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Let $E$ Lebesgue measurable set. Let $\lambda\in\mathbb{R},\lambda\neq 0$. Let $\lambda E=\left\{\lambda a:a\in E\right\}$.

Show that $\lambda E$ is measurable.

I have this.

Let $f:E\to \lambda E$

$f(a)=\lambda a$ continuos and bijective.

Now $g:\lambda E\to E$ with $g=f^{-1}$. i.e. $g(b)=\frac{b}{\lambda}$ Then $\lambda E=g^{-1}(E)$ and $g$ continuous, therefore $\lambda E$ is measurable set. It is correct?

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    $\begingroup$ You are mixing up Borel measurability and Lebesgue measurability. Inverse image of a Lebesgue measurable set under a continuous map need not be Lebesgue measurable. In this case $g$ is also Lipschitz and you can use this to complete the argument. $\endgroup$ – Kavi Rama Murthy Dec 3 '18 at 23:28
  • $\begingroup$ If $f:E\to E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable? $\endgroup$ – eraldcoil Dec 3 '18 at 23:41
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    $\begingroup$ Yes, because Lipschitz functions map null sets to null sets. $\endgroup$ – Kavi Rama Murthy Dec 3 '18 at 23:43
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First we show that if $A$ has measure zero, then $\lambda A$ has measure zero.

Let $A \subset \mathbb{R}$ have measure zero. Let $\epsilon > 0$ be given. Then there exists a countable collection of intervals $\{(a_i, b_i)\}_{i = 1}^{n}$ such that $A \subset \bigcup_{i = 1}^{\infty}(a_i, b_i)$ and $\sum_{i = 1}^{\infty}(b_i - a_i) < \frac{\epsilon}{|\lambda|}$. Let $a \in A$. Then $a \in (a_i, b_i)$ for some $i \in \mathbb{N}$. Now consider 2 cases:

Case 1 ($\lambda < 0$): If $\lambda < 0$ then $a \in (\lambda b_i, \lambda a_i)$. Thus $\lambda A \subset \bigcup_{i = 1}^{\infty}(\lambda b_i, \lambda a_i)$. We see

$$\sum_{i = 1}^{\infty}\lambda a_i - \lambda b_i = -\lambda \sum_{i = 1}^{n}\left(b_i - a_i\right) < \epsilon.$$

Case 2 ($\lambda > 0$): If $\lambda > 0$ then $a \in (\lambda a_i, \lambda b_i)$. Thus $\lambda A \subset \bigcup_{i = 1}^{\infty}(\lambda a_i, \lambda b_i)$. We see

$$\sum_{i = 1}^{\infty}\lambda b_i - \lambda a_i = \lambda \sum_{i = 1}^{n}\left(b_i - a_i\right) < \epsilon.$$

Let $m^{\ast}$ denote the Lebesgue outer-measure on $\mathbb{R}$. We see $$ m^{\ast}(\lambda A) = \inf\left\{\sum_{i = 1}^{\infty} b_i - a_i : A \subset \bigcup_{i = 1}^{\infty}(a_i, b_i)\right\} < \epsilon.$$

Thus $m^{\ast}(\lambda A) = 0$. Since Lebesgue measure is complete, $\lambda A$ is Lebesgue measurable.

Let $E \subset \mathbb{R}$ be Lebesgue measurable. There is a Borel set $G$ such that $E \subset G$ and $G \backslash E$ has measure zero. Define $g: \mathbb{R} \to \mathbb{R}$ by $g(x) = \frac{1}{\lambda}x$. Clearly $g$ is continuous. Since continuous functions are Borel measurable, $\lambda G = g^{-1}(G)$ is measurable. Since $G \backslash E$ has measure zero, $\lambda G\backslash E$ is Lebesgue measurable by the argument above. Thus $\lambda E = \lambda G \backslash (\lambda G \backslash E)$ is Lebesgue measurable.

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    $\begingroup$ Why $\lambda E=g^{-1}(G)$? $\endgroup$ – eraldcoil Dec 7 '18 at 2:24
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    $\begingroup$ Must be $\lambda G=g^{-1}(G)$ $\endgroup$ – eraldcoil Dec 7 '18 at 4:25
  • $\begingroup$ Yes, you're absolutely right! $\endgroup$ – Sean Haight Dec 7 '18 at 5:58
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If $E$ measurable then for any $\epsilon>0$ exists $G$ open set such that $E\subset G$ and $m(G-E)<\epsilon$. Now, $E\subset G$ then $\lambda E\subset\lambda G$ and $m(\lambda G-\lambda E)=m(\lambda (G-E))=|\lambda|m(G-E)<|\lambda|\epsilon$

Therefore $\lambda E$ is measurable.

It is correct?

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  • $\begingroup$ Why is $\lambda G - \lambda E$ measurable? $\endgroup$ – Sean Haight Dec 6 '18 at 22:35
  • $\begingroup$ Oh. Sorry. Should be $m^{\ast}$ outer measure $\endgroup$ – eraldcoil Dec 6 '18 at 22:37

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