Measure of self-sums of a particular meager set of measure zero.

Question

The following question is from Makarov, B.M. et. al., Selected Problems in Real Analysis, AMS translations.

Define $$E=\Big\{\sum^\infty_{k=1}\frac{\varepsilon_k}{k!}: \varepsilon_k\in\{0,1\}\Big\}$$ Show that $E$ is a nowhere dense closed set of (Lebesgue) measure zero. Is it true that the set $E+\ldots + E$ ($m$ times) has measure zero for all $m$?

I have solved the first part as follows: For each $n$ define $E_n=\Big\{\sum^{n}_{k=1}\frac{\varepsilon_k}{k!}: \varepsilon_k\in\{0,1\}\Big\}$. $E_n$ has $2^n$ elements, say $0=x^{(n)}_1<x^{(n)}_2=\frac{1}{n!} <\ldots <x^{(n)}_{2^n}=\sum^n_{k=1}\frac{1}{k!}$ Since

$$\Big|\sum^\infty_{k=0}\frac{\varepsilon_k}{k!}-\sum^n_{k=0}\frac{\varepsilon_k}{k!}\Big|\leq\sum^\infty_{k=n+1}\frac{1}{k!}<\frac{1}{(n+1)!}\sum^\infty_{k=0}3^{-k}=\frac32\frac{1}{(n+1)!}$$

The set $F_n=\bigcup_{x\in E_n}\big[x,x+\frac{3}{2}\frac{1}{(n+1)!}\big]$ defines a sequence of closed sets that contain $E$. Also,

$$E_n=E_{n-1}\cup(E_{n-1}+\frac{1}{n!})$$

Hence, for $n\geq 2$, if $x,x'\in E_n$, $x<x'$, are consecutive points in $E_n$, then either $x\in E_{n-1}$ and $x'=x+\frac{1}{n!}$, or $x=x''+\frac{1}{n!}$ and $x'=x''+\frac{1}{(n-1)!}$ for some $x''\in E_{n-2}$. In the former case $$|x'-x|=\frac{1}{n!}>\frac32\frac1{(n+1)!}$$ and in the latter $$|x'-x'|=\frac{1}{(n-1)!}-\frac{1}{n!}=\frac{(n-1)}{n!}>\frac{3}{2}\frac{1}{(n+1)!}$$ The components of $F_n$ are pairwise disjoint intervals and from that,

Claim: $$E=\bigcap^\infty_{n=1}F_n$$ This implies that $E$ is a compact set (closed in particular) and that $\mu(E)\leq \mu(F_n)=\frac32\frac{2^n}{n!}\rightarrow0$ as $n\rightarrow\infty$. Thus, $E$ is also nowhere dense.

Problem(s):

1. My claim that $E=\bigcap_nF_n$ I think is correct but I don't have a precise argument. I would appreciate if someone can help me prove (or disprove) this claim.
2. I have no definite solution for the last part of the problem. My guess is that there is $m$ such that $\lambda(E^{\oplus m})>0$ ($E^{\oplus m}=\{x_1+\ldots + x_m: x_j\in E\}$). My rational for this is that every number in $x\in [0,e-1]$ has a unique representtion of the form $$x-\lfloor x\rfloor=\sum^\infty_{j=2}\frac{a_j}{j!}$$ where for infinitely many $j$'s, $0\leq a_j< j$, but I am at a loss here.

Thank you!

Answer 1

1:

You have an easier claim: $$E\subseteq\bigcap_{n\in\Bbb N}F_n$$From which $\lambda(E)=0$ and "$E$ is nowhere dense" still follows, as $\overline{E}\subseteq\bigcap_{n\in\Bbb N}F_n$ too. I suspect that $E$ is a proper subset of the right hand side, but whether or not this is true is irrelevant for your problem.

