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If $A$ is a measurable set, $x_n$ is a sequence converging to $0$ (all $x_n$ different from $0$), and $x_n +A = A$ for all $n$, then $\lambda(A) = 0$ or $\lambda(A^c) = 0$ where $\lambda$ is the Lebesgue measure.

I have been trying this problem for quite some time but made no progress. Any hints?

Mittens
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    It seems to be false, since it implies every measurable set has measure $0$ on taking constant $0$ sequence. – Martund May 02 '21 at 09:08
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    My first comment is still valid, since it implies every measurable set has the property that either itself or its complement has measure zero. – Martund May 02 '21 at 09:46
  • That still isn't enough. You need that the sequence is not eventually constant, so by considering subsequences, $x_n \neq 0$ for all $n$. – Brian Moehring May 02 '21 at 10:34
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    Any working hypothesis should be able to reduce to this case of $x_n > 0$ for all $n$ by using the [readily provable] fact the set of all $x \in \mathbb{R}$ such that $A + x = A$ forms a subgroup of $(\mathbb{R}, +)$. If this subgroup has a least positive element $L$, then the statement you're asked to prove may be false, for instance for $$A = A_L = \bigcup_{n\in\mathbb{Z}} \Big(Ln, L(n+0.5)\Big).$$ If it doesn't have a least positive element, then you can extract a positive sequence converging to $0$ from it as long as $A+x = A$ is true for at least one nonzero $x$. – Brian Moehring May 02 '21 at 11:20
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    As a hint: if you can use Lebesgue's density theorem, that may be the easiest method. Assuming $\lambda(A) > 0$ and $x_n > 0$ for all $n$, one can see that there's a dense set of points $X \subseteq [0,1] \cap A$ with density $1$ such that the convergence rate of $\lim_{h\to\infty} \frac{\lambda((a-h,a+h) \cap A)}{2h} = 1$ is uniform for $a \in X$. You can use this to show $\lambda([0,1] \cap A) = 1$ so $\lambda([0,1] \cap A^c) = 0$ and therefore $\lambda(A^c) = 0$. – Brian Moehring May 02 '21 at 12:33

2 Answers2

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Assume $\lambda(A) > 0$, so that we want to prove $\lambda(A^c) = 0$.

First I'll reduce the problem to a [notationally] simpler case:

  • By replacing $x_n$ with $|x_n|$ (checking $x_n \to 0 \iff |x_n|\to 0$ and $x+A=A \iff |x|+A=A$), we may without loss of generality assume $x_n > 0$ for all $n$.
  • By replacing $A$ with $\frac{1}{x_0}A$ and $x_n$ with $\frac{x_n}{x_0}$, we may assume without loss of generality that $x_0 = 1$. [Checking that $\lambda(\frac{1}{x_0}A) > 0$, that $\frac{x_n}{x_0} + \frac{1}{x_0}A = \frac{1}{x_0}A$, that $\frac{x_n}{x_0} > 0$, that $\frac{x_n}{x_0} \to 0$, and that $\lambda\left(\left(\frac{1}{x_0}A\right)^c\right) = 0 \implies \lambda(A^c) = 0$]
  • By Lebesgue's density theorem, there is some $a_0 \in A$ with density $1$. Since $\lambda$ is translation invariant and $x+A = A \iff x+(A-a_0)=(A-a_0)$, we may replace $A$ with $A-a_0$ to assume, without loss of generality, that $a_0=0$.

Since $1 + A = A$, we have $n+A = A$ and therefore $$[n,n+1)\cap A = [n,n+1)\cap(n+A) = n+[0,1)\cap A$$ for every integer $n$. Therefore $$\lambda(A^c) = \sum_{n\in\mathbb{Z}}\lambda([n,n+1) \cap A^c) = \sum_{n\in\mathbb{Z}}\left(1 - \lambda([n,n+1)\cap A)\right) = \sum_{n\in\mathbb{Z}}(1 - \lambda([0,1)\cap A)),$$ so it suffices to show $\lambda([0,1) \cap A) = 1$.

To this end, let $N$ be such that $n \geq N \implies x_n < 1/3$. Then for $n \geq N$, $$\bigcup_{k=1}^{\lfloor (1-x_n)/(2x_n)\rfloor}\Big(2kx_n + (-x_n, x_n)\cap A\Big) \subseteq [0,1) \cap A \subseteq [0,1)$$ where sets in the union are disjoint, so by applying $\lambda$, $$\left(\frac{1-3x_n}{2x_n}\right)\lambda((-x_n,x_n)\cap A) \leq \left\lfloor\frac{1-x_n}{2x_n}\right\rfloor\lambda((-x_n,x_n)\cap A) \leq \lambda([0,1)\cap A) \leq 1$$

Now we may finish by appealing to the density of $0 \in A$.

Let $\varepsilon \in (0,1)$. Then since $0 \in A$ has density $1$ and $x_n > 0$ tends to $0$, there is an $M$ such that $$n \geq M \implies \lambda\Big((-x_n, x_n) \cap A\Big) \geq 2x_n(1-\varepsilon)$$

Therefore, if $n \geq \max(M,N)$, $$(1-3x_n)(1-\varepsilon) = \left(\frac{1-3x_n}{2x_n}\right)(2x_n(1-\varepsilon)) \leq \lambda([0,1) \cap A) \leq 1$$

Let $n \to \infty$ to show $$1-\varepsilon \leq \lambda([0,1) \cap A) \leq 1$$ Then let $\varepsilon \to 0^+$ to show $\lambda([0,1) \cap A) = 1$ which is what we needed to show. It follows that $\lambda(A^c) = 0$.

Brian Moehring
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    There may be a more direct method (I was never known for writing the shortest proofs), but this was the method I kept coming back to as I was thinking about the problem, and it's essentially the one I was hinting towards in my comment. – Brian Moehring May 02 '21 at 21:20
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Sketch of a solution:

I am assuming the ambient space is $(\mathbb{R},\mathscr{B}(\mathbb{R}),\lambda)$.

Here we prove a little stronger result:

If $f$ is a measurable function and $f(x+x_n)=f(x)$ for all $x\in\mathbb{R}$ and $n\in\mathbb{N}$, then there is $c\in\mathbb{R}$ such that $f(x)=c$ $\lambda$--a.s. By considering bounded functions first, can be extended to any measurable function.

As pointed out in the comments, the set $$G:=\{a\in\mathbb{R}:f(x+a)=f(x) \,\text{for all}\,x\in\mathbb{R}\}$$ is a additive subgroup of $\mathbb{R}$. By assumption, $\{x_n:n\in\mathbb{N}\}\subset G$. The assumption $x_n\neq0$ for all $n$ and $x_n\rightarrow0$, implies $G$ is not of the form $h\mathbb{Z}$, but a dense additive subgroup of $\mathbb{R}$.

Consider the function $\phi(x)=\frac{1}{2}\mathbb{1}_{[-1,1]}(x)$, and for any $\varepsilon>0$ let $\phi_\varepsilon(x)=\varepsilon^{-1}\phi(x\varepsilon^{-1})$ Suppose $f\in L_\infty(\mathbb{R})$. Then, $f\in\mathcal{L}^1_{loc}(\mathbb{R})$, $$f*\phi_\varepsilon(x)=\int \phi_\varepsilon(y)f(x-y)\,dy$$ is a continuous function (in fact uniformly continuous by Young's convolution theorem, but that is not that important for our purposes), and $f*\phi_{\varepsilon}(x)\xrightarrow{\varepsilon\rightarrow0}f(x)$ at every Lebesgue point of $f$ (and hence a.s.).

For any $a\in G$ and $x\in\mathbb{R}$ $$ f*\phi_\varepsilon(x+a)=\int \phi_\varepsilon(y)f(x+a-y)\,dy=\int \phi_\varepsilon(y)f(x-y)\,dy = f*\phi_\varepsilon(x) $$ As $G$ is dense in $\mathbb{R}$, this means that $f*\phi_\varepsilon$ is a constant function, say $c_\varepsilon$. From this, it follows that $f$ is constant a.s. (one can for example substitute $f$ by $f\mathbb{1}_K$ for some compact to get that $(f\mathbb{1}_K)*\phi_\varepsilon\xrightarrow{\varepsilon\rightarrow0}f\mathbb{1}_K$ in $L_1$, and then through a subsequence $\varepsilon_n\rightarrow0$ get convergence a.s.)

To conclude, if $f=\mathbb{1}_A$ with $A+x_n=A$ for all $n\in\mathbb{N}$, then $\mathbb{1}_A$ is constant a.s. Hence, either $\lambda(A)=0$ or $\lambda(A^c)=0$.

Edit: Doing a quick review of the problem again in MSE I found that there are at least three postings 1, 2, and 3 that addressed this questions. The solutions in those postings are different to the ones presented here so far, but very interesting too. I also just learnt that the OP is appears as an exercise in Real and Complex Analysis, Rudin, W. chapter 7.

Mittens
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  • +1: This is the proof I originally wanted to write because it made intuitive sense that something like this should hold, but I kept on failing when I tried to write it. I eventually settled for the one I wrote. Thanks for filling out the details (I'm not sure I'd call it a "sketch" except to a logician, to whom all our proofs are "sketches") – Brian Moehring May 03 '21 at 01:41