Hi, dicepaulga:
Since this may be your first time using Google Answers, let me point
out that you can "Request Clarification" of my Answer using the button
on this page. I've assumed familiarity on your part with such notions
as the triangle inequality, pointwise convergence, and ?algebras, but
I'll be happy to fill in any gaps I have unintentionally left in the
presentation.
* * * * * * * * * * * * * * * * * * * * *
Let's review the definitions of measurable function and simple
measurable function for the sake of convenient reference, but only
briefly since these are already familiar to you.
A function from a measurable space X to a measurable space Y is
measurable iff the inverse image of each measurable subset in Y is a
measurable subset of X. If Y is a topological space (like R or C)
then unless otherwise specified, the Borel subsets of Y are assumed to
be the relevant ?algebra there.
If Y is also a field, such as the real numbers or the complex numbers,
then a simple measurable function is one that can be written as a
linear combination of (finitely many) characteristic functions ?(A)
for subsets A of X.
The problem of constructing a sequence of simple measurable functions
{s_n} such that the limit s_n(x) = f(x) for every x in X can be
reduced from handling complex values for f, to handling real values,
and even to handling only nonnegative real values. Restricting our
attention to the simplest case may make it easier to visualize the
construction, and from there we will build back out to the original
complex case.
Case: f is a nonnegative real measurable function
=================================================
So let's start with a measurable function f:X > [0,?) and construct
the required sequence of simple measurable functions.
For fixed positive integer n, define:
s_n(x) = (2^(n)) * floor( (2^n) * min(f(x),n) )
Note that s_n(x) = k/(2^n) for nonnegative integer k depending on x:
k = floor( (2^n) * min(f(x),n) )
? (2^n) * n
precisely when x belongs to the measurable subset A(k,n) of X defined by:
A(k,n) = { x in X  f(x) in (2^(n))*[k,k+1) } if k < (2^n)*n
A(k,n) = { x in X  f(x) ? n } if k = (2^n)*n
Therefore s_n = ? (k/(2^n)) ?(A(k,n)) over k = 0,..,(2^n)*n.
Since each A(k,n) is measurable (as the inverse image of a Borel set),
it follows that s_n is a simple measurable function (because we've
expressed it as a finite linear combination of such characteristic
functions).
It remains to be shown that for each x in X:
f(x) = LIMIT s_n(x) AS n > ?
Let ? > 0 be given. Choose N such that N > f(x) and 2^(N) < ?.
Then for every n ? N, we have min(f(x),n) = f(x) and thus:
s_n(x) = (2^(n)) * floor( (2^n) * f(x) )
(2^n) s_n(x) = floor( (2^n) * f(x) )
0 ? (2^n) * (f(x)  s_n(x)) < 1
0 ? f(x)  s_n(x) < (2^(n)) < ?
We have shown that s_n(x) converges to f(x), since ? > 0 was
arbitrary, and that in fact the convergence is monotonically
increasing. In other words the simple measurable functions {s_n}
converge pointwise to f(x) monotonically.
Case: f is any real measurable function
=======================================
It is now easy to extend this result to the case that f is an
arbitrary (not necessarily nonnegative) real function. Any measurable
realvalued function:
f: X > R
can be written as the difference of two nonnegative realvalued functions:
f(x) = g(x)  h(x)
where:
/ f(x) when f(x) > 0
g(x) = <
\ 0 otherwise,
and:
/ f(x) when f(x) < 0
h(x) = <
\ 0 otherwise.
A moment's thought is needed to verify that g is a measurable function
(resp. h is a measurable function) since (for example) the inverse
image of (a,?) under g agrees with the inverse image under f when a >
0 or is all of X when a ? 0.
So apply the earlier result to g and h, obtaining a sequence {s_n} of
simple measurable functions that converge pointwise to g and another
sequence {t_n} that converge pointwise to h. Then clearly {s_n  t_n}
is also a sequence of simple measurable functions, one that converges
pointwise to f = g  h by a standard application of the triangle
inequality (see final case for more details).
[Note that the convergence is no longer montonically increasing, since
we converge monotonically up on the support of g and down on support
of h.]
