Page 330 - DMTH401_REAL ANALYSIS

Real Analysis

Notes If we can prove that
g := f : X \A  
X\A
is continuous, then we have proven our theorem with C = X \ A (modulo the closedness condition).
Since { } is a basis for the topology of  to prove that g is continuous all we have to do is prove
k
–1
that for each k, g ( ) is open in X \A. To prove this, we shall prove that
k
–1
(3.1) g ( ) = (X\A)   ;
k k
–1
n
then, since  is an open subset of  , it follows that g ( ) is open in X\A and we’re done. Now
k k
to prove the desired equality note that, by definition of g, we have
g ( ) = (X\A)  f ( )  (X \A)   ,
–1
–1
k k k
–1
since f ( )   . On the other hand, observe that
k k
x  (X\A)    x ÏA, x  
k k
 x Ï ( \f ( )), x  
–1
k k k
 x  f ( ).
–1
k
In the second implication we used that A =   j 1 (\f ()) so x ÏA implies, in particular, that
–1
=
j
j
x Ï ( \f ( )). Therefore,
–1
k k
(X\A)    (X\A)  f ( ),
–1
k k
which completes the proof of (3.1).
Step 2: We now require that C be closed. Given  > 0 by Step 1 we can choose a measurable set
B  X such that m(X\B) < /2 and f is continuous on B. By Littlewood’s First Principle we can
n
choose a closed set C   such that C  B and m(B\C) < /2. Since
X\C = (X \B)  (B \C),
we have
m(X \C) £ m(X \B) + m(B \C) < .
Also, since C  B and f is continuous on B, the function f is automatically continuous on the
smaller set C. This completes the proof of our theorem.
n
We shall see that Luzin’s theorem holds not just for  but for topological spaces as well.
27.4 Borel Measurability on Topological Spaces
Recall that the collection of Borel subsets of a topological space is the -algebra generated by the
open sets. For a measurable space (T, S ) where T is a topological space with S its Borel subsets,
we call a measurable function f: T   Borel measurable to emphasize that the -algebra S is
the one generated by the topology and it is not just any -algebra on T. For example, a Borel
–1
n
n
n
measurable function on  is a function f :    such that f (a, ]  B for all a  .
Proposition: Any continuous real-valued function on a topological space is Borel measurable.
n
The proof of this proposition follows word-for-word the  case in Example, so we omit its
proof. A nice thing about Borel measurability is that it behaves well under composition.
Proposition: If f :    is Borel measurable and g : X   is measurable, where X is an
arbitrary measurable space, then the composition,
f o g : X  

is measurable.

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