Let $\delta =(\delta_1,\delta_2,\cdots,\delta_d)$ where $\delta_i>0$, and $E$ is a subset of $\mathbb{R}^d$. Define $\delta E$ by $\delta E = \{(\delta_1x_1,\delta_2x_2,\cdots, \delta_dx_d)|(x_1,x_2,\cdots,x_d)\in E\}$ Prove that if $E$ is meaurable, then $\delta E$ is measurable and $m(\delta_E)= \delta_1\cdots\delta_d m(E)$.
It is the same question as this link: If E is measurable, then $\delta E$ is measurable.
However, the answer on that link works with a linear map, $T$, which has not come in my class. I believe I am supposed to do this by showing $m^{\mathrm{*}}(O - \delta E) < \epsilon$ where $O$ is an open set. Can anyone help me get started on this one? Edit: We also have not covered Borel sets or Sigma Algebras
To ease discussion, I consider the case $d=2$ only. The general case can be treated in exactly the same way. Define a function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ by $f(x,y)=(\frac{1}{\delta_{1}}x,\frac{1}{\delta_{2}}y).$ Firstly, we show that $f$ is measurable. Let $\pi_{1},\pi_{2}:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be the canonical projections onto the first coordinate and the second coordinate respectively. By universal property of product $\sigma$-algebra, to show that $f$ is $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$, it suffices that $\pi_{1}\circ f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ and $\pi_{2}\circ f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ are $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable. Observe that $\pi_{1}\circ f(x,y)=\frac{1}{\delta_{1}}x$, so for any interval $(\alpha,\beta)$, we have $(\pi_{1}\circ f)^{-1}\left((\alpha,\beta)\right)=(\delta_{1}\alpha,\delta_{2}\beta)\times\mathbb{R}\in\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$. Since $\{(\alpha,\beta)\mid\alpha<\beta\}$ is a generator of the $\sigma$-algebra $\mathcal{B}(\mathbb{R})$, it follows that $\pi_{1}\circ f$ is $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable. Similarly, $\pi_{2}\circ f$ is $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable. Therefore, $f$ is $\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})\otimes\mathcal{B}(\mathbb{R})$.
Let $E\subseteq\mathbb{R}^{2}$ be a Borel set. Note that $f^{-1}(E)=\{(\delta_{1}x,\delta_{2}y)\mid(x,y)\in E\}$. For, let $(x',y')\in\mbox{RHS}$, then $(x',y')=(\delta_{1}x,\delta_{2}y)$ for some $(x,y)\in E$. Now $f(x',y')=(\frac{1}{\delta_{1}}x',\frac{1}{\delta_{2}}y')=(x,y)\in E$, so $(x',y')\in f^{-1}(E)=\mbox{LHS}$. On the other hand, let $(x',y')\in\mbox{LHS}$, then $f(x',y')\in E$, i.e., $(\frac{1}{\delta_{1}}x',\frac{1}{\delta_{2}}y')\in E$. Let $x=\frac{1}{\delta_{1}}x'$ and $y=\frac{1}{\delta_{2}}y'$, then $(x,y)\in E$ and $(x',y')=(\delta_{1}x,\delta_{2}y)$. This shows that $(x',y')\in\mbox{RHS}$. It follows that $\{(\delta_{1}x,\delta_{2}y)\mid(x,y)\in E\}$ is a Borel set.
We go to show that $m_{2}(f^{-1}(E))=\delta_{1}\delta_{2}m_{2}(E)$ for any Borel set $E\subseteq\mathbb{R}^{2}$, where $m_{2}$ is the usual Lebesgue measure on $\mathbb{R}^{2}$. Define measures $\mu_{1},\mu_{2}:\mathcal{B}(\mathbb{R}^{2})\rightarrow[0,\infty]$ by $\mu_{1}(E)=m_{2}(f^{-1}(E))$ and $\mu_{2}(E)=\delta_{1}\delta_{2}m_{2}(E)$. Let $n\in\mathbb{N}$ be arbitrary. We go to show that $\mu_{1}(E)=\mu_{2}(E)$ for any $E\in\mathcal{B}((-n,n]\times(-n,n])$ by invoking Dynkin $\pi$-$\lambda$ theorem. Let $\mathcal{P}=\{(-n,a]\times(-n,b]\mid a,b\in(-n,n]\}$ and $\mathcal{D}=\{E\in\mathcal{B}((-n,n]\times(-n,n])\mid\mu_{1}(E)=\mu_{2}(E)\}$. We verify that $\mathcal{P}$ is a $\pi$-class and $\mathcal{D}$ is a $\lambda$-class. Let $A_{1}=(-n,a_{1}]\times(-n,b_{1}]$ and $A_{2}=(-n,a_{2}]\times(-n,b_{2}]$ for some $a_{1},a_{2},b_{1},b_{2}\in(-n,n],$ then $A_{1}\cap A_{2}=(-n,a_{1}\wedge a_{2}]\times(-n,b_{1}\wedge b_{2}]\in\mathcal{P}$, so $\mathcal{P}$ is a $\pi$-class (Here, $x\wedge y=\min(x,y)$). Clearly $\emptyset\in\mathcal{D}$. It can be verified directly that $(-n,n]\times(-n,n]\in\mathcal{D}$. Let $A\in\mathcal{D}$. Since $\mu_{1}$ and $\mu_{2}$ are finite when restricted on $\mathcal{B}((-n,n]\times(-n,n])$, we have that \begin{eqnarray*} \mu_{1}(A^{c}) & = & \mu_{1}((-n,n]\times(-n,n])-\mu_{1}(A)\\ & = & \mu_{2}((-n,n]\times(-n,n])-\mu_{2}(A)\\ & = & \mu_{2}(A^{c}). \end{eqnarray*} This shows that $A\in\mathcal{D}\Rightarrow A^{c}\in\mathcal{D}$. Finally, let $A_{1},A_{2},\ldots\in\mathcal{D}$ be pairwisely disjoint. We have that \begin{eqnarray*} \mu_{1}(\cup_{n}A_{n}) & = & \sum_{n=1}^{\infty}\mu_{1}(A_{n})\\ & = & \sum_{n=1}^{\infty}\mu_{2}(A_{n})\\ & = & \mu_{2}(\cup_{n}A_{n}) \end{eqnarray*} and hence $\cup_{n}A_{n}\in\mathcal{D}$. This shows that $\mathcal{D}$ is a $\lambda$-class. It can be verified directly that $\mathcal{P}\subseteq\mathcal{D}$. Now, by Dynkin's $\pi$-$\lambda$ theorem, we conclude that $\sigma(\mathcal{P})\subseteq\mathcal{D}$. However, it is well-known that $\sigma(\mathcal{P})=\mathcal{B}((-n,n]\times(-n,n])$. Hence, we actually have $\sigma(\mathcal{P})=\mathcal{D}.$ That is, for any Borel subset $E\subseteq(-n,n]\times(-n,n]$, we have that $\mu_{1}(E)=\mu_{2}(E)$. Finally, if $E\subseteq\mathbb{R}^{2}$ is a Borel set, by continuity of measure, we have that \begin{eqnarray*} \mu_{1}(E) & = & \lim_{n\rightarrow\infty}\mu_{1}(E\cap(-n,n]\times(-n,n])\\ & = & \lim_{n\rightarrow\infty}\mu_{2}(E\cap(-n,n]\times(-n,n])\\ & = & \mu_{2}(E). \end{eqnarray*}