Case: f is a complex measurable function
========================================
Finally we extend the result to original case, that f is a complex
measurable function. The technique is again to express f as a linear
combination of two real measurable functions using its real and
imaginary parts:
f = Re(f) + i*Im(f)
Note that the inverse image of (a,?) under Re(f) is the inverse image of:
{ z in C  Re(z) in (a,?) }
under f, so that f measurable implies Re(f) measurable. A similar
argument shows that Im(f) is a real measurable function. Find simple
measurable function sequences {s_n} converging pointwise to Re(f) and
{t_n} converging pointwise to Im(f), and take the appropriate linear
combinations of these:
{ s_n + i*t_n }
The pointwise convergence of these linear combinations to f = Re(f) +
i*Im(f) is then a standard argument (pick N large enough so that s_n
is within ?/2 of Re(f) and t_n within ?/2 of Im(f), for all n ? N, and
apply the triangle inequality).
regards, mathtalkga 
Request for Answer Clarification by
dicepaulga
on
12 Sep 2004 10:40 PDT
I read the proof, but i have many questions : First iam asking
question from the First part of proof.
Q1.How did u define s_n(x) = (2^(n)) * floor( (2^n) * min(f(x),n) ) ??
how did the idea come to you ??
Q2. How did you define
A(k,n) = { x in X  f(x) in (2^(n))*[k,k+1) } if k < (2^n)*n
A(k,n) = { x in X  f(x) ? n } if k = (2^n)*n ??
Depending on two cases: when k < (2^n)*n and k = (2^n)*n
Q3. How you eventually wrote
s_n = ? (k/(2^n)) ?(A(k,n)) over k = 0,..,(2^n)*n ??
Q4. How you wrote
0 ? (2^n) * (f(x)  s_n(x)) < 1 ??
I still have to read and understand "II and III part" of the proof and
as suggested by you , i need to write the proof for "part III" myself
by using the sketch provided by you .
But before doing that i need your help to understand the "part I" completely.
I think my questions may sound trivial , but i will be grateful if you
can help me getting the concepts cleared.
This is because i did my Masters in Bangladesh three years back, the
methodology and the level was much different from what iam doing
currently here in usa.
I did not know when i signed up for these subjects that it will be
nearly impossible for me to do well.
Now i have to somehow survive through this semester, my two subjects
are Complex analysis and ordinary differential equations.
Iam looking for somebody who can guide me in doing the assignments as
well as help me in learning the skills for solving these kind of
problems.
I can afford to pay upto $15/hr.
If you are interested or if you know somebody as good as you who can
help me out, Please get in touch via eaddr "dicepaul" a yah
account.
Please let me know how i can contact you?
Thanks for your time.
dice

Clarification of Answer by
mathtalkga
on
12 Sep 2004 11:34 PDT
Hi, dice:
Well, let me put on a pot of coffee and see if I can work through your
points. Please note that under the Google Answers Terms of Service, I
will be unable to communicate with you outside of this site.
Q1: How did I come up with the idea of defining:
s_n(x) = (2^(n)) * floor( (2^n) * min(f(x),n) ) ?
The idea is suggested by a mental picture of a graph of a nonnegative
real function and a grid that divides each unit interval of vertical
distance into 2^n equal subintervals. If you had asked for hints,
rather than a detailed proof, I would have tried to elicit from you
this same idea.
To insure that we have only finitely many "subsets" A(k,n) to deal
with (for any fixed step n), the values of s_n are "capped" at n. The
discrete values that s_n can take are then 0, 1/(2^n), 2/(2^n), ... ,
n*(2^n)/(2^n) = n. As n increases, not only will the maximum possible
values increase, but the granularity of the approximation will shrink.
Although not essential to our proof, the use of the "binary"
divisions 1/(2^n) ensures that the pointwise convergence is
consistently increasing.
Q2. Why did I define:
A(k,n) = { x in X  f(x) in (2^(n))*[k,k+1) } if k < (2^n)*n
A(k,n) = { x in X  f(x) ? n } if k = (2^n)*n
using these two cases, when k < (2^n)*n or k = (2^n)*n ?
The last case, k = n*(2^n) is a bit different from the rest as far as
the inverse image is concerned. As we see from the above, we limit
the size of s_n by taking min(f(x),n) in the definition. As a result
everything in X that is mapped "above" n by f gets lumped together by
s_n. So the form of the set A(k,n) is these points when k = n*(2^n)
at the end. When k < n*(2^n) the function s_n takes the value k on a
subset A(k,n) that is the inverse image under f of a "halfopen"
interval [k/(2^n),(k+1)/(2^n)).
The goal here is to define s_n in a way that makes it a simple
measurable function. So we want to devise some measurable sets A(k,n)
so that a linear combination of their characteristic functions gives a
"good" approximation to f.
Q3. How did I eventually write:
s_n = ? (k/(2^n)) ?(A(k,n)) over k = 0,..,(2^n)*n ??
