When does $\lim_{n\to\infty}f(x+\frac{1}{n})=f(x)$ a.e. fail

Question

We know that if $f\in L^1(\mathbb{R})$, then $\|f(\cdot+1/n)-f(\cdot)\|_{L^1}\to 0$ as $n\to \infty$, which implies that there exists a subsequence $f_{n_k}=f(x+\frac{1}{n_k})$ such that $f_{n_k}\to f$ a.e.

My problem is that Is there any $f\in L^1(\mathbb{R})$ function such that the almost convergence result $$ f(x)=\lim_{n\to\infty}f(x+1/n)\quad \text{a.e.} $$ is NOT valid? I intend to believe that there exits such function. Otherwise, there would be a beautiful statement such that $f(x)=\lim_{n\to\infty}f(x+1/n)$ a.e. for any $f\in L^1(\mathbb{R})$.

Thank you for the long commenting below @David C. Ullrich. This question has been reduce to find a compact set $K$ such that for any $x\in K$, there are infinite $n$ such that $x+\frac{1}{n}\not\in K$.

Answer 1

Not an answer - two long comments for the benefit of anyone who's looking for that counterexample that "must" exist:

First, I realized the other day that the sort of construction I'd been trying cannot possibly work. To save others from attempting the impossible:

Say $f$ is a counterexample. Say $(n_j)$ is a bad sequence for $x$ if $$|f(x)-f(x+1/n_j)|>\epsilon$$for all $j$. The sorts of constructions I'd been thinking about, fat Cantor sets, etc., would have had this property if successful:

There exists a single sequence $S=(n_j)$ such that $S$ is a bad sequence for $x$, for all $x$ in some set of positive measure.

That's impossible. Given $S$, the set of $x$ such that $S$ is bad for $x$ must have measure zero. Because there does exist a subsequence of $S$ giving convergence almost everywhere.

(To be specific: I noticed that if $C$ is the standard Cantor set then for all but countably many $x\in C$ we have $x+1/3^n\notin C$. So $\chi_C$ would be a counterexample, except $m(C)=0$. This got me thinking about fat Cantor sets. The above shows that if a fact Cantor set does give a counterexample the explanation cannot be that simple; it cannot be that "uniform".)

In the other direction:

If there is a counterexample then there is a counterexample of the form $\chi_K$ for some compact $K$.

This follows from Lusin's theorem. Say $f$ is a counterexample. There exists $[a,b]$ such that if $E$ is the set of bad points in $[a,b]$ then $m(E)>0$. Suppose that $K\subset[a,b]$ is compact, $m(K)+m(E)>b-a$, $g$ is continuous, and $g=f$ on $K$. If $x\in K\cap E$ then we must have $x+1/n\notin K$ for infinitely many $n$. Since $m(K\cap E)>0$ this shows that $\chi_K$ is a counterexample.

Answer 2

This is also not an answer but too long for a comment as well:

I dont know whether you can always find a measurable set $C$ with $\mu(C) > 0$ such that for almost every $x \in C$ , for infinite $n$ , $x+\frac{1}{n} \notin C$ where $n \in N$

But any constructive proof relying on removal of countable union of intervals $\displaystyle I=\bigcup_{i\in N} I_i$ from [a, b] to construct $C$ is not going to work because at each step we dont know which elements we have to remove from $[a,b]$ because we havent defined $C$ yet.

This means the only chance we have is to take a measurable set $C$ that has already been defined and check whether $\mu(C) > 0$ and for almost every $x \in C$ , for infinite $n$ , $x+\frac{1}{n} \notin C$ where $n \in N$

Answer 3

This question has been reduce to find a compact set $K$ such that for any $x\in K$, there are infinite $n$ such that $x+\frac{1}{n}\not\in K$".

As I have understood David C. Ullrich's answer, it is also required that $m(K)>0$.

I just received a letter from Taras Banakh with the following

Corollary. For any decreasing sequence $(a_n)_{n=1}^\infty$ of positive real numbers with $\lim_{n\to\infty}a_n=0$ and $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$, there exists a compact subset $K\subset\mathbb R$ of positive Lebesgue measure such that $\bigcap_{n=m}^\infty(K-a_n)=\emptyset$.

