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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 $yin mathbb{R}$, egin{equation} onumber 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 egin{equation} onumber W = left{displaystylesum_{n=1}^{infty}l(I_n)mid Asubseteqdisplaystyle igcup_{n=1}^{infty}I_n ight} end{equation} Then the outer measure of $A$ is, egin{equation} onumber mu^{*}(A)=inf,{W}. end{equation} Now consider $xin W$, then there is a particular collection $hat{I}_n$ that covers $A$, such that $displaystylesum_{n=1}^{infty}l(hat{I}_n)=x$, and that of course is the $inf,{W}$. Further, we see that the collection ${hat{I}_n+y}$ covers $A+y$, that is, $A+ysubseteq displaystyle igcup_{n=1}^{infty}{hat{I}_n + y}$. And from this, we obtain the following outer measure: egin{equation} onumber egin{aligned} mu^{*}(A+y)&=displaystylesum_{n=1}^{infty}l(hat{I}_n+y) &=displaystylesum_{n=1}^{infty}l(hat{I}_n),; ext{since};l; ext{is translation invariant}. &=x. end{aligned} end{equation} And therefore, $Wsubseteqleft{displaystylesum_{n=1}^{infty}I_nmid A+ysubseteq displaystyle igcup_{n=1}^{infty}I_n ight}$, 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 egin{equation} onumber V = left{displaystylesum_{n=1}^{infty}l(I_n)mid A+ysubseteqdisplaystyle igcup_{n=1}^{infty}I_n ight} end{equation} Then the outer measure of $A$ is, egin{equation} onumber mu^{*}(A+y)=inf,{V}. end{equation} Now consider $xin V$, then there is a particular collection $hat{I}_n$ that covers $A+y$, such that $displaystylesum_{n=1}^{infty}l(hat{I}_n)=x$, and that of course is the $inf,{V}$. Further, we see that the collection ${hat{I}_n+(-y)}$ covers $A$, that is, $Asubseteq displaystyle igcup_{n=1}^{infty}{hat{I}_n + (-y)}$. And from this, we obtain the following outer measure: egin{equation} onumber egin{aligned} mu^{*}(A)&=displaystylesum_{n=1}^{infty}l(hat{I}_n+(-y)) &=displaystylesum_{n=1}^{infty}l(hat{I}_n),; ext{since};l; ext{is translation invariant}. &=x. end{aligned} end{equation} And therefore, $Vsubseteqleft{displaystylesum_{n=1}^{infty}I_nmid Asubseteq displaystyle igcup_{n=1}^{infty}I_n ight}$, implying $mu^{*}(A+y)leq mu^{*}(A)$.
   
   Since we have shown both cases, then $mu^{*}(A)=mu^{*}(A+y).hspace{3.7cm} lacksquare$

Reference

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