# 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?

• 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. Dec 3, 2018 at 23:28
• If $f:E\to E'$ Lipschitz function (with same sigma algebra in Domain and Codomain) and $E$ measurable then $E'$ is measurable? Dec 3, 2018 at 23:41
• Yes, because Lipschitz functions map null sets to null sets. Dec 3, 2018 at 23:43

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.

• Why $\lambda E=g^{-1}(G)$? Dec 7, 2018 at 2:24
• Must be $\lambda G=g^{-1}(G)$ Dec 7, 2018 at 4:25
• Yes, you're absolutely right! Dec 7, 2018 at 5:58

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)

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?

• Why is $\lambda G - \lambda E$ measurable? Dec 6, 2018 at 22:35
• Oh. Sorry. Should be $m^{\ast}$ outer measure Dec 6, 2018 at 22:37