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

**parg** · November 20th 2010, 09:35 PM

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]?

**emakarov** · November 21st 2010, 12:50 AM

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.

**parg** · November 21st 2010, 01:18 PM

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.

**emakarov** · November 22nd 2010, 06:39 AM

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.

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