## Sunday, 7 September 2014

### 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 $y\in \mathbb{R}$, $$\nonumber \mu^{*}(A)=\mu^{*}(A+y)$$ 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 $$\nonumber W = \left\{\displaystyle\sum_{n=1}^{\infty}l(I_n)\mid A\subseteq\displaystyle\bigcup_{n=1}^{\infty}I_n\right\}$$ Then the outer measure of $A$ is, $$\nonumber \mu^{*}(A)=\inf\,\{W\}.$$ 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: \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} 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 $$\nonumber V = \left\{\displaystyle\sum_{n=1}^{\infty}l(I_n)\mid A+y\subseteq\displaystyle\bigcup_{n=1}^{\infty}I_n\right\}$$ Then the outer measure of $A$ is, $$\nonumber \mu^{*}(A+y)=\inf\,\{V\}.$$ 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: \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} 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 $\mu^{*}(A)=\mu^{*}(A+y).\hspace{3.7cm}\blacksquare$