Corollary follows from Theorem, see below.

Let $\mathcal K(\mathbb R)$ be the space of non-empty compact subsets of the real line, endowed with the Vietoris topology (which is generated by Hausdorff metric). It is well-known that the space $\mathcal K(\mathbb R)$ is locally compact (more precisely, each closed bounded subset of $\mathcal K(\mathbb R)$ is compact.

For every $c>0$ let $\mathcal K_c(\mathbb R)=\{K\in\mathcal K(\mathbb R):\lambda(K)\ge c\}$ be the subspace of consisting of compact sets $K$ of Lebesgue measure $\lambda(K)\ge c$. The regularity (or countable additivity) of the Lebesgue measure $\lambda$ implies that $\mathcal K_c(\mathbb R)$ is a closed subspace in $\mathcal K(\mathbb R)$. Consequently the space $\mathcal K_c(X)$ is locally compact and Polish.

Theorem. Let $(a_n)_{n=1}^\infty$ be a decreasing sequence of positive real numbers such that $\lim_{n\to\infty}a_n=0$ and $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$. Then for every $c>0$ the set $$\{K\in\mathcal K_c(X):\forall m\in\mathbb N\;\bigcap_{n=m}^\infty(K-a_n)=\emptyset\}$$ is dense $G_\delta$ in $\mathcal K_c(X)$.

Proof. It suffices to prove that for every $m\in\mathbb N$ the set $$\mathcal K_m=\{K\in\mathcal K_c(X):\bigcap_{n=m}^\infty(K-a_n)=\emptyset\}$$ is open and dense in $\mathcal K_c(X)$.

To see that $\mathcal K_n$ is open, take any compact set $K\in\mathcal K_m$. By the compactness of $K$, there exists $l>m$ such that $\bigcap_{n=m}^l(K-a_n)=\emptyset$. Then for every $x\in \mathbb R$ there exists $n_x\in[m,l]$ such that $x\notin K-a_{n_x}$. So, we can find a symmetric neighborhood $U_x\subset[-1,1]$ of zero such that $x+U_x$ is disjoint with $U_x+K-a_{n_x}$. By the compactness of the set $L=[-1,1]+(K-a_m)$, there exists a finite subset $F\subset L$ such that $L\subset\bigcup_{x\in F}(x+U_x)$. We claim that the open neighborhood $\mathcal U=\{K'\in\mathcal K_c(X):K'\subset\bigcap_{x\in F}(K+U_x)\}$ of $K$ in $\mathcal K_c(X)$ is contained in the set $\mathcal K_m$. Assuming that $\mathcal U\not\subset\mathcal K_m$, we can find a compact set $K'\in\mathcal U\setminus \mathcal K_m$. Since $K'\notin\mathcal K_m$, there exists a point $z\in \bigcap_{n=m}^\infty (K'-a_n)$. It follows that $z\in K'-a_m\subset [-1,1]+K-a_m=L$ and hence $z\in x+U_x$ for some $x\in F$. Then $$z\in (x+U_x)\cap\bigcap_{n=m}^\infty K'-a_n\subset (x+U_x)\cap (K'-a_{n_x})\subset (x+U_x)\cap (U_x+K-a_{n_x})=\emptyset,$$ which is a desired contradiction, showing that $\mathcal U\subset\mathcal K_m$ and the set $\mathcal K_m$ is open in $\mathcal K_m$.

Next, we show that $\mathcal K_m$ is dense in $\mathcal K_c(\mathbb R)$. Given any $K\in\mathcal K_c(\mathbb R)$ and $\varepsilon>0$ we need to find a set $K'\in\mathcal K_m$ on the Hausdorff distance $d_H(K',K)<\varepsilon$ for $K$.

