Dilatation of a measurable set is a measurable set. 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?
 A: 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. 
A: Let $\{I_k\}_{k=1}^{\infty}$ be a countable collection of sets. Then $\{I_k\}_{k=1}^{\infty}$ covers $E$ if and only if $\{\lambda I_k\}_{k=1}^{\infty}$ covers $\lambda E$. Moreover, if each $I_k$ is an open interval, then each $\lambda I_k$ is an open interval and $l(\lambda I_k)=\mid\lambda\mid l(I_k)$(lengths, $\mid\lambda\mid$ : absolute value) and so $\sum\limits_{k=1}^{\infty}l(\lambda I_k)=\mid\lambda\mid\sum\limits_{k=1}^{\infty}l(I_k)$. So $m^*(\lambda E)=\mid\lambda\mid m^*(E)$.
Let $A\subset\mathbb{R}$. Then $m^*(A)=\mid\lambda\mid m^*(\frac{1}{\lambda} A)=\mid\lambda\mid(m^*(\frac{1}{\lambda}A\cap E)+m^*(\frac{1}{\lambda}A\cap E^\complement))=\mid\lambda\mid\cdot\frac{1}{\mid\lambda\mid}(m^*(A\cap \lambda E)+m^*(A\cap {(\lambda E)}^\complement))=m^*(A\cap \lambda E)+m^*(A\cap {(\lambda E)}^\complement)$ 
so $\lambda E$ is measurable. (the second equality follows from the measurability of $E$
and the third follows from $\lambda(\frac{1}{\lambda}A\cap E)=A\cap\lambda E$ and $\lambda(\frac{1}{\lambda}A\cap E^\complement)=A\cap{(\lambda E)}^\complement$.)
REAL ANALYSIS by H.L. Royden, P.M. Fitzpatrick, Fourth Edition page 33(Proposition 2), 39(Proposition 10)
A: 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?