I apologize if the notation is unclear. I'm writing out explicitly
s_n as a linear combination of the characteristic functions ?(A(k,n)),
using as index of summation k = 0,..,(2^n)*n. Although I failed to
point it out, the sets {A(k,n)} for each fixed n ? 1 partition X.
Therefore the value of s_n(x) is given by the coefficient k/(2^n) for
the unique term A(k,n) containing x.
Q4. How did I conclude:
0 ? (2^n) * (f(x)  s_n(x)) < 1 ??
Here we are showing that pointwise the functions s_n converge to f.
So we can fix a single point x in X and work with that.
Looking back to the steps leading up to this one, I'd restricted n to
be greater than some N > f(x). So the comparison min(f(x),n) beyond
that point gives f(x).
So the definition of s_n(x) is accordingly simplified slightly, to:
s_n(x) = (2^(n)) * floor( (2^n) * f(x) )
Now I multiplied both sides by (2^n):
(2^n) s_n(x) = floor( (2^n) * f(x) )
Because the righthand side is the "floor" of (2^n)*f(x), it is true that:
(2^n) s_n(x) ? (2^n) f(x)
and hence by collecting terms on one side:
0 ? (2^n)*( f(x)  s_n(x) )
But also by definition of the "floor" function, the value (2^n)*f(x)
cannot exceed floor( (2^n)*f(x) ) by as much as 1:
(2^n)*f(x)  floor( (2^n)*f(x) ) < 1
Substituting (2^n) s_n(x) for its equivalent floor( (2^n)*f(x) ) gives:
(2^n)*f(x)  (2^n)*s_n(x) < 1
Combining this with the other half of the "twosided" inequality just
proven yields the result you ask about:
0 ? (2^n) (f(x)  s_n(x)) < 1
Then dividing by 2^n shows that s_n(x) differs from f(x) by at most 2^(n):
0 ? f(x)  s_n(x) < 1/(2^n)
Thus as n increases without limit, s_n(x) approaches f(x).
regards, mathtalkga

Request for Answer Clarification by
dicepaulga
on
12 Sep 2004 14:43 PDT
Thanks for the explanation.
We understand the rules and regulations and i would be glad if you can
help me solve problems through Google Answers Site only.
Iam trying my level best to follow the explanation but looks like it
will take some time for me to understand it fully and crack this
problem,
So for now if you can help me with the Final proof itself.
I have question from the II and III part of the proof,
Q1. How to verify that g is a measurable function(res. h is a measurable) ??
Q2. Mainly For the III part,
a)How to show that Re(f) and Im(f) is measurable ??
b)How to construct simple
measurable function sequences {s_n} converging pointwise to Re(f) and
{t_n} converging pointwise to Im(f)??
c)How to do this :The pointwise convergence of these linear
combinations to f = Re(f) + Im(f) ??
Thanks for your time
dice

Clarification of Answer by
mathtalkga
on
12 Sep 2004 17:44 PDT
Hi, dice:
Let me make a suggestion that I think will, in the long run be helpful
to your progress in this subject (and in graduate school generally).
You've asked some very cogent questions about what I wrote, which
demonstrates I think that you have a basic understanding of the flow
of the argument. Before reading my responses to these points, give
yourself a chance to fill in the "gaps" for yourself. Often I find
that after asking the right questions, which you've definitely done
here, the answers will more or less spontaneously occur to you after a
bit of "unconscious" thinking about the problems.
* * * * * * * * * * * * * * * *
Q1. Verify (in Part II) that g is a measurable function (resp. h measurable).
What I said about this previously is that the inverse image under g of
subset (a,?) is either:
i) the same as the inverse image under f when a > 0, OR
ii) all of X when a ? 0.
The point is that every subset of R the form { y > a } has a
measurable inverse image under g, and this is sufficient to show that
g is a measurable function.
Now by the definition g is a measurable function iff the inverse image
of every measurable subset of R is a measurable subset of X. However
taking complements and countable unions in the ?algebra on X allows
one to deduce this from just the facts that each "basis" inverse image
{ x  g(x) > a} is measurable.
Since you advised me you were familiar with the definition of
measurable function (and since in the realvalued case this is often
taken as the definition), I will leave it at that unless further
clarification is needed.
The situation for h is similar, in that the inverse image under h of
(a,?) is either:
i) the inverse image of (?,a) under f when a > 0, OR
ii) all of X when a ? 0.
Again either of these inverse images are measurable subsets of X.
