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Measurable Functions

Date: 09/13/2004 at 15:59:23
From: Paul
Subject: Measurable Functions

I am currently reading a book on Lebesgue measure, I have finished the 
introduction and understand the properties of a measurable set: the 
unique extension of length measure to Lebesgue via the outer measure, 
etc.

I have just got to the chapter on integration, which starts by 
defining a measure space.  That's fine. Then it goes on to define a 
measurable function as a function f: X - [-infinity, +infinity] such 
that the set {x e X : f(x)>a} is measurable for all a e R.

I just don't understand the motivation behind this definition.  I was 
hoping you could provide some insight as to why these particular 
functions are called measurable, or at least give some motivation for 
this definition.  


Date: 09/13/2004 at 17:04:05
From: Doctor Vogler
Subject: Re: Measurable Functions

Hi Paul,

Thanks for writing to Dr. Math.  Have you done any topology?  A 
function is continuous if the inverse image of an open set is open. 
Well, a function is measurable if the inverse image of an open set is
measurable.  But if the inverse image of

  (a, infinity)

is measurable for any real a, then so is the inverse image of

  (-infinity, a]

by taking complements, or

  (-infinity, a)

by taking a countable union of

  (-infinity, a-(1/n)]

or

  [a, infinity)

by taking a countable intersection of

  (a-(1/n), infinity),

and so too

  (a, b)
  [a, b)
  (a, b]
  [a, b]

by taking intersection of two of the above sets, including the 
single-point set

  {a} = [a, a].

So inverse images of any normal set (any real Borel set) are all
measurable.  So the definition says only the minimum needed, but it
actually tells you a whole lot about measurable functions.  In fact,
changing the definition to

  { x e X : f(x) >= a } is measurable for all a e R

or

  { x e X : f(x) < a } is measurable for all a e R

or

  { x e X : f(x) <= a } is measurable for all a e R

doesn't change anything.  However, it is *not* strong enough to say

  { x e X : f(x) = a } is measurable for all a e R

since this does not imply the above.

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 09/15/2004 at 18:13:01
From: Paul
Subject: Measurable Functions

I haven't taken a topology course but I know the usual interpretation 
for a continuous function in real analysis.  This definition is very 
difficult to get my head around. (As an aside, these two definitions 
remind me of Hall's marriage theorem, or daughters of the general 
theorem--because you have conditions that seem nesscessary but not 
sufficient to give nice mathematical objects, or results in terms of 
the theorems.)

So the set in question is the inverse image of a Borel set.  We want 
all collections of sets that can be mapped to a Borel set to be 
measurable.  This seems resonable because then f will not produce a 
Borel set when mapping from a collection that includes an unmeasurable 
set.  And Borel sets are measurable, so if the x's are measurable then 
so will the set of f(x)'s.

Is this all the definition is saying?  If so--why Borel sets?  Why not
say that f will not produce a measurable set when mapping from a
collection of sets that include an unmeasurable member, i.e., that f
is measurable if

  f^-1 (M) = {x e X : f(x) e M}  where M is a measurable set

Also, is this defintion specific to Lebesgue measure, or is it for any 
general measure?  I get the feeling we are using Borel sets due to the 
special relationship between Borel sets and Lebesgue measurable sets, 
i.e., that there exists B such that m(B) = m(L) for all Lebesgue sets 
and such that m(B-L) = 0.

From your definition of continuous I assume I can take it that all 
continuous functions are measurable?

One last thing: I feel the necessity behind of the definition of a 
measurable function but what exactly is useful about "preserving" the 
measure in this way? Can you give me a practical example of a theorem 
or lemma where if f didn't have this property it wouldn't hold?

I appreciate your help.

Regards,

Paul


Date: 09/16/2004 at 10:45:35
From: Doctor Vogler
Subject: Re: Measurable Functions

Hi Paul,

The main point of measure theory and Lebesgue measure is to say how to
integrate functions in such a way that you can take limits of 
functions and get the limit of the integral.  That is,

          b              b
   lim   int f (x) dx = int   lim   f (x) dx
  n->inf  a   n          a   n->inf  n

You see, with standard Riemann integrals, you might often expect that
this should be true, but the limit function on the right is not
integrable.  So we have to be more careful about what we mean by an
integral.  Well, we really just need to be able to get a length of the
intervals on which f(x) is near some value.  That's what Riemann
integrals really are.  

But what if we don't require those sets to be intervals?  Now we 
really just need to be able to give a "length" or measure for the 
*sets* on which f(x) is near some value. 

Unfortunately, it is impossible (and this can be proven) that our
measure be consistent and still work for all sets.  So we have to limit 
ourselves to "measurable" sets, which means that when we look
at the set on which f(x) is near some value, it had better be measurable.

So I think we really only need the inverse image of an open interval
like (a, b) to be measurable.  But that small thing actually gives us
a lot more, some of which is useful, but most of which is not
necessary.  But it comes with the territory.  If the inverse image of
(a, b) is going to be measurable, then we can start taking limits and
such, and we soon find that the inverse image of *any* Borel set is
measurable.  Do we care?  Not often.  But that's why it's not really
important exactly what kind of set you use to define a measurable
function.

Does that straighten anything out?  Does it make more sense now?

- Doctor Vogler, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 09/22/2004 at 18:12:12
From: Paul
Subject: Thank you (Measurable Functions)

Thanks for your time on this. I at least understand or have a better 
intuitive grasp behind the motivation for the definition and what we 
are actually doing.
Associated Topics:
High School Functions
High School Sets

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