1. ## Measurability/Dilations

Hello. I wasn't going to ask this question on here, since I'm sure I've been over-thinking and just need to take a minute to reevaluate my solution attempt, but I need to get this done ASAP.

Problem Statement: Let $\delta = (\delta_{1}, ..., \delta_{d}), \delta_{i} > 0$. Prove that if $E \subset R^{d}$ is measurable, then so is $\delta E$ where $\delta E =: {(\delta_{1}x_{1}, ..., \delta_{d}x_{d}) : (x_{1}, ..., x_{d}) \in E}$ and $m(\delta E) = \delta_{1}\delta_{2}...\delta_{d}m(E)$.

Attempt at solution: The first thing I proved is that the statement is true for open and closed cubes; this is straight forward. Then we assume $E \subset R^{d}$ is measurable, and so $\forall \epsilon > 0 \exists O \subset R^{d} s.t. E \subset O and m(O-E) < \epsilon$.

Now I'm not sure exactly what to do (in a succinct manner) to prove the result; I've basically scribbled down like 3 pages of scratchwork trying to make everything workout, but I keep getting the wrong inequalities, or I'm not quite sure what I'm doing is legit (very very new to measure theory here).

Obviously if the above is satisfied, then we also have $\delta E \subset \delta O$ and $m(\delta O -\delta E) < \epsilon$. This is clear from the definitions involved. So I guess this proves that $\delta E$ is measurable. (Maybe this requires a more rigorous justification?).

From here, I need to prove that the equality holds, but I'm not really sure how to do this in a rigorous fashion (I've attempted a few time son scratchwork, but I can't get a succinct proof to fall out). I was trying to use the above epsilon-inequality, since I can work with the term $m(\delta O)$ since I have the cube-decomposition lemma for open sets in $R^{d}$ and the proof that the theorem holds for cubes, but I can't work with $m(\delta E)$ in any tangible way (or can I?).

By the way, it is correct to interpret $m(O-E) < \epsilon$ as $m(O) - m(E) < \epsilon$?

Finally, here was my other attempt at the solution, which doesn't really seem rigorous:

$m(\delta E) = \inf m(\delta O) = \delta_{1}...\delta_{d} \inf m(O) = \delta_{1}...\delta{d} m(E)$. The first $\inf$ is of course with respect to coverings of $\delta E$ per the definition of Lebesgue measure (its agreement with the outer measure), and the second is for coverings of just $E$.

Any help would be appreciated,

Thanks!

2. ## Re: Measurability/Dilations

Suppose that $E \subset R^{d}$ is a Lebesgue measureable set. Then $\forall \epsilon > 0$ $\exists$ $O_{\epsilon}$ $s.t.$ $\mu (O_{\epsilon} - E) < \epsilon$ which is to say $\mu (O_{\epsilon}) - \mu (E) < \epsilon$ and since $\epsilon$ is arbitrary, we must have that $E = \inf_{O \supset E} \mu (O)$. Note that $O \subset R^{d}$ is an open set such that $O \supset E$ (but becomes equal to E in the limit represented by the infimum!).
Correspondingly, we have by the above definitions: $\mu (\vec{\delta} O_{\epsilon} - \vec{\delta} E) < \epsilon^{\prime} = \epsilon \cdot max(\delta_{1}, ..., \delta_{d})$. But the $\delta_{max}$ term is finite, and so $\epsilon^{\prime}$ can $\rightarrow 0$. It follows $\vec{\delta} E$ is measurable.
\begin{align*} \mu(\vec{\delta} E) &= \inf_{O \supset \vec{\delta} E} \mu (O) \\ &= \inf_{O \supset E} \mu (\vec{\delta} O) \\ &= \inf_{\{Q_{j}\} \supset E} \mu (\vec{\delta} \bigcup\limits_{j=1}^{\infty} Q_{j}) \\ &= \inf_{\{Q_{j}\} \supset E} \mu (\bigcup\limits_{j=1}^{\infty} \vec{\delta} Q_{j}) \\ &= \inf_{\{Q_{j}\} \supset E} \sum\limits_{j=1}^{\infty} |\vec{\delta} Q_{j}| \\ &= \inf_{\{Q_{j}\} \supset E} \sum\limits_{j=1}^{\infty} \prod\limits_{i=1}^{d} (\delta_{i} b_{i} - \delta_{i} a_{i}) \\ &= \inf_{\{Q_{j}\} \supset E} \delta_{1}\delta_{2}\cdot\cdot\cdot\delta_{d} \sum\limits_{j=1}^{\infty}|Q_{j}| \\ &=\delta_{1}\delta_{2}\cdot\cdot\cdot\delta_{d} \inf_{\{Q_{j}\} \supset E}\sum\limits_{j=1}^{\infty} \mu (Q_{j}) \\ &=\delta_{1}\delta_{2}\cdot\cdot\cdot\delta_{d} \inf_{O \supset E} \mu (O) \\ &=\delta_{1}\delta_{2}\cdot\cdot\cdot\delta_{d} \, m^{*}(E) \\ &= \delta_{1}\delta_{2}\cdot\cdot\cdot\delta_{d}\, \mu(E). \end{align*}.