Page 313 - DMTH401_REAL ANALYSIS

Unit 26: Lebesgue Measure

æ
æ
é
é
ö ù
m* A   ¥ = ë E i ÷  m* A   n E i ÷ ö ù Notes
ç
ç
ê
ê
è
è
úø û
úø û
= ë
i 1
i 1
n
= å m* (A  E ).
i
=
i 1
Letting n  ¥ lead to
æ é ¥ ö ù ¥
m* A   E i ÷  å m* (A  E ).
ç
è ê i 1 úø û i 1 i
= ë
=
The opposite inequality follows from countable subadditivity.
Theorem 4: Let { } n be a sequence of measurable sets, then E =  ¥ i 1 E is also measurable.
E
=
i
i
Moreover, if E ,E ,... are disjoint then m(E) = å ¥ i 1 m(E ).
1 2 = i
This is called the countable additivity which can be proved by putting A = 
Proof: We first assume E ,E ,...are disjoint. Then for all A  R, n   we have
1 2
æ
ö ù
é
æ
m*(A) = m* A   n E i ÷ + m* A   n E c ö
ç
i
è ê i 1 úø û ç è ( ) ÷ ø
= ë
i 1
=
n
c
+
 å m* (A  E ) m* (A  E ).
i
i 1
=
Letting n  ¥,
¥
c
+
m*(A)  å m* (A  E ) m* (A  E )
i
=
i 1
c
= m*(A  E) +m*(A  E ).
This proved E is measurable.
Now, if E ,E ,... are not disjoint, we let
1 2
F = E , F = E \F , F = E \(F  F ),
1 1 2 2 1 3 3 1 2
and in general F = E \  i 1 F for k > 2. Then F , F ,... are disjoint and  ¥ F =  ¥ i 1 E . Since M is
k 1
-
=
k k = i 1 2 i 1 i = i
an algebra, F , F ,... are all measurable. So E =  i 1 E =  ¥ F is measurable.
¥
=
1 2 = i i 1 i
Notes M is proved to be a -algebra. The next result shows that all Borel sets are
measurable. Recall that the family of Borel sets in  is, by definition, the smallest
-algebra containing all open subsets of .
Theorem 5: M contains all Borel subsets of .
Proof: It suffices to show that (a, ¥)  M for all a   (why?). Let A  . We need to show that
m*(A)  m*(A  (–¥, a]) + m*(A  (a, ¥)).
Without loss of generality, we may assume m*(A) < ¥. For convenience, let A = A  (–¥, a] and
1
A = A  (a, ¥). Then we need to show
2
m*(A) +   m*(A ) + m*(A ) for all  > 0.
1 2
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