Page 362 - DMTH401_REAL ANALYSIS

Real Analysis

Notes

Notes (i) By considering the complements of U ’s in the above, we get the following
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conclusion: if (X,) is a complete metric space, then X cannot be written as a countable
union of nowhere dense (closed) subsets of X. That is, X is of second category in itself.
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(ii) Since  is a complete metric space with respect to the Euclidean metric,  cannot be
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written as a countable union of nowhere dense (closed) subsets of  . (iii) From a topological
point of view, a first category subset is considered as a small set and a dense G subset is

considered as a big set because of Baire Category Theorem. However, a set that is
topologically big (small) need not be measure theoretically big (small). (iv) The
uncountability of the middle-third Cantor set can be proved with the help of Baire Category
Theorem also.
We observe in the following the distinction between topological bigness (smallness) and measure
theoretical bigness (smallness).

Task For any Y   , the set Y\Y is discrete and hence countable. In particular, every
(1)
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uncountable subset of  has a limit point in  . [Hint: Let y  Y\Y . If B(y, 1/n)  Y
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(1)
contains a point other than y for every n  , then y  Y , a contradiction.]
(1)
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(i) For every  > 0, there is a dense open set U   such that m * L,d [U] < .
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(ii) There is a dense G subset Y   with m * [Y] = 0.
 L,d
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(iii) There is an F set X   of first category with m * [X] = ¥ and m * [ \X] = 0.
 L,d L,d
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(iv) For every closed d-box A and every  > 0, there is anywhere dense compact set K   such
that K  A and m * L,d [K] > Vol (A) – .
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Proof: (i) Write  = {x ,x ,…}. For each n  , let A be an open d-box with x  A and Vol (A )
1 2 n n n d n
< /2 . Put U =  ¥ n 1 A .
n
=
n
d
(ii) Let U   be a dense open subset with m * [U ] < 1/n and put Y =  ¥ U .
=
n L,d n n 1 n
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(iii) Let Y be as in (ii) and take X =  \Y.
(iv) LetUbeasin(i)andletK = A\U.
The next result shows that the Lebesgue outer measure does not satisfy finite additivity (and
hence it does not satisfy countable additivity), even though it satisfies countable subadditivity.
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Let X =   [0, 1] . Then, there is a subset Y  [0, 1] satisfying the following:
(i) The translations Y + x are pairwise disjoint for x  X.
[
(ii) There exist finitely many distinct elements x ,…,x  X such that m * L,d  n (Y + x )] 
=
1 n i 1 i
n
å i 1 m * L,d [Y + x ].
=
i
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Proof: Define an equivalence relation on [0, 1] by the condition that a ~ b iff a – b   . By the
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axiom of choice, we can form a set Y  [0, 1] whose intersection with each equivalence class is
a singleton.
(i) We verify that (Y + r)  (Y + s) = Ø for any two distinct r, s  X. If (Y + r)  (Y + s) = Ø for
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r, s  X, then there are a, b  Y such that a + r = b + s. Now we have a – b = s – r   , and
hence a ~ b. So we must have a = b by the definition of Y, and then necessarily r = s.
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