Hi Madukar-ga,
First, let us review some definitions. Recall that the Borel
sigma-field (or sigma-algebra) 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 sigma-fields on R (i.e. collections of subsets
that are closed under complementation and countable intersection) that
contain the intervals; you get a sigma-field, 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 sigma-field.
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 sigma-field containing the intervals.
Therefore it must contain the Borel sets, which are the minimal
sigma-field 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
Borel-measurable. So f is Borel measurable, Q.E.D. (for the lemma).
To finish our proof: If f is monotone, then, as mathtalk-ga 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 Borel-measurable.
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
dannidin-ga
on
03 Dec 2002 23:48 PST
Madukar-ga,
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, half-open, or
closed interval, depending on whether f is right-continuous at a and
left-continuous at b. But it's always an interval.
Cheers,
dannidin
|
