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The following is an exercise from Bruckner's Real Analysis:

5:12.2 Let $E$ be a Lebesgue measurable set of positive measure, and let ${\{x_n}\}$ be any sequence of points in the interval $[0, 1]$. Show that there must exist a point $y$ and a subsequence ${\{x_{n_k}}\}$ so that $y + x_{n_k} \in E$ for all $k$. [Hint: Consider the functions $f_n(t)=χ_E(t − x_n)$ and their integrals.]

I tried the hint of the book but it seems to be not useful: $\int_{[0, 1]} χ_E(t − x_n) d \mu = \mu (x_n+E) = \mu(E)$. By LDCT, $\lim_n \int_{[0, 1]} χ_E(t − x_n) d \mu = \int_{[0, 1]} \lim_n χ_E(t − x_n) d \mu = \int_{[0, 1]} χ_E(t − x) = \mu (x+E) = \mu(E)$; as expected. But how does it guide to the existence of $y$ and a sub-sequence ${\{x_{n_k}}\}$ such that $y + x_{n_k} \in E$ for all $k$? Also how there can be a $y \in [0, 1]$ if the set $E$ 'is spread enough' and includes both points ${\{0,1}\}$?

Matematleta
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  • It would help a little more to have the number of the problem as usually there is association with the material cover in the book. I proposed a solution based in continuity of translations of integrable functions, which has to do with translation invariance of Lebesgue measure and the density of certain nice functions in $L_1$. It seems to me you must be preparing for some exam, is this a qualifying test? – Mittens Apr 22 '21 at 17:31
  • @OliverDiaz, sorry for the slow reply.. yes, midterm exam of MSc Real Analysis course which is from the book Bruckner. I can solve about 85-90% of questions and I upload impenetrable ones. Possibility of any question from the book in the exam is not low and your answer seems to be out of reach to me! Also even when answers are readable reading them takes more than the time I'll have to answer in the exam! .. –  Apr 22 '21 at 20:28

2 Answers2

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I will leave some details to you.

For any $f\in L_1$, define $\tau_hf(\cdot)=f(\cdot-h)$.

Without loss of generality assume $x_n\xrightarrow{n\rightarrow\infty}x$ for some $x\in[0,1]$, and that $0<m(E)<\infty$. As in the hint, consider $\phi=\mathbb{1}_E$ and $\phi_n(\cdot)=\tau_{-x_n}\phi= \phi(\cdot+x_n)$ (Here, and for the rest of my answer $m$ is Lebesgue's measure).

A well known result in integration theory states:

For any $f\in L_1(m)$, the map $t\mapsto \tau_tf(\cdot)=f(\cdot-t)$ is uniformly continuous in $L_1(m)$. That is $$\lim_{h\rightarrow0}\|\tau_{h}f- f\|_1=0 $$ (Uniformity in fact follows from translation invariance since $$\|\tau_{y+h}f-\tau_yf\|_1=\|\tau_hf=f\|_1$$)

Hence $$\lim_n\|\tau_{-x_n}\phi-\tau_{-x}\phi\|_1=0$$

Then, there exists a subsequence $n_k$ such that $\phi_{n_k}\xrightarrow{k\rightarrow\infty}\tau_{-x}\phi=\mathbb{1}_{E+x}$ pointwise almost surely. Since $m(E)>0$ and $\int\tau_{-x}\phi=m(E)$, there $y\in\mathbb{R}$ such that $y\in E-x$ and $\phi_{n_k}(y)\xrightarrow{k\rightarrow\infty}\tau_{-x}\phi(y)=1$. (Here you may need to take a further subsequence to have $y+x_{n'}\in E$ for all $x_{n'}$)

Addendum: If you are not aware of the result I mentioned earlier, here a proof of it that is is more or less standard. the link I provided above has another approach.

The following proof works more generally for $L_p(\mathbb{R},m)$ ($p\geq1$).

We first prove this lemma for continuous functions of compact support $\mathcal{C}_{00}(\mathbb{R})$. Suppose that $g\in\mathcal{C}_{00}(\mathbb{R})$ and that its support $\overline{\{\phi\neq0\}}=\operatorname{supp}(g) \subset B(0,a)$. Then $g$ is uniformly continuous. Given $\varepsilon > 0$, by uniform continuity of there is a $0<\delta<a$ such that $|s-t|<\delta$ implies $$\begin{align} |g(s) - g(t)| &< (\lambda(B(0,3a)))^{-1/p}\varepsilon. \end{align} $$ Hence, by translation invariance of Lebesgue's measure $$\begin{align} \int |g(x-t) - g(x-s)|^p \, dx =\|\tau_t g - \tau_s g\|^p_p = \|\tau_{t-s}g -g\|^p_p < \varepsilon^p. \end{align} $$ Therefore $t\mapsto \tau_tg$ is uniformly continuous. For general $f\in L_p(\mathbb{R},m)$, the conclusion follows from the density of ${\mathcal C}_{00}(\mathbb{R})$ in $L_p(\mathbb{R},m)$.

Mittens
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  • 1- Isn't the result from the link you provided same as : https://math.stackexchange.com/questions/1391528/if-f-is-lebesgue-integrable-on-r-then-lim-h-to-0-int-r-fxh-fxdx ? If so, I already studied it as another exercise of the book! 2- How $\lim_n|\tau_{-x_n}\phi-\tau_{-x}\phi|1=0$ implies that there exists a subsequence $n_k$ such that $\phi{n_k}\xrightarrow{k\rightarrow\infty}\tau_{-x}\phi$ pointwise a.e.? 3- Even if $m(E)>0$, $E$ may not contain an interval about $y+x$ such that a further subsequence to have $y+x_{n'}\in E$ for all $x_{n'}$ (?) –  Apr 22 '21 at 21:11
  • @L.G. Like a said, it is a standard result (it appears in any textbook in Lebesgue integration). I am not surprise that this has been posted several times in MSE. As for your last question, convergence in $L_1$ implies convergence almost surely through a subsequence. That is also a standard result in integration. – Mittens Apr 22 '21 at 21:15
  • The 3rd question? –  Apr 22 '21 at 21:18
  • The set $E-x$ has positive measure ($m(E)=m(E-x)$) and $\phi_{n_k}$ converges to $\mathbb{1}{E-x}$ so there must be $y\in E-x$ for which $\mathbb{1}{E-x_{n_k}}(y)=\phi_{n_k}(y)\xrightarrow{k\rightarrow\infty}\mathbb{1}_{E-x}(y)=1$ – Mittens Apr 22 '21 at 21:23
  • Maybe I'd understand better if I know where the positivity of measure is used? If measure were zero, why the conclusion fails? –  Apr 22 '21 at 21:32
  • For that y, y+x is in E. But why even some subsequence y+x_n exists, all lie in E? There maybe no interval inside E around the point y+x! –  Apr 22 '21 at 21:38
  • $y$ may not be in $E$, in fact, if you read the problem carefully, nowhere does it say that $y\in E$. The positivity $m(E-x)=m(E)>0$ is used to ascertain that there is a point $y\in E-x$, for which $\mathbb{1}{E-x{n_k}}(y)=\phi_{n_k}(y)\xrightarrow{k\rightarrow\infty}\phi(y)=\mathbb{1}{E-x}(y)=1$. This means that for some $K$ large enough, $y\in E-x{n_k}$ for all $k\geq K$. I think you need to take some time and think about this. – Mittens Apr 22 '21 at 21:49
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Assume Lebesgue measure on $\mathbb R.$ Note the following facts:

$(1).\ $ Translation is continuous. That is, if $t\mapsto f_t=f(x-t)$ then $\underset{h\to 0}\lim\|f_{t+h}-f_t\|_{L^1}=0.$

$(2).\ $ If $f_n\to f$ in norm, then there is a subsequence $(f_{n_k})$ that converges pointwise a.e. to $f.$

$(3).\ $ There is a subsequence of $(x_n),$ which we still call $(x_n)$ for convenience, and an $x\in [0,1]$ such that $x_n\to x.$

Now, $(1)$ implies that $\int |\chi_E(t-x_n)-\chi_E(t-x))|dt\to 0$, and it follows from $(2)$ that there is a subsequence of $(\chi_E(\cdot-x_n))$ which we still call $(\chi_E(\cdot-x_n))$ for convenience, such that $\chi_E(t-x_n)\to \chi_E(t-x)$ a.e.

Since $|E|>0,$ there must be a $t\in \mathbb R$ such that $\chi_E(t-x)=1$ and such that $\chi_E(t-x_n)\to \chi_E(t-x),$ so if $n$ is large enough, say $n=N$ then $\chi_E(t-x_n)=1$ also. It follows that $t\in x_n+ E$ for the subsequence $\{x_N,\cdot,x_{N+1},\cdots,\}.$

Matematleta
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