Measurable FunctionsDate: 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. |
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