Let $m(\cdot)$ denote the Lebesgue measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Let $T:\mathbb{R}\to\mathbb{R}$ be the map $T(x) = x^{2}$. Evaluate the induced measure $mT^{-1}(A)$ of the set $A$, where
(a) $A = [0,t]$, $t > 0$.
(b) $A = (-\infty,0)$
(c) $A = \{1,2,3,\ldots\}$
(d) $\displaystyle A = \bigcup_{i=1}^{\infty}\left(i^{2},\left(i + \frac{1}{i^{2}}\right)^{2}\right)$
(e) $\displaystyle A = \bigcup_{i=1}^{\infty}\left(i^{2},\left(i + \frac{1}{i}\right)^{2}\right)$
MY ATTEMPT
(a) In this case, $T^{-1}(A) = [-\sqrt{t},\sqrt{t}]$. Thus $mT^{-1}(A) = 2\sqrt{t}$.
(b) In this case, $T^{-1}(A) = \varnothing$. Thus $mT^{-1}(A) = 0$.
(c) In this case, $T^{-1}(A) = \{\pm 1,\pm\sqrt{2},\pm\sqrt{3},\ldots\}$. Thus $mT^{-1}(A) = 0$ due to the countable additivity property of the lebesgue measure and its continuity.
(d) In this case, one has that
\begin{align*} T^{-1}(A_{i}) = \left(-i - \frac{1}{i^{2}},-i\right)\cup\left(i,i + \frac{1}{i^{2}}\right) \Rightarrow mT^{-1}(A_{i}) = \frac{2}{i^{2}} \Rightarrow mT^{-1}(A) = \sum_{i=1}^{\infty}\frac{2}{i^{