Existence of a subsequence in a set of positive measure for any sequence in [0,1]

Question

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 ${\{0,1}\}$?

Answer 1

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 $x_{n'}$)

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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 $L_p(\mathbb{R},m)$.

Answer 2

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 $\{x_N,\cdot,x_{N+1},\cdots,\}.$