Choose $k\ge m$ so large that $\frac1k<\frac12\varepsilon$. Consider the cover $\mathcal I_k=\big\{\big[\frac{n}k,\frac{n+1}k\big]:n\in\mathbb N\big\}$ of $\mathbb R$ by closed intervals of length $\frac1n$. The choice of $k$ ensures that the compact set $\tilde K=\bigcup\{I\in\mathcal I_n:I\cap K\ne\emptyset\}$ has $d_H(\tilde K,K)\le \frac1k<\frac12\varepsilon$. Also it is clear that $\tilde K$ has Lebesgue measure $\lambda(\tilde K)>\lambda(K)\ge c$. Choose $p\ge k$ so large that $(1-\frac1p)\cdot\lambda(\tilde K)\ge c$.

Since $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1$, there exists $q>p$ such that $\frac{a_{n+1}}{a_{n}}>1-\frac1{4p}$ for all $n\ge q$. Finally, take $l>q$ such that $\frac2{lk}<a_q$. Consider the open set $$Z:=\bigcup_{z\in\mathbb Z}{\big]}\tfrac{z}{kl},\tfrac{z}{kl}+\tfrac1{pkl}{\big[}$$ and observe that for every interval $I\in\mathcal I_k$ the set $I\setminus Z$ has Lebesgue measure $\lambda(I\setminus Z)=(1-\frac1p)\lambda(I)$ and Hausdorff distance $d_H(I,I\setminus Z)<\frac12\varepsilon$. Consequently, the compact set $K'=\tilde K\setminus Z$ has Lebesgue measure $\lambda(K')=(1-\frac1p)\cdot\lambda(\tilde K)\ge c$ and $d_H(K',\tilde K)=\frac1{2pkl}<\frac12\varepsilon$. Then $d_H(K',K)\le d_H(K,\tilde K)+d_H(\tilde K,K)<\varepsilon$. It remains to prove that $K'\in\mathcal K_m$. Assuming that $K'\notin\mathcal K_m$, we could find a point $x\in\bigcap_{n=m}^\infty (K'-a_n)$. Then $x+a_n\in K'\subset\mathbb R\setminus Z$ for all $n\ge m$. Let $z\in Z$ be the unique integer number such that $\frac{z-1}{lk}<x\le \frac{z}{lk}$. The choice of $l$ guarantees that $\frac2{lk}<a_q$. Then $\frac{z+1}{lk}=\frac{z-1}{lk}+\frac2{lk}<x+a_q$. Let $i\ge q$ be the smallest number such that $x+a_i\ge \frac{z}{lk}+\frac1{plk}$. Then $x+a_{i+1}<\frac{z}{lk}+\frac1{plk}$ and hence $a_{i+1}<\frac{z}{lk}+\frac1{plk}-x<\frac{z}{lk}+\frac1{plk}-\frac{z-1}{lk}= \frac1{lk}+\frac1{plk}<\frac2{lk}$.

On the other hand, $$a_i-a_{i+1}=a_{i+1}\tfrac{a_i}{a_{i+1}} (1-\tfrac{a_{i+1}}{a_i})<\tfrac2{lk}(1-\tfrac1{4p})^{-1}\tfrac1{4p}<\tfrac1{lkp}$$ and hence $x+a_{i+1}>x+a_i-\frac1{lkp}\ge\frac{z}{lk}+\frac1{plk}-\frac1{plk}=\frac{z}{kl}$. Then $\frac{z}{kl}<x+a_{i+1}<\frac{z}{lk}+\frac1{lkp}$ implies that $x+a_{i+1}\in Z$, which is a desired contradiction, completing the proof of $K'\in\mathcal K_m$.$\square$

PS. Also Taras Banakh sent to me the reference to the talk “Theorems of H. Steinhaus, S. Picard and J. Smital” by W. Wilczyński from Ger-Kominek Workshop in Mathematical Analysis and Real Functions 20-21.11.2015.

At the last page are stated the following results.

Theorem. If $A\subset\Bbb R$ is a measurable set, then for each sequence $\{x_n\}_{n\in\Bbb N}$ convergent to $0$ the sequence of characteristic functions $\{\chi_{A+x_n}\}_{n\in\Bbb N}$ converges in measure to $\chi_A$.

