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

1. ## 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]?

2. 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.

3. 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.

4. Originally Posted by parg
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.