Page 361 - DMTH401_REAL ANALYSIS

Unit 30: Riemann's and Lebesgue

Notes

d
d
Task If Y is a discrete subset of  , then Y is countable. [Hint: Let  = {B(x, 1/n): x   ,
n  }. Then,  is countable and for each y  Y, there is B   such that B  Y = {y}.]
Let K  [0,1] be the middle-third Cantor set. Then, K is an uncountable, nowhere dense compact
set with µ , [K] = m * [K] = 0. Moreover, K has no isolated points.
J 1 L,1
Proof: We recall the construction of K. Let K = [0,1], K = [0,1/3]  [2/3,1], K = [0,1/9]  [2/9,1/3]
0 1 2
 [2/3, 7/9]  [8/9,1], and so on. That is, K is the disjoint union of 2 closed subintervals of [0,1],
n
n
n
each having length 1/3 , and K is obtained from K by removing the middle-third open
n+1 n
n
intervals from each of these 2 closed intervals. The middle-third Cantor set K is defined as
K =  ¥ n 0 K . Being the intersection of compact sets, K is compact. Since the maximal length of an
=
n
interval contained in K is (1/3) , K does not contain any open interval, and hence K is nowhere
n
n
* *
dense. Also, since K  K , the above description yields m [K]  (2/3) . So m [K] = 0 and hence
n
n J,1 J,1
m * L,1 [K] = 0 also.
¥
n
It may be verified that K = {å x /3 : x  {0, 2}}. That is, K is precisely the set of those x  [0,1]
=
n 1 n n
whose ternary expansion (i.e., base 3 expansion) x = 0.x x … contains only 0’s and 2’s. Hence K
1 2

is bijective with {0, 2} which is uncountable.
We show K has no isolated point. Let x  K and let U be a neighborhood of x. Choose n large
enough so that one of the 2 closed intervals constituting K , say J , satisfies x  J  U. Let y  J \{x}
n
n n n n
be an end point of J . This end point is never removed in the later construction, so y  K for
n m
every m  n. Thus y  K  (U\{x}).
Notes It may be noted that for x  K, the base 3 expansion x = 0.x x … is eventually
1 2
constant iff x is an end point of a removed open interval. This helps to see that K contains
points other than the end points of the (countably many) removed open intervals.
The following theorem is relevant while considering big and small sets in a topological sense.
d
Task If Y is contained in a vector subspace W of  with dim(W)  d – 1, then Y is a
d
nowhere dense subset of  . [Hint: W is closed in  ( fix a basis for W and argue with the
d
d
coefficients of each basis vector separately) and W does not contain any open ball of  .]
Baire Category Theorem: Let (X, ) be a complete metric space and let U  X be open and dense
n
in X for n  . Then,  ¥ n 1 U is also dense in X. In particular,  ¥ n 1 U  Ø.
=
=
n
n
Proof: Let V  X be a nonempty open set. It suffices to show V  (  ¥ n 1 Un)  Ø. Since U is open
=
1
and dense, U  V is a nonempty open set. Let B be an open ball in X such that B  U  V and
1 1 1 1
diam[ B ] < 1. Since U is open and dense, B  U = Ø. Let B  X be an open ball with B  B  U
1 2 1 2 2 2 1 2
and diam[B ] < 1/2. In general, let B  X be an open ball with B  B  U and diam[B ]
+
2 n+1 n 1 n n+1 n+1
< 1/(n +1). If x is the center of the ball B , then we note that for every n, m  k we have x ,x  B
n n n m k
and hence  (x ,x )  diam[B ] < 1/k. So (x ) is a Cauchy sequence. Since (X, ) is complete, there
n m k n
is x  X such that (x )  x. Now, for any n, we have x  B for m  n and hence x  B . Thus
n m n n
x   ¥ n 1 B  V  (  ¥ n 1 U ).
n
=
=
n
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