$E$ Lebesgue measurable implies $E^2$ Lebesgue measurable?

Question

Suppose $E \subset \mathbb{R}$ is Lebesgue measurable. Define $$ E^2 = \{x^2 : x \in E\}. $$ Is $E^2$ Lebesgue measurable as well?

I believe the answer is yes, but I am struggling to prove it. I tried forming interval covers of $E$ for which the sum of the lengths of the intervals is very close to $m(E)$, then considering the square of these intervals. However, if $E$ is unbounded, there is no guarantee these intervals will remain "small".

Try thinking of functions which are measurable that would take $E$ to $E^2$, and then use the properties of measurable functions to get your result. — siegehalver, Feb 11 '16 at 23:48

score 4 · Accepted Answer · answered Feb 11 '16 at 23:57

4

Note for example (using your notation) that $$ E^2=\{x^2:x\in E\cap\mathbb [0,+\infty)\}\cup \{x^2:x\in E\cap\mathbb (-\infty,0]\}. $$ The last two sets are measurable because they are, respectively, inverse images with respect to the continuous functions $[0,+\infty)\ni x\mapsto\sqrt x$ and $[0,+\infty)\ni x\mapsto-\sqrt x$.

answered Feb 11 '16 at 23:57

John B

16,854

So you are saying that because $\sqrt{x}$ and $-\sqrt{x}$ are continuous, they are measurable functions. And then... the inverse image of a measurable set with respect to a measurable function is measurable? @Jonas – camel_case Feb 12 '16 at 00:15
Precisely. I thought that you knew the two. – John B Feb 12 '16 at 00:15
Ah, I did not know those two facts. Thank you so much! I can look in my textbook to understand why those facts are true. – camel_case Feb 12 '16 at 00:17
shouldn't the second function be $\sqrt{-x}$ ? – Asinomás Feb 15 '16 at 02:55
@The Kind Cat No, you don't to consider complex numbers. – John B Feb 15 '16 at 15:43
Oh yeah, thanks, – Asinomás Feb 15 '16 at 20:12
Actually the inverse image of a Lebesgue measurable set with a continuous function need not be Lebesgue measurable. – Aug 02 '17 at 19:51
@ByronSchmuland I acknowledge the mistake, thanks. – John B Aug 02 '17 at 21:58
@JohnB Everybody makes mistakes! – Aug 02 '17 at 22:07
@JohnB: It's easy to see with your argument that it maps Borels to Borels. You still need to show that it maps sets of measure $0$ to sets of measure $)$. That can also be shown, since $f$ is Lipschitz on bounded intervals. – orangeskid Sep 12 '17 at 19:47

score 1 · Answer 2 · answered Aug 02 '17 at 19:47

The square root function $s(x)=\sqrt{x}$ is continuous, and so $s:([0,\infty),{\cal B})\mapsto ([0,\infty),{\cal B})$ is measurable, where ${\cal B}$ is the Borel $\sigma$-field on $[0,\infty)$.

Let $\lambda$ be Lebesgue measure on $([0,\infty),{\cal B})$ and $\mu$ the image of $\lambda$ under the map $s$. Then change of variables tells us that $\mu$ has a density with respect to $\lambda$, i.e., $\mu(B)=\int_B 2x\,\lambda(dx)$. In particular, if $\lambda(B)=0$ then $\mu(B)=0$.

Because $A:=E\cap [0,\infty)$ is Lebesgue measurable there exist $A_1,A_2\in{\cal B}$ so that $A_1\subseteq A\subseteq A_2$ and $\lambda(A_2\setminus A_1)=0.$ In particular, $\mu(A_2\setminus A_1)=0$.

Since $s^{-1}(A_i)\in {\cal B}$ for $i=1,2$ and $s^{-1}(A_1)\subseteq s^{-1}(A) \subseteq s^{-1}(A_2)$ this shows that $s^{-1}(A)$ is Lebesgue measurable since $$\lambda(s^{-1}(A_2)\setminus s^{-1}(A_1))=\lambda(s^{-1}(A_2\setminus A_1)) =\mu(A_2\setminus A_1)=0.$$

Similarly, $s^{-1}(E\cap (-\infty,0))$ is Lebesgue measurable, and taking the union we find that $E^2$ itself is Lebesgue measurable.

$E$ Lebesgue measurable implies $E^2$ Lebesgue measurable?

2 Answers2

Linked

Related