Translation Invariant of Lebesgue Outer Measure

Another proving problem, this time on Real Analysis.

Problem

1. Prove that the Lebesgue outer measure is translation invariant. (Use the property that, the length of an interval $l$ is translation invariant.)

Solution

1. Proof. The outer measure is translation invariant if for $y\in \mathbb{R}$, $$\begin{equation}\nonumber \mu^{*}(A)=\mu^{*}(A+y) \end{equation}$$ Hence, we need to show that Case 1: $\mu^{*}(A)\leq \mu^{*}(A+y)$; and Case 2: $\mu^{*}(A+y)\leq \mu^{*}(A)$.
   
   Case 1: Consider a countable collection $\{I_n\}_{n=1}^{\infty}$, and let $$\begin{equation}\nonumber W = \left\{\displaystyle\sum_{n=1}^{\infty}l(I_n)\mid A\subseteq\displaystyle\bigcup_{n=1}^{\infty}I_n\right\} \end{equation}$$ Then the outer measure of $A$ is, $$\begin{equation}\nonumber \mu^{*}(A)=\inf\,\{W\}. \end{equation}$$ Now consider $x\in W$, then there is a particular collection $\hat{I}_n$ that covers $A$, such that $\displaystyle\sum_{n=1}^{\infty}l(\hat{I}_n)=x$, and that of course is the $\inf\,\{W\}$. Further, we see that the collection $\{\hat{I}_n+y\}$ covers $A+y$, that is, $A+y\subseteq \displaystyle\bigcup_{n=1}^{\infty}\{\hat{I}_n + y\}$. And from this, we obtain the following outer measure: $$\begin{equation}\nonumber \begin{aligned} \mu^{*}(A+y)&=\displaystyle\sum_{n=1}^{\infty}l(\hat{I}_n+y)\\ &=\displaystyle\sum_{n=1}^{\infty}l(\hat{I}_n),\;\text{since}\;l\;\text{is translation invariant}.\\ &=x. \end{aligned} \end{equation}$$ And therefore, $W\subseteq\left\{\displaystyle\sum_{n=1}^{\infty}I_n\mid A+y\subseteq \displaystyle\bigcup_{n=1}^{\infty}I_n\right\}$, implying $\mu^{*}(A)\leq \mu^{*}(A+y)$.
   
   Case 2: Using the same flow of reasoning as in Case 1, consider a countable collection $\{I_n\}_{n=1}^{\infty}$, and let $$\begin{equation}\nonumber V = \left\{\displaystyle\sum_{n=1}^{\infty}l(I_n)\mid A+y\subseteq\displaystyle\bigcup_{n=1}^{\infty}I_n\right\} \end{equation}$$ Then the outer measure of $A$ is, $$\begin{equation}\nonumber \mu^{*}(A+y)=\inf\,\{V\}. \end{equation}$$ Now consider $x\in V$, then there is a particular collection $\hat{I}_n$ that covers $A+y$, such that $\displaystyle\sum_{n=1}^{\infty}l(\hat{I}_n)=x$, and that of course is the $\inf\,\{V\}$. Further, we see that the collection $\{\hat{I}_n+(-y)\}$ covers $A$, that is, $A\subseteq \displaystyle\bigcup_{n=1}^{\infty}\{\hat{I}_n + (-y)\}$. And from this, we obtain the following outer measure: $$\begin{equation}\nonumber \begin{aligned} \mu^{*}(A)&=\displaystyle\sum_{n=1}^{\infty}l(\hat{I}_n+(-y))\\ &=\displaystyle\sum_{n=1}^{\infty}l(\hat{I}_n),\;\text{since}\;l\;\text{is translation invariant}.\\ &=x. \end{aligned} \end{equation}$$ And therefore, $V\subseteq\left\{\displaystyle\sum_{n=1}^{\infty}I_n\mid A\subseteq \displaystyle\bigcup_{n=1}^{\infty}I_n\right\}$, implying $\mu^{*}(A+y)\leq \mu^{*}(A)$.
   
   Since we have shown both cases, then $\mu^{*}(A)=\mu^{*}(A+y).\hspace{3.7cm}\blacksquare$

Reference

1. Royden, H.L. and Fitzpatrick, P.M. (2010). Real Analysis. Pearson Education, Inc.