2:

For any $m\in\Bbb N$ we have: $$E^{\oplus m}=\left\{\sum_{j=1}^\infty\frac{\epsilon_j}{j!}:\epsilon_j\in\{0,1,\cdots,m\}\right\}$$

We can just mimic your previous work. The proof is really the same. For any $n\in\Bbb N$ and sequence $(\epsilon_j)_{j=1}^\infty$ we have: $$0\le\sum_{j=1}^\infty\frac{\epsilon_j}{j!}-\sum_{j=1}^n\frac{\epsilon_j}{j!}\le m\sum_{j=n+1}^\infty\frac{1}{j!}\le\frac{3m}{2}\cdot\frac{1}{(n+1)!}$$

And you can define $E^m_n$ to be the partial sum set, which now has at most $(m+1)^n$ elements, which is still finite so that: $F^m_n=\bigcup_{x\in E_n}\left[x,x+\frac{3m}{2}\cdot\frac{1}{(n+1)!}\right]$ is still closed, and $E^{\oplus m}\subseteq F_n$ still holds for any $n$, so: $$E^{\oplus m}\subseteq\bigcap_{n\in\Bbb N}F^m_n$$Still holds. Again, it will suffice to show the right hand side is null.

Your work about the intervals being pairwise disjoint was not necessary. I do not care if the intervals are disjoint in this new $F^m_n$, I can still make the bound: $$\lambda(F^m_n)\le(m+1)^n\cdot\frac{3m}{2}\cdot\frac{1}{(n+1)!}\overset{n\to\infty}{\longrightarrow}0$$

Which is a sufficient condition for $\bigcap_{n\in\Bbb N}F^m_n$ to have null measure. Done! $E^{\oplus m}$ is always nowhere dense and of measure zero.

Your idea about any number being expressible as $\sum_j\frac{a_j}{j!}$ is reasonable but the issue is that $(a_j)_j$ may be an unbounded sequence, which would make it not of the form $(\epsilon_j)_j$ for any threshold integer $m$.

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To see why $E$ is closed: suppose $(x_n)_{n\in\Bbb N}$ is a Cauchy sequence of elements of $E$. With no loss of generality we may suppose that for all $k,m\in\Bbb N$, $|x_m-x_{m+k}|<\frac{m+2-e}{(m+1)!}$ holds. Note $\sum_{j\ge m+1}1/j!\le\frac{e-1}{(m+1)!}$ for any such $m$.

Let ‘digits’ conveniently refer to the epsilon-sequence of coefficients for elements of $E$. By induction it follows that for any $m$ and $k$, $x_{m+k}$ shares its first $m$ ‘digits’ with $x_m$. Define now a binary sequence $(\varepsilon_j)_{j\in\Bbb N}$ where $\varepsilon_j$ is defined to be the $j$th ‘digit’ of $x_j$.

It follows that the sequence $(x_n)_{n\in\Bbb N}$ converges to the element of $E$ given by the specified sequence $(\varepsilon_j)_{j\in\Bbb N}$. Hence, $E$ is closed.

The exact same argument can be adapted to show that $E^{\oplus m}$ is closed for any $m\in\Bbb N$ too.

Answer 2

Here is a sketch of a proof that $E=\bigcap_nF_n$.

1. Suppose $x\in \bigcap^\infty_{n=1}F_n$, and let $$x=\sum^\infty_{k=1}\frac{a_k}{k!}$$ where for $k\geq 2$, $0\leq a_k\leq k-1$ for and for infinitely many $k$, $0\leq a_k<k-1$, be its unique factorial expansion.

2. Starting with $n=1$, either $x\in \Big[0,\frac32\frac{1}{2}\Big]$ or $x\in\Big[1,1+\frac32\frac{1}{2}\Big]$. If the former case it follows that $a_1=0$ and in the later case $a_1=1$.

3. By induction, suppose that for al $1\leq n\leq N$, if $x\in\Big[x_n,x_n+\frac32\frac{1}{(n+1)!}\Big]\subset F_n$, where $x_n=\sum^n_{k=1}\frac{\varepsilon_k}{k!}$, $x_k\in\{0,1\}$, then $a_k=\varepsilon_k$ for all $1\leq k\leq n$.

4. As $x\in F_{n+1}$ and the intervals that make up $F_n$ are pairwise disjoint, $x$ is contained in either $\big[x_n,x_n+\frac32\frac{1}{(n+2)!}\big]$, or in $\big[x_n+\frac{1}{(n+1)!},x_n+\frac{1}{(n+1)!}+\frac32\frac{1}{(n+2)!}\big]$ for $x_n+\frac{a}{(k+1)!}\notin F_{n+1}$ when $2\leq a \leq n$ as one can see by noticing that for such $a$, $\frac{a}{(n+1)!}\geq\frac32\frac{1}{(n+2)!}$. It follows that $a_{n+1}=0$ or $a_{n+1}=1$.

As a consequence, $x\in E$.