(x )
decreases. It is also bounded below by
0, so the limit must exist and equals inf
{ωf,r (x ) : r > 0}
. (Why?) Then the following
simple lemma whose proof follows directly from the definition of continuity gives the
reason for this definition.
Lemma C.4.3A function f is continuous at x if and only if ω_{f}
(x)
= 0.
This concept of oscillation gives a way to define how discontinuous a function is at a
point. The discussion will depend on the following fundamental lemma which gives the
existence of something called the Lebesgue number.
Definition C.4.4Letℭbe a set whose elements are sets of ℝ^{n}and letK ⊆ ℝ^{n}. The set ℭ is called a cover of K if every point of K is contained in someset of ℭ. If the elements of ℭ are open sets, it is called an open cover.
Lemma C.4.5 Let K be sequentially compact and let ℭ be an open cover ofK. Then there exists r > 0 such that whenever x ∈ K,B(x,r) is contained in someset of ℭ.
Proof:Suppose this is not so. Then letting r_{n} = 1∕n, there exists x_{n}∈ K such that
B
(xn,rn)
is not contained in any set of ℭ. Since K is sequentially compact, there is a
subsequence, x_{nk} which converges to a point x ∈ K. But there exists δ > 0 such that
B
(x,δ)
⊆ U for some U ∈ℭ. Let k be so large that 1∕k < δ∕2 and
|xnk − x |
< δ∕2 also.
Then if z ∈ B
(xnk,rnk)
, it follows
|z − x| ≤ |z− x |+ |x − x| < δ + δ = δ
nk nk 2 2
and so B
(xnk,rnk)
⊆ U contrary to supposition. Therefore, the desired number exists
after all. ■
Theorem C.4.6Let f be a bounded function which equals zero off a boundedset and let W denote the set of points where f fails to be continuous. Then f∈ℛ
n
(ℝ )
if W has content zero. That is, for all ε > 0 there exists a grid G suchthat
∑
v(Q) < ε (3.10)
Q∈GW
(3.10)
where
GW ≡ {Q ∈ G : Q ∩ W ⁄= ∅} .
Proof: Let W have content zero. Also let
|f (x)|
< C∕2 for all x ∈ ℝ^{n}, let
ε > 0 be given, and let G be a grid which satisfies (3.10). Since f equals zero
off some bounded set, there exists R such that f equals zero off of B
( )
0,R2
.
Thus W ⊆ B
( )
0,R2
. Also note that if G is a grid for which 3.10 holds, then this
inequality continues to hold if G is replaced with a refined grid. Therefore, you
may assume the diameter of every box in G which intersects B
(0,R )
is less
than
R3-
and so all boxes of G which intersect the set where f is nonzero are
contained in B
(0,R )
. Since W is bounded, G_{W} contains only finitely many boxes.
Letting
n
Q ≡ ∏ [ai,bi]
i=1
be one of these boxes, enlarge the box slightly as indicated in the following
picture.
PICT
The enlarged box is an open set of the form,
^ ∏n
Q ≡ (ai − ηi,bi + ηi)
i=1
where η_{i} is chosen small enough that if
n ( )
∏ ( bi + ηi − (ai − ηi)) ≡ v ^Q ,
i=1
and
^
GW
denotes those
^
Q
for Q ∈G which have nonempty intersection with W,
then
∑ ( )
v ^^Q < ε (3.11)
^Q∈G^W
(3.11)
where
^^Q
is the box,
n∏
((ai − 2ηi), bi + 2ηi).
i=1
For each x ∈ ℝ^{n}, let r_{x}< min
(η1∕2,⋅⋅⋅,ηn∕2)
be such that
ωf,rx (x) < ε + ωf (x). (3.12)
(3.12)
Now let ℭ denote all intersections of the form
^Q
∩B
(x,rx)
such that x ∈B
(0,R)
so that
ℭ is an open cover of the compact set B
(0,R )
. Let δ be a Lebesgue number for this open
cover of B
(0,R)
and let ℱ be a refinement of G such that every box in ℱ has
diameter less than δ. Now let ℱ_{1} consist of those boxes of ℱ which have nonempty
intersection with B
(0,R ∕2)
. Thus all boxes of ℱ_{1} are contained in B
(0,R )
and each one is contained in some set of ℭ. Let ℭ_{W} be those open sets of ℭ,
^
Q
∩ B
(x,rx)
, for which x ∈ W. Thus each of these sets is contained in some
^^
Q
where
Q ∈G_{W}. Let ℱ_{W} be those sets of ℱ_{1} which are subsets of some set of ℭ_{W}.
Thus
∑
v(Q) < ε. (3.13)
Q∈ℱW
(3.13)
because each Q in ℱ_{W} is contained in a set
^^
Q
described above and the sum of the
volumes of these is less than ε by 3.11. Then
∑
U ℱ (f)− ℒℱ (f) = (MQ (f) − mQ (f))v (Q )
Q∈ℱW
∑
+ (MQ (f) − mQ (f))v (Q ).
Q∈ℱ1∖ℱW
If Q ∈ℱ_{1}∖ℱ_{W}, then Q must be a subset of some set of ℭ ∖ℭ_{W} since it is not in any set
of ℭ_{W}. Say Q ⊆
^Q
1
∩ B
(x,r )
x
where x
∕∈
W. Therefore, from 3.12 and the observation
that x
Consider the inside sum with the aid of the following picture.
PICT
In this picture, the little rectangles represent the boxes Q_{j}×
[ai,ai+1]
for
fixed j. The part of P having x contained in Q_{j} is between the two surfaces,
x_{n+1} = g
(x)
and x_{n+1} = f
(x)
and there is a zero placed in those boxes for
which
MQ ×[a ,a ](XP) − mQ ×[a,a ](XP ) = 0.
j i i+1 j i i+1
You see, X_{P} has either the value of 1 or the value of 0 depending on whether
(x,y)
is
contained in P. For the boxes shown with 0 in them, either all of the box is contained in
P or none of the box is contained in P. Either way,
MQj ×[ai,ai+1](XP )− mQj×[ai,ai+1](XP ) = 0
on these boxes. However, on the boxes intersected by the surfaces, the value
of
MQj× [ai,ai+1](XP)− mQj ×[ai,ai+1](XP )
is 1 because there are points in this box which are not in P as well as points
which are in P. Because of the construction of G^{′} which included all values of
ε
MQj (fXE )+ -----,MQj (f XE),
M (gX ),m4mK(fX ),m (gX )
Qj E Qj E Qj E