Remark. There exists a measurable set $A\subset\Bbb R$ and a sequence $\{x_n\}_{n\in\Bbb n}$ convergent to $0$ such that $\{\chi_{A+x_n}\}_{n\in\Bbb N}$ does nor converge almost everywhere to $\chi_A$.

Theorem. If $A$ has the Baire property, then $\{\chi_{A+x_n}\}_{n\in\Bbb N}$ converges to $\chi_A$ except on a set of the first category.

Answer 4

This is similar to Alex Ravsky's answer, but just answering exactly the question and nothing more. Define

$$K_d=\bigcup_{k=0}^{2^d-1} (k2^{-d}+2^{2-2d},(k+1)2^{-d})\subset [0,1]\qquad\text{for }d\geq 10$$ $$K=\bigcap_{d\geq 10}K_d.$$

It is easy to compute $\mu([0,1]\setminus K_d)=2^{2-d}$ which gives $\mu(K)\geq 1/2.$ I claim that for all $x\in[0,1]$ and for all $d\geq 10$ there is some $n$ with $2^{d-1}\leq n\leq 2^d$ such that $x+1/n\notin K_d$ - this would imply that $x+1/n\notin K$ for infinitely many $n,$ which suffices to answer the question.

Geometrically/intuitively: There is a distance of less than $2^{2-2d}$ between consecutive elements in the list $x+1/2^d, x+1/(2^d-1), \dots, x+1/2^{d-1}.$ This means that as we go along the list we cannot jump over a gap of length $2^{2-2d}.$ And the difference between the first and last elements of the list is $2^{-d}.$ Since $K_d$ has gaps of size $2^{2-2d}$ at every multiple of $2^{-d},$ not every element in the list can be in $K_d.$

Algebraically: set $n=\lfloor 1/(k2^{-d}-x)\rfloor$ where $k=\lceil x2^d\rceil+1.$ Note that $2^{d-1}\leq n\leq 2^d,$ because $1\leq k-x2^d\leq 2.$ Since $1/(n+1)\leq k2^{-d}-x\leq 1/n$ and $1/n-1/(n+1)=1/n(n+1)<1/n^2\leq 2^{2-2d},$ we have $k2^{-d}-x\leq 1/n\leq k2^{-d}+2^{2-2d}-x.$ So $x+1/n\notin K_d$ as required.

Answer 5

I submit that we can show that the a.e. convergence property $$\tag{FALSE} f(x+\tfrac1n)\to f(x)\qquad x\text{-a.e.}\quad \forall\, f\in L^1(\mathbb T). $$ does NOT hold by using Stein's maximal principle (thank you David C.Ullrich for pointing this out). This has the advantage of a simpler proof at the expense that we won't produce an explicit counterexample. (EDIT. See David C.Ullrich's answer for such counterexample).

Definitions. For $f\in L^1(\mathbb T)$ we define the maximal function $$ f^\star(x)=\sup_{n\in \mathbb N} |f(x+\tfrac1n)|.$$ To show that Stein's maximal principle does not hold we need to construct a sequence $(f_k)_{k\in \mathbb N}$ in $L^1(\mathbb T)$ such that $$ \frac{\|f_k^\star\|_{1,\infty}}{\|f_k\|_1}\to +\infty, $$ where the weak $L^1$ norm is defined as $$ \|g\|_{1,\infty} =\sup_{t>0} t|\{x\in\mathbb T\ :\ |g(x)|>t\}|,$$ and $|A|$ for a set $A\subset \mathbb T$ denotes Lebesgue measure.

Proof. The sequence $f_k$ is defined as $$ f_k(x)=k^2\mathbf 1_{\{|x|\le \frac1{k^2}\}}.$$ We have $$f_k^\star(x)\ge \sup_{1\le n \le k} |f_k(x+\tfrac1n)|=\sum_{n=1}^k k^2\mathbf1_{\{|x+\tfrac1n|\le\frac1{k^2}\}}=:k^2\mathbf1_{A_k}.$$ Notice that $A_k$ is the union of $k$ disjoint intervals of length $1/k^2$. Using this observation we can bound from below the weak $L^1$ norm as $$ \|f_k^\star\|_{1,\infty}\ge \|k^2\mathbf1_{A_k}\|_{1,\infty}=k. $$ Since $\|f_k\|_1=1$ for all $k$, the proof is complete. $\square$

Remark. Here we considered functions on the torus $\mathbb T=\mathbb R/\mathbb Z$, because the Stein maximal principle needs a compact group structure. This is not a real limitation, because the statement (FALSE) is local in nature. If such a statement were true for $f\in L^1(\mathbb R)$ it would also be true for functions supported on bounded intervals, and this we disproved.

Answer 6

The solution I have is a bit lengthy, so I will define f and leave out details. For $0<x<1$ $a_n(x)=[(n!)x]-n[(n-1)!x]$ ,[.] being the greatest integer function.The $x=\sum \frac {a_n(x)} {n!}$ and this representation is unique. Let $\epsilon >0$, $\frac 1 {2^{N}} <\epsilon$ and $B_k=\{x:a_{n_k} =0\}$ where $n_k =2^{N+k}$. Let $f=I_A$ where A is the union of the sets $B_k$, $k \geq 1$. Then $f(x-1/n)$ fails to converge to f(x) on $A^{c}$ whose measure is at least $1-\epsilon$.

Answer 7

Heh, people banging head against wall for days, then overnight three apparently correct answers appear. On the one hand, Giuseppe Negro's answer seems the simplest, but on the other hand it depends on Stein's theorem about maximal functions.

The point to this answer is to show that that other answer can easily be converted to an explicit counterexample, based on a vague recollection of how the proof of Stein's theorem goes.

Say $X_1,\dots$ are iid random variables, uniformly distributed on $[0,1]$. Define $$I_k=[X_k,X_k+1/k^2].$$Choose $N$ so $$\sum_N^\infty\frac1{k^2}<1,$$and let $$E=\bigcup_N^\infty I_k$$and $$f=\chi_E.$$

Let $$A_k=\bigcup_{\frac k2\le n\le k}\left(I_k-\frac1n\right).$$

As Giuseppe explains, $$m(A_k)\ge c/k.$$ So Borel-Cantelli shows that almost surely almost every point lies in infinitely many $A_k$. In particular, if $B$ is the set of $x\in[0,1]\setminus E$ such that $x$ lies in infinitely many $A_k$ then $m(B)>0$. But if $x\in B$ then $f(x)\ne\lim f(x+1/n)$.

(In case it's not clear, the point to adding the restriction $n\ge k/2$ to GN's definition of $A_k$ is to ensure that if $x\in A_k$ for infinitely many $k$ then $x\in I_k-1/n$, where $n$ is unbounded.)

Note the appplication of Borel-Cantelli is a little iffy, because $A_k$ need not be a subset of $[0,1]$. If that bothers you, regard the above as a contruction on the torus $\Bbb T=\Bbb R/\Bbb Z$. As Guiseppe points out, a counterexample on $\Bbb T$ implies a counterexample on $\Bbb R$.

Edit: It's been objected that the events $A_k$ are not independent. I didn't intend to give the impression I was saying that they were. In fact $A_k\subset\Bbb T$; if we regard $\Bbb T$ as a probability space then the $A_k$ are in fact "events", but regardless they're not the independent events that I was applying Borel-Cantelli to.

With apologies for not explaining the following at the start - I thought it was well known that if $\sum m(A_k)=\infty$ and you consider "random" translates then you almost surely cover almost every point infinitely often:

For each $k$ and each $x\in\Bbb T$ let $E_{k,x}$ be the event $$x\in A_k.$$That's an honest event, a subset of the probability space where the $X_k$ live. Now if $x$ is fixed it's clear that the events $E_{k,x}$ are independent (since, again for fixed $x$, there exists a set $S_k$ such that $x\in A_k$ if and only if $X_k\in S_k$). And since $X_k$ is uniformly distributed on $\Bbb T$ it's clear that $P(E_{x,k})=m(A_k)$. So for each $x$ it is almost surely true that $x$ lies in infinitely many $A_k$, by Borel-Cantelli. Now Fubini shows that it is almost surely true that almost every $x$ lies in infinitely many $A_k$.