Prove $\lim_{h\rightarrow0}m(E\Delta(E+h)) = 0$ for measurable set $E$ with finite measure

Question

Here is my attempt:

Define $f_n=\chi_{E\Delta(E+ \frac{1}{n})}$. Then $f_n$ decreses with regard to $n$. Since $$m(E\Delta(E+\frac{1}{n})) = \int_\mathbb{R}\chi_{E\Delta(E+ \frac{1}{n})}dm,$$ it suffices to show $$lim_{n\to\infty}\int_\mathbb{R}\chi_{E\Delta(E+ \frac{1}{n})}dm = 0.$$ According to Lebesgue dominated convergence theorem, $$lim_{n\to\infty}\int_\mathbb{R}\chi_{E\Delta(E+ \frac{1}{n})}dm = \int_\mathbb{R}\lim_{n\to\infty}\chi_{E\Delta(E+ \frac{1}{n})}dm.$$ Thus we only need to show $$\chi_{E\Delta(E+ \frac{1}{n})}\overset{a.e.}\to0.$$

EDIT:

By the Approximation Theorem of Measure Theory, $\forall \epsilon > 0$ there exist a finite number of disjoint intervals $\{I_k\}_{k=1}^N$ such that $m(E\Delta(\cup_{k=1}^NI_k)) < \epsilon$. Assume $F = \cup_{k=1}^NI_k$, then $m(E\Delta F) < \epsilon$. Define $f_n = \chi_{E\Delta (E+1/n)}$, $g_n = \chi_{F\Delta (F+1/n)}$.

Step 1. I will show $\int \mid f_n - g_n\mid dm < 2\epsilon$. Since $$(E\Delta (E+1/n))\Delta (F\Delta (F+1/n))\subseteq (F\Delta E)\cup ((F + 1/n)\Delta (E + 1/n))$$ we have $$\int \mid f_n - g_n\mid dm = m((E\Delta (E+1/n))\Delta (F\Delta (F+1/n))) \leq m(F\Delta E) + m((F + 1/n)\Delta (E + 1/n)) < 2\epsilon$$

Step 2. I will show $\lim_{n\to \infty}\int \mid g_n\mid dm = 0$. $\{I_k\}_{k=1}^N$ can be written as $\{[a_k,b_k)\}_{k=1}^N$, then $$\int \chi_{F\Delta (F+1/n)}dm = m(\cup_{i=1}^N([a_i,a_i+1/n)\cup [b_i,b_i+1/n)))\leq \frac{2}{n}N$$ Therefore $$\lim_{n\to \infty}\mid g_n\mid dm = \lim_{n\to \infty} g_n dm = 0$$

Step 3. $$\int \mid f_n - g_n\mid dm < 2\epsilon$$ $$\implies \int \mid f_n\mid dm - \int \mid g_n\mid dm < 2\epsilon$$ $$\implies \lim_{n\to \infty}\int \mid f_n \mid dm < 2\epsilon$$ Let $\epsilon \to 0$, we get $\lim_{n\to \infty}\int\mid f_n\mid dm=0$. Since $f_n$ is non-negative, $\lim_{n\to \infty}\int f_n dm=0$

Answer 1

It is neither true that $f_n $ is decreasing nor is it true that $f_n \to 0$a.e.

By the Approximation Theorem of Measure Theory (Ref. Halmos's book) we can find a finite disjoint union $F$ of intervals of the type $[a_i,b_i), 1 \leq i \leq N$ such that $m (E\Delta F) <\epsilon$. Let $g_n= \chi_{F\Delta (F+\frac 1 n)}$. I will let you verify that $\int |f_n-g_n| <2 \epsilon$ and $\int g_n \leq \frac 2 n N \to 0$.

Answer 2

The key is to exploit translation invariance of Lebesgue's measure. To make some arguments simpler, I use the observation that $\mathbb{1}_{A\Delta B}=|\mathbb{1}_A-\mathbb{1}_B|$ and so, $\lambda(A\Delta B)=\|\mathbb{1}_A-\mathbb{1}_B\|_1$, were $\|\;\|_1$ is the $L_1$ norm.

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A simple proof may be obtained from the following well known result:

Theorem: Suppose $1 \leq p < \infty$, and let $f\in\mathcal{L}_p(\mathbb{R}^n,\lambda_n)$. Then, the mapping $\tau:\mathbb{R}^n\longrightarrow \mathcal{L}_p(\mathbb{R}^n,\lambda_n)$, given by $t \mapsto \tau_t f=f(\cdot-t)$ is uniformly continuous.

Here is a short proof of this Theorem:

First consider continuous functions of compact support. Suppose that $g\in\mathcal{C}_{00}(\mathbb{R}^n)$ and that $\operatorname{supp}(g) \subset B(0,a)$ then, $g$ is uniformly continuous. Given $\varepsilon > 0$, by uniform continuity of there is a $0<\delta<a$ such that $|s-t|<\delta$ implies $$ |g(s) - g(t)| < (\lambda(B(0,3a)))^{-1/p}\varepsilon. $$ Hence, $$ \int |g(x-t) - g(x-s)|^p \, dx =\|\tau_t g - \tau_s g\|^p_p = \|\tau_{t-s}g -g\|^p_p < \varepsilon^p. $$ Therefore $t\mapsto \tau_tg$ is uniformly continuous. For general $f\in\mathcal{L}_p$, the conclusion follows from the density of ${\mathcal C}_{00}(\mathbb{R}^d)$ in $\mathcal{L}_p$.

• For the problem at hand, let $f=\mathbb{1}_{E}$, where $E$ is a measurable set of finite measure. Then $\|\tau_h\mathbb{1}_E-\mathbb{1}_E\|_1=\lambda_d((E+h)\Delta E)\xrightarrow{h\rightarrow0}0$.

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Another proof may be obtained by using the inner regularity of the Lebesgue measure. To make some arguments simpler, I use the observation that $\mathbb{1}_{A\Delta B}=|\mathbb{1}_A-\mathbb{1}_B|$ and so, $\lambda(A\Delta B)=\|\mathbb{1}_A-\mathbb{1}_B\|_1$, were $\|\;\|_1$ is the $L_1$ norm.

For any $\varepsilon>0$ there is a compact set $K\subset E$ such that $\|\mathbb{1}_E-\mathbb{1}_K\|_1<\varepsilon/3$.

By the translation invariance of the Lebesgue measure $$ \|\mathbb{1}_{E+h}-\mathbb{1}_E\|_1\leq \|\mathbb{1}_{E+h}-\mathbb{1}_{K+h}\| +\|\mathbb{1}_{K+h}-\mathbb{1}_{K}\|_1 + \|\mathbb{1}_{K}-\mathbb{1}_E\|_1 = \frac{2\varepsilon}{3}+\|\mathbb{1}_{K+h}-\mathbb{1}_K\|_1$$

Since $K+h\subset K^{|h|}=\{x:d(x,K)\leq|h|\}$, $K^{|h|}$ is compact, and $K^{|h|}\searrow K$ as $|h|\rightarrow0$, it follows that $$\lambda((K+h)\setminus K)\leq \lambda(K^{|h|}\setminus K)=\|\mathbb{1}_{K^{|h|}}-\mathbb{1}_K\|_1\xrightarrow{|h|\rightarrow0}0$$ Here we are using monotone convergence along with the fact that $K\subset K^{\delta}\subset K^{\delta'}$ whenever $0<\delta<\delta'$ which implies that $\bigcap_nK^{\delta_n}= K$ for any positive decreasing sequence $\delta_n\searrow0$.

By translation invariance of Lebesgue's measure $$ \lambda(K\setminus(K+h))=\lambda((K-h)\setminus K)\leq\lambda(K^{|h|}\setminus K)\xrightarrow{|h|\rightarrow0}0 $$ Here we have used the fact that $(A\setminus B)+x = (A+x)\setminus(B+x)$ for all set $A$ and $B$, and points $x$. Consequently, there is $\delta>0$ such that $|h|<\delta$ implies that $\|\mathbb{1}_{K+h}-\mathbb{1}_K\|_1<\frac{\varepsilon}{3}$ and so, $$ \|\mathbb{1}_{E+h}-\mathbb{1}_E\|_1<\varepsilon\quad\text{whenever}\quad|h|<\delta $$

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Answer 3

For any measurable set $E$ and $h,x\in\mathbb{R}$, we have \begin{eqnarray*} 1_{(E+h)\Delta E}(x) & = & \left|1_{E+h}(x)-1_{E}(x)\right|\\ & = & |1_{E}(x-h)-1_{E}(x)|. \end{eqnarray*} We go to prove a general case: For any integrable function $ f:\mathbb{R}\rightarrow\mathbb{R}$, $$\lim_{h\rightarrow0}\int|f(x-h)-f(x)|dx=0.$$

1. If $f=1_{[a,b]}$ the result follows from direct calculation.

2. If $f=\sum_{k=1}^{n}\alpha_{k}g_{k}$, where $g_{k}=1_{[a_{k},b_{k}]}$ (i.e., $f$ is a step function), then \begin{eqnarray*} & & \int|f(x-h)-f(x)|dx\\ & \leq & \sum_{k=1}^{n}|\alpha_{k}|\int|g_{k}(x-h)-g_{k}(x)|dx\\ & \rightarrow & 0 \end{eqnarray*} as $h\rightarrow0$.

3. $f$ is an integrable function. Let $\varepsilon>0$. Choose a step function $g$ (i.e., function of the form in (2)) such that $||f-g||_{1}<\varepsilon$. (We assume the fact that the class of step function is $||\cdot||_{1}$-dense in $L^{1}(\mathbb{R})$.) Choose $\delta>0$ such that $\int|g(x-h)-g(x)|dx<\varepsilon$ whenever $h\in(-\delta,\delta)$. Now, for any $h\in(-\delta,\delta)$, we have \begin{eqnarray*} & & \int|f(x-h)-f(x)|dx\\ & \leq & \int|f(x-h)-g(x-h)|dx+\int|g(x-h)-g(x)|dx+\int|g(x)-f(x)|dx\\ & = & \int|f-g|+\int|g(x-h)-g(x)|dx+\int|f-g|\\ & \leq & 3\varepsilon. \end{eqnarray*} This shows that $\lim_{h\rightarrow0}\int|f(x-h)-f(x)|dx=0$.