Hi Madukarga,
First, let us review some definitions. Recall that the Borel
sigmafield (or sigmaalgebra) on R is defined to be the minimal
collection of subsets of R which contains the intervals and is closed
under complementation and under countable intersection operations.
This definition contains a statement that needs to be proven, that
such a minimal collection exists: Well, take the collection which is
the intersection of all sigmafields on R (i.e. collections of subsets
that are closed under complementation and countable intersection) that
contain the intervals; you get a sigmafield, containing the
intervals, which is obviously the minimal one having these properties,
since it is contained in any other one having these properties.
A subset of R is called a Borel set if it is in the Borel sigmafield.
Next, a function f:R>R is called Borel measurable if the inverse
image of any Borel set under f is a Borel set.
I will now prove to you that if f:R>R is monotone, then f is Borel
measurable. What we need is the following simple
Lemma: If f:R>R has the property that the inverse image of any
INTERVAL is a Borel set (compare the def. of Borel measurable
function), then f is Borel
measurable.
Proof: Consider the collection C of all subsets B of R such that
f^(1)(B)
(the inverse image of B under f) is a Borel subset. What do we know
about C?
1. C contains the intervals (by hypothesis)
2. C is closed under complementation: if B is in C, then f^(1)(B) is
Borel; then f^(1)(complement(B)) = complement(f^(1)(B)) is also
Borel, therefore complement(B) is in C.
3. C is closed under countable intersection: if B_1, B_2, B_3, ... are
in C,
then for any n, f^(1)(B_n) is Borel; then
f^(1) (intersection(B_n)) = intersection(f^(1)(B_n)) is also Borel
(since countable intersection of Borels is Borel). Therefore
intersection(B_n) is in C.
(an abbreviated way of explaining 2. and 3. above is: inverse image
preserves set operations such as complementation, intersection)
We have shown that C is a sigmafield containing the intervals.
Therefore it must contain the Borel sets, which are the minimal
sigmafield with these properties. In other words, for any Borel set
B, f^(1)(B) is Borel. But this is exactly the definition of f being
Borelmeasurable. So f is Borel measurable, Q.E.D. (for the lemma).
To finish our proof: If f is monotone, then, as mathtalkga indicated,
the inverse image of any interval under f is also an interval and in
particular a Borel set. Thus, by the lemma, f is Borelmeasurable.
Q.E.D. (for your question)
Hope this helps. If there is anywhere in the proof where you want me
to explain in more detail, please ask!
Regards,
dannidin 
Clarification of Answer by
dannidinga
on
03 Dec 2002 23:48 PST
Madukarga,
Here's a graphical proof that the inverse image of an interval under a
monotone function is an interval:
y axis 

 ****
inverse /  * 
image /  **  *** = graph of function
f
of _  * 
interval   * 
= interval \  * 
\ *** 
 *** 
*  
> x axis
(interval)
Formally, it's clear that f^ ((a,b)) = (f^(a),f^(b)) if f is monotone
increasing (f^ denotes the inverse of f) or f^((a,b)) = (f^(b),f^(a))
if f is decreasing. (f^(a),f^(b)) may be either an open, halfopen, or
closed interval, depending on whether f is rightcontinuous at a and
leftcontinuous at b. But it's always an interval.
Cheers,
dannidin