For an illustration of defining measurable function using only these
restricted subsets, see the MathWorld entry on this topic:
[Measurable Function  from MathWorld]
http://mathworld.wolfram.com/MeasurableFunction.html
* * * * * * * * * * * * * * * *
Q2. For Part III:
a) Show that Re(f) and Im(f) are measurable.
b) Construct simple measurable function sequences {s_n} converging
pointwise to Re(f) and {t_n} converging pointwise to Im(f).
c) Prove the linear combinations {s_n + i*t_n} converge pointwise to:
f = Re(f) + i*Im(f).
Here's what I said previously about Re(f) being measurable, that the
inverse image of (a,?) under Re(f) is the inverse image of:
{ z in C  Re(z) in (a,?) }
under f, so that f measurable implies Re(f) measurable. In other
words I'm again simplifying the "test" for realvalued Re(f) to be
measurable to the criteria involving sets of the form (a,?). Since
the subset of C shown above is open (hence measurable in the sense of
belonging to the Borel ?algebra on C), its inverse image under f is a
measurable subset of X. That's enough (as we've just discussed in
connection with Part II) to show Re(f) is a measurable function.
The argument that Im(f), also a realvalued function, is measurable is
similar. In this case the inverse image of (a,?) under Im(f) is equal
to the inverse image of:
{ z in C  Im(z) in (a,?) }
under f, and again the openness of the above subset in C plus f being
measurable implies that the inverse image in X is measurable.
The construction of the sequences {s_n} and {t_n} converging pointwise
to Re(f) and Im(f) (respectively) is now a direct application of what
we showed in Part II (just as the sequences of simple measurable
functions in Part II were by an application of Part I). That is,
Re(f) and Im(f) have been shown to be realvalued measurable functions
on X, so by Part II there exists:
i) a sequence {s_n} of simple measurable functions that converges
pointwise to Re(f), and
ii) a sequence {t_n} of simple measurable functions that converges
pointwise to Im(f).
As I "hinted" in my original Answer, proving that the combined sequence:
{ s_n + i*t_n }
converges pointwise to f = Re(f) + i*Im(f) is an immediate consequence
of the triangle inequality.
First, though, note that because s_n and t_n are simple measurable
functions, so too is their linear combintation s_n + i*t_n a simple
measurable function (ie. a finite linear combination of characteristic
functions of measurable subsets of X).
To prove pointwise convergences, let x in X be fixed. Let any ? > 0 be
chosen. Choose integer M > 0 s.t. for all n ? M:
 Re(f(x))  s_n(x)  < ?/2
and integer N > 0 s.t. for all n ? N:
 Im(f(x))  t_n(x)  < ?/2
by virtue of the pointwise convergence of {s_n} to Re(f) and {t_n} to Im(f).
Then:
 f(x)  (s_n(x) + i*t_n(x)) 
=  (Re(f(x))  s_n(x)) + i*(Im(f(x))  t_n(x)) 
?  Re(f(x))  s_n(x)  +  Im(f(x))  t_n(x) 
< (?/2) + (?/2) = ?
where I've made use of the triangle inequality and i = 1.
* * * * * * * * * * * * * * * *
Hopefully this gives you enough details to be able to thoroughly
understand the proof that every complex measurable function can be
approximated by simple measurable functions, in the sense of pointwise
convergence. By careful handling of the monotone convergence
property, mentioned but not strictly required by your exercise, one
can show that integrals involving f are also limits of the
counterparts written in terms of the approximating simple measurable
functions. The difficulty here is that "monotonicity" only makes
sense in working with realvalues; the complex field is not ordered,
so to apply the Monotone Convergence Theorem, one must "break down"
the analysis to realvalued components.
Knowing this was one of the reasons I gave the treatment I did here,
to help you see the connection between the real and imaginary parts of
a measurable complex function with the measure/integration theory of
realvalued functions.
The essence of simple measurable functions is that they are
"constants" on some family of measurable subsets (of X).
One can "skip" directly to approximating measurable complex functions
with simple measurable functions by partitioning the complex plane
into "semiopen" rectangles:
{ z  Re(z) in [a,b) and Im(z) in [c,d) }
of small size (growing smaller as n tends to infinity). Although the
"proof" would be short, the "bookkeeping" of:
what points in x are mapped to what values a + ic in C
by the approximating functions s_n would be somewhat complicated. I
think the hierarchical approach here has the best chance of shedding
some light for you on the purpose of the constructions, as opposed to
just reaching the conclusion in the minimum numbers of steps.
regards, mathtalkga
