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Measurable Functions

  1. Jun 17, 2010 #1
    Rudin Theorem 1.14: If fn: X → [-∞,∞] is measurable, for n = 1,2,3,..., and

    [tex]g = \sup_{n \geq 1} f_n, \qquad h = \limsup_{n \rightarrow \infty} f_n,[/tex]​

    then g and h are measurable. (I can post more details if necessary.)

    I find it amazing that measurable functions handle these limiting processes so well; of course, that's what gives the theory so much power. Logically, the proof follows from the properties of sigma algebras. However, I am wondering if anyone has a more intuitive way of understanding why this theorem is true.

    Put another way, it seems to me very non-obvious that the motivating definitions behind sigma algebras would lead to this result. Am I missing a better way of understanding this concept? Is this just the culminating work after lots of dead ends that somehow led to this? Is this something that I will just have to learn to live with and eventually incorporate from frequent exposure?

    By the way, this could be the start of many questions, as I am planning on slowly working through Rudin. Thanks for any comments, whatever they may be.
     
  2. jcsd
  3. Jun 17, 2010 #2

    mathman

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    Intuitively the operations you described involved countability. That is what makes the sigma in sigma-algebras, so it is not surprising that measurability holds.
     
  4. Jun 18, 2010 #3

    Gib Z

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    BIG Rudin? Good luck! :(
     
  5. Jun 18, 2010 #4
    Is the proof left as an exercise? If not, then I don't know how helpful this is, but for any [tex]a \in \mathbb{R}[/tex], you have, by definition of a measureable function, for every n,

    [tex]f_n^{-1}( (a, \infty] )[/tex]

    is a measurable set. And,

    [tex]g^{-1}( (a, \infty] ) = \{ x \in X : g(x) > a \} = \{ x \in X : \sup_{n \in \mathbb{N}} f_n(x) > a \} = \bigcup_{n \in \mathbb{N}} \{x \in X : f_n(x) > a \} = \bigcup_{n \in \mathbb{N}} f_n^{-1}( (a, \infty] ) [/tex] .

    Since countable unions of measurable sets are also measurable, the term on the right is measurable, and it follows g is a measurable function. (I couldn't have gone from that 3rd expression to the forth if the greater-than were replaced with greater-than-or-equal, convincing yourself why might be helpful, if need be.)

    What Rudin book do you have? If you want to go through exercises and compare solutions every now and then, send me a private message. I've done measure theory but could want to stay sharp with that stuff.


    ...P.S., also, you can see my above argument wouldn't apply to arbitrary (not neccesarily countable) familes of functions.
     
    Last edited: Jun 18, 2010
  6. Jun 18, 2010 #5
    That's the proof, except I'm sure Rudin left out all those equalities between the first and the last, which is fairly reasonable regardless of whether you've read anything by Rudin before. Once you've filled in those steps as some_dude has, the proof of the theorem should at least seem fairly intuitive. But perhaps our notions of what is regarded as "intuitive" differ.
     
  7. Jun 18, 2010 #6
    Thanks for all the responses. I will need luck, but when I say very slowly, I mean at the pace of sub-snail, at least for this summer. After that, I will be taking a course using this book (Real and Complex Analysis), so I'm working on developing the necessary maturity for it right now. Basing intuition off of countability does seem a good idea, as it is essential.

    The proof was not left as an exercise, but it followed the exact same reasoning as yours, some_dude, so good job. I felt a little uncomfortable with it at first, because I wondered to myself how one should know to focus attention on sets of the form (a,∞]; to me, it is a clever idea. I do see that the proof would fail for [a,∞] directly.

    Here is something for you guys to ponder until I find my next point of confusion (or maybe it has already been solved). Clearly complements is very important; it is the reason why proving g is measurable on (a,∞] is enough. I imagine that lack of this is a main reason why this theorem doesn't hold for continuous functions (easy example: fn = sin(nx) at 2π). Is there an easy way to see that this is due to lack of complements, perhaps in the context of that example? Maybe, maybe not. Of course, the theorem does hold for some sequences of continuous functions. Are there certain criteria that determine when or when not?
     
  8. Jun 18, 2010 #7
    I don't get your new question. Continuous functions are measurable, so sups and infs of sequences of continuous functions are also measurable...

    Do you mean when are sups and infs of continuous sequences also continuous? Probably not of interest since the cases where it would be true would be so specific (maybe it holds for monotone, uniformly convergent sequences of functions?). But there's that "uniform-limit-of-continuous-functions-is-continuous" theorem that addresses limits.
     
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