I would like to show that the set constructed from this algorithm:

  1. from $[0,1]$ I remove an open interval of length $\frac{1}{4}$ center in its middle point $ x=\frac{1}{2} $;
  2. in both remaining closed intervals I remove an open interval of length $\frac{1}{4^2}$ centered in each of the middle points of the 2 closed intervals;
  3. in each of the $2^2$ remaining closed intervals I remove an open interval of length $\frac{1}{4^3}$ centered in... ...

is not a set of measure zero. How could I show this fact, knowing only the definition of a measure zero set?

The definition is: A set $X$ is of measure zero if $\forall \epsilon >0$ there exists a cover of $X$ made by an at most countable system of intervals $I_n$ such that:


I can't find the right way.

Could you give me just an hint, please?

Edit. We could consider as known the following properties:

  1. A single point is a set of measure zero;
  2. A countable union of measure zero set is a measure zero set;
  3. A subset of a measure zero set is of measure zero;
  4. Any closed interval $[a,b]$ with $a<b$ is not a set of measure zero.
  • 2
    $\begingroup$ Think first about how much measure you are removing from the set at each step. $\endgroup$ – rubikscube09 Apr 8 '19 at 20:55
  • $\begingroup$ Yes, I know, but I, in principle, don't know what is the 'measure' of the intervals removed at each step. I could just say that every interval that I remove is not a measure zero set. But this is not useful for us. $\endgroup$ – Nameless Apr 8 '19 at 20:57
  • $\begingroup$ Knowing only the measure-zero definition implies that I don't know what 'this set has measure x' means (if $x\neq 0$). $\endgroup$ – Nameless Apr 8 '19 at 20:58
  • $\begingroup$ @Nameless: the measure of the intervals removed at each step is the sum of their lengths. At step $n$ you remove $2^{n-1}$ intervals of length $\frac 1{4^n}$, so the total length is ???. Sum these up and it is less than $1$ $\endgroup$ – Ross Millikan Apr 8 '19 at 20:59
  • $\begingroup$ Thanks for your reply, but please read my last comment. I already know that proof, but I'm searching another kind of proof. $\endgroup$ – Nameless Apr 8 '19 at 21:01

The set you presented is alike the Cantor set. It is an uncountable union of measure zero sets, so it's not covered by your properties, in particular, it does not follow property 2. It seems to me that, with only those properties, we can't state if it is of measure zero or not.

Edit 2:

Let $S$ be your set.

I will take a look at the complementary set $\bar{S}$ in the interval $[0; 1]$, i.e., $\bar{S}$ is the union of the removed intervals.

Let $I_{j,k}$ be the $k^\textrm{th}$ interval removed in step $j$.

We have $2^{j-1}$ intervals $I_{j,k}$ at step $j$, hence, $k \in \{ 1, 2, \ldots, 2^{j-1} \}$.

Each interval $I_{j,k}$ has lenght $4^{-j}$.

From property 4 we only know that the measure $\mu$ of the interval is not zero, so, $\mu(I_{j,k}) > 0$.
It might be possible that measure could differ with the position of the interval. So, from what I understood, it might be possible, for instance, that $\mu(I_{3,1}) \neq \mu(I_{3,2})$.

So, I will add another property:

  1. The measure of intervals is proportional to their length.

With this property, the initial interval will have measure $c > 0$, i.e., $\mu([0; 1]) = c $. Hence, the measure of the intervals $I_{j,k}$ will be $\mu(I_{j,k}) = c \frac{1}{4^j} $.

So, the measure of $\bar{S}$ is the sum of the measures of the intervals, since they are disjoint. I.e., it is proportional to the total length of the removed intervals:

$$ \mu(\bar{S}) = \sum_{j=1}^{\infty} \sum_{k=1}^{2^{j-1}} c \frac{1}{4^j} = \sum_{j=1}^{\infty} 2^{j-1} c \frac{1}{4^j} = \sum_{j=1}^{\infty} \frac{c}{2} \times \frac{1}{2^j} = \frac{c}{2} \times 1 $$

From this we would get that summing measure of your set with the measure of the complementary set we will have the measure of the interval:

$$ \mu(S) + \mu(\bar{S}) = \mu([0; 1]) $$

Hence, $$ \mu(S) = \frac{c}{2} $$

So, unlike the Cantor set that has measure $0$, the set $S$ has positive measure.

  • $\begingroup$ Thanks for your reply. So, according to you, what are the key properties that will solve the problem? Can they be demonstrated in a relatively simple way? $\endgroup$ – Nameless Apr 9 '19 at 21:31
  • $\begingroup$ @Nameless: I added another property and a computation of the measure. Did you intended something like this? $\endgroup$ – Ertxiem - reinstate Monica Apr 10 '19 at 0:03
  • $\begingroup$ Unfortunately, this proof uses the concept of 'measure' as a function that maps sets in real numbers. I can know only the definition of 'measure zero set' without knowing what a measure is. Anyway I thank you for time you spent. $\endgroup$ – Nameless Apr 10 '19 at 20:27

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