How can I prove that $X$ is $\sigma(X^2)$ measurable?

Question

I have given a random variable $X\sim U[-1,1]$ and want to prove if it is $\sigma(Y)$ measurable where $Y=X^2$

In order to understand the definition of $\sigma(Y)$-measurable a bit better I wanted to work with the definition. I somehow get a bit colfused with the definition. Because normally we write $f:(E,\epsilon)\rightarrow (E',\epsilon')$ is $\epsilon$-$\epsilon'$-measurable and do not only indicate one $\sigma$-algebra.

I know that $\sigma(Y)=\{Y^{-1}(A):A\in \mathfrak{B}([0,1])\}$ where $\mathfrak{B}$ denotes the Borel-$\sigma$-algebra.Now I would say that $X$ is $\sigma(Y)$ measurable if for all $B\in \sigma(Y)$ $$X^{-1}(B)\in \sigma(Y)$$ I would prove this as follows:

Proof Let $B\in \sigma(Y)$, i.e. $B=Y^{-1}(A)$ for some $A\in \mathfrak{B}([0,1])$. Then $X^{-1}(B)=X^{-1}(Y^{-1}(A))=(Y\circ X)^{-1}(A)$

But now I do not see where to conclude.

Could maybe someone help me further?

Thanks a lot

Answer 1

$X$ is not $\sigma (X^2)$-measurable... Having the information on $X^2$ don't allows to have information on $X$... Indeed, $X=f(X^2)$ for some function $f$, then $f$ is not well defined since you would have for example $$-1=f((-1)^2)=f(1)=f(1^2)=1.$$

However, it would have been the case if for exemple $X\sim \mathcal U[0,1]$ or $X\sim U[-1,0]$.