Results 1 to 4 of 4

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

  1. #1
    Newbie
    Joined
    Nov 2010
    Posts
    2

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

    Hello,

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

    The full question I'm looking at asks to show that
    $\displaystyle \limsup_n_\rightarrow_\infty \frac{1}{n} \sum_{1}^{k} X_n(x)$ 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 $\displaystyle \limsup_n_\rightarrow_\infty$ 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]?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790
    The set $\displaystyle \{x\in[0,1)\mid X_2(x)=1\}=[1/4,1/2)\cup[3/4,1)$. In general, the $\displaystyle X_n^{-1}(0)$ and $\displaystyle X_n^{-1}(1)$ are some unions of parts that you get by splitting [0, 1] into $\displaystyle 2^n$ equal segments.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2010
    Posts
    2
    Thanks for the reply! So the preimages of the sequence of $\displaystyle X_n (x)$ for any $\displaystyle x \in [0,1]$ 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 $\displaystyle x$=0, does this mean that the function is continuous as long as $\displaystyle x \neq 0$ (almost everywhere)?

    Wouldn't this also mean that $\displaystyle \int_0^1 f(x)dx$ is just 1/2? That seems to make sense, but I'm still very new to binary digit expansions and want to make certain.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,577
    Thanks
    790
    Quote Originally Posted by parg View Post
    Since the only empty set occurs when $\displaystyle x$=0, does this mean that the function is continuous as long as $\displaystyle x \neq 0$ (almost everywhere)?
    I am not sure I understand. What do you mean that the empty set occurs? What function is continuous? The function $\displaystyle X_n(x)$ is not continuous; it is equal to 1 and 0 on alternating segments. I believe $\displaystyle X_n(x)$ has n - 1 points on [0, 1) where it is not continuous.

    Wouldn't this also mean that $\displaystyle \int_0^1 f(x)dx$ is just 1/2?
    I think so.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Digit sum & digit product of number x
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Jan 19th 2011, 08:07 AM
  2. Binary Digit & Decimal value
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Feb 22nd 2010, 07:22 PM
  3. finite and infinite binary expansion of 3/16
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Feb 5th 2010, 09:25 AM
  4. Real Number Analysis: About Metric Spaces
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 6th 2008, 07:39 PM
  5. Replies: 2
    Last Post: Aug 6th 2008, 02:58 AM

Search Tags


/mathhelpforum @mathhelpforum