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Math Help - Real Analysis: Is n-binary digit expansion of a number in [0,1] measurable?

  1. #1
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    Real Analysis: Is n-binary digit expansion of a number in [0,1] measurable?

    Hello,

    If I have a function X_n(x) where x \in [0,1] and X_n(x) is the n-binary digit expansion of x, how can I show that X_n(x) is measurable?

    The full question I'm looking at asks to show that
    \limsup_n_\rightarrow_\infty \frac{1}{n} \sum_{1}^{k} X_n(x) is measurable.

    I know that when we have a sequence of measurable functions, their sum will be measurable, and dividing a measurable function by n will still leave it measurable as well. Finally, when dealing with a sequence of measurable functions than their \limsup_n_\rightarrow_\infty will also be measurable.

    I have most of the puzzle figured out, but have never dealt with n-binary digits before. If I can just show that this expansion is measurable, everything else will fall into place.

    My textbook tells me that if a sequence of functions converges, then it's limit will be measurable. Doesn't that apply in this case, since the n-binary digit expansion is converging to some point in [0,1]?
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  2. #2
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    The set \{x\in[0,1)\mid X_2(x)=1\}=[1/4,1/2)\cup[3/4,1). In general, the X_n^{-1}(0) and X_n^{-1}(1) are some unions of parts that you get by splitting [0, 1] into 2^n equal segments.
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  3. #3
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    Thanks for the reply! So the preimages of the sequence of X_n (x) for any x \in [0,1] will be unions of measurable sets, thus making the function itself measurable? This explanation was much simpler than what I had in mind.

    Since the only empty set occurs when x=0, does this mean that the function is continuous as long as x \neq 0 (almost everywhere)?

    Wouldn't this also mean that \int_0^1 f(x)dx is just 1/2? That seems to make sense, but I'm still very new to binary digit expansions and want to make certain.
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  4. #4
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    Quote Originally Posted by parg View Post
    Since the only empty set occurs when x=0, does this mean that the function is continuous as long as x \neq 0 (almost everywhere)?
    I am not sure I understand. What do you mean that the empty set occurs? What function is continuous? The function X_n(x) is not continuous; it is equal to 1 and 0 on alternating segments. I believe X_n(x) has n - 1 points on [0, 1) where it is not continuous.

    Wouldn't this also mean that \int_0^1 f(x)dx is just 1/2?
    I think so.
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