Hey there, let's dig into this construction and the questions surrounding it. First, let's anchor ourselves on the key properties that define the Vitali set's paradoxical behavior—this is the bar we're trying to hit with our own sets:

• The set S must have measure $0$
• Countably infinite disjoint copies of S cover a set of finite positive measure

This leads to the classic contradiction: to get a positive-measure union, S would need to have non-zero measure (since summing infinitely many zeros gives zero), but if S had any positive measure, summing infinitely many copies would blow up to infinity—not a finite positive value. But the question here is: can we build any family of sets that satisfy these two criteria, not just the original Vitali set?

Step 1: Base Construction Setup #

Let's start with the unit interval $I=[0,1]$. For any integer $n$, we split $I$ into $n$ equal disjoint subintervals, where the $i^\text{th}$ subinterval $I_i$ corresponds to a "copy" of $I$. For example, if $n=5$:
$$
\begin{align*}
I_1&=[0,\frac{1}{5}), \
I_2&=[\frac{1}{5},\frac{2}{5}), \
I_3&=[\frac{2}{5},\frac{3}{5}), \
I_4&=[\frac{3}{5},\frac{4}{5}), \
I_5&=[\frac{4}{5},1].
\end{align*}
$$

Step 2: Iterative Interval Swapping #

Now we perform a Cantor-set-style iterative operation, but instead of removing intervals, we shift them cyclically:

1. In each $I_i$, remove the middle $\frac{1}{5}$ of every subinterval present.
2. Shift that removed interval to $I_{i+1}$ (with the removed piece from $I_n$ wrapping around to $I_1$).
3. Repeat this process infinitely many times.

After infinite iterations:

• Each resulting set (let's call them $I_i^\infty$) remains bounded within $[0,1]$.
• Measure is preserved through every shift, so each $I_i^\infty$ still has measure $\frac{1}{n}$.
• The sets stay disjoint (since we only shift pieces between disjoint sets), and their union is still the full interval $[0,1]$ (measure $1$).

Step 3: Adjusting to Get Congruent Sets $S_n$ #

The problem with the $I_i^\infty$ sets is they aren't congruent—they start at different positions in the cyclic shift. To fix this, we:

1. Take each $I_i^\infty$ and shift it so its "starting" interval overlaps with $I_1^\infty$'s original position.
2. Union these shifted sets to form $S_n$.

For example, with $n=5$, this gives $S_n$ a measure of $\frac{9}{25}$, and 5 disjoint copies of $S_n$ will cover $[0,1]$.

Concrete Small Example: $n=2$ #

Let's make this tangible with $n=2$, denoting $I_i^k$ as the state of $I_i$ after $k$ iterations:

• Initial state ($k=0$):
  $$I_1^0=[0,\frac{1}{2}), \quad I_2^0=[\frac{1}{2},1]$$
• First iteration ($k=1$): Remove the middle half of each interval and shift to the other set:
  $$
  \begin{align*}
  I_1^1&=[0,\frac{1}{8})\cup[\frac{3}{8},\frac{1}{2})\cup [\frac{5}{8},\frac{7}{8}), \
  I_2^1&=[\frac{1}{8},\frac{3}{8})\cup[\frac{1}{2},\frac{5}{8})\cup[\frac{7}{8},1].
  \end{align*}
  $$
• Second iteration ($k=2$): Repeat the shift on every subinterval:
  $$
  \begin{align*}
  I_1^2&=[0,\frac{1}{32})\cup [\frac{3}{32},\frac{1}{8})\cup[\frac{5}{32},\frac{7}{32})\cup[\frac{3}{8},\frac{13}{32})\cup[\frac{15}{32},\frac{1}{2})\cup[\frac{17}{32},\frac{19}{32})\cup[\frac{5}{8},\frac{21}{32})\cup[\frac{23}{32},\frac{7}{8})\cup[\frac{29}{32},\frac{31}{32}), \
  I_2^2&=[\frac{1}{32},\frac{3}{32})\cup[\frac{1}{8},\frac{5}{32})\cup[\frac{7}{32},\frac{3}{8})\cup[\frac{13}{32},\frac{15}{32})\cup[\frac{1}{2},\frac{17}{32})\cup[\frac{19}{32},\frac{5}{8})\cup[\frac{21}{32},\frac{23}{32})\cup[\frac{7}{8},\frac{29}{32})\cup[\frac{31}{32},1].
  \end{align*}
  $$

As $k\rightarrow\infty$, these become "dust-like" point sets, but their union still has measure $1$. To get $S_2$, we reverse $I_2^\infty$ and overlap it with $I_1^\infty$ on their original intervals. This $S_2$ is symmetric about its midpoint, bounded in $[-\frac{1}{2},1]$, and has measure $\frac{3}{4}$.

Step 4: The Limit Set $S$ #

Now we extend this to all $n=3,4,5,\dots$ and define:
$$S=\lim_{n\rightarrow\infty}S_n$$

We can calculate the measure of $S_n$ as $\frac{2n-1}{n^2}$ (the reasoning: combining $I_1$ and a shifted $I_n$ gives two sets of measure $\frac{1}{n}$, but their overlap subtracts $\frac{1}{n^2}$). Taking the limit:
$$|S|=\lim_{n\rightarrow\infty}|S_n|=\lim_{n\rightarrow\infty}\frac{2n-1}{n^2}=0$$

So S has measure $0$, yet countably many disjoint copies of S cover $[0,1]$ (positive measure). Also, S is bounded within $[-1,2]$ since the cyclic shifts only move intervals by at most ~1 asymptotically.

Key Question: Is $S$ Unmeasurable? #

Wait a second—this seems to hit the Vitali set's paradoxical properties, but I'm wondering if I'm missing a basic piece here. Does this construction actually produce an unmeasurable set? And if it works, could we generate infinitely many such sets just by varying how we remove and swap intervals during the iterative steps?

备注：内容来源于stack exchange，提问作者Locke Demosthenes