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

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. Commented Feb 11, 2016 at 23:48

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 $[0,+\infty)\ni x\mapsto-\sqrt x$.

• 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 Commented Feb 12, 2016 at 0:15
• Precisely. I thought that you knew the two. Commented Feb 12, 2016 at 0:15
• 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. Commented Feb 12, 2016 at 0:17
• shouldn't the second function be $\sqrt{-x}$ ? Commented Feb 15, 2016 at 2:55
• @The Kind Cat No, you don't to consider complex numbers. Commented Feb 15, 2016 at 15:43

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.