# Measure of a Subset of [0, 1]

• April 19th 2012, 10:39 PM
TaylorM0192
Measure of a Subset of [0, 1]
Hello.

Problem Statement: Compute the measure of $E\subset [0, 1]$ obtained by deleting all elements with a digit 4 appearing in their decimal expansions.

Solution Attempt: The first thing I noticed is that this very similar to a Cantor set, where the subset is formed by deleting all elements with the digit 2 appearing in their ternary expansions. This leads me to believe (although it seems unintuitive) that the set has measure zero (although I could be wrong, since I just finished an exercise on "Fat" Cantor sets).

In the first stage of the construction, we remove the interval $I = (0.4, 0.5]$.
In the second stage, we remove the intervals $I = (0.04, 0.05]; (0.14, 0.15]; (0.24, 0.25]; (0.34, 0.35]$ and so on...

So it seems that by the kth stage, what we have left is $9^{k}$ intervals of length $10^{-k}$ which implies that the measure is zero by taking the limit.

But is this rigorous? It seems similar to the cantor set where you get $(\frac{2}{3})^{k} \rightarrow 0$.

I'd like to also be able to show that the limit of the total length removed is equal to one as well.

Any help would be appreciated - thanks! And please...no probabilistic arguments (I'm not at that level yet!)
• April 20th 2012, 02:09 PM
TaylorM0192
Re: Measure of a Subset of [0, 1]
Well, I thought about it more and here is my solution:

We will construct a subset $E \subset R^{d}$ as follows: At each kth step, restrict the subset to values which do not have a digit 4 appearing in the kth digit slot of their respective decimal expansions. Thus, after k steps, what remains is $9^{k}$ segments of length $l = 10^{-k}$. On the other hand, at each individual jth step, we remove $9^{j-1}$ segments of length $10^{-j}$, and therefore by the kth step we have removed a total of $\sum\limits_{j=1}^{k} 9^{j-1}$ segments.

We condense these facts into the two following observations in the limit as $k \rightarrow \infty$:

$(1) \lim_{k \to \infty} (\frac{9}{10})^{k} = 0$ (Total length remaining).
$(2) \lim_{k \to \infty} \sum\limits_{j=1}^{k} \frac{9^{j-1}}{10^{j}} = (\frac{1}{9})\sum\limits_{k=1}^{\infty} (\frac{9}{10})^{k} = \frac{\frac{1}{9}}{1-\frac{1}{9}} = 1$ (Total length removed).

We conclude therefore that $\mu (E) = 0$.

Indeed, it is evident this is true for any similar set since $(\frac{d-1}{d})^{k} \rightarrow 0, k \rightarrow \infty$ where d = # of digits in chosen expansion (binary, ternary, decimal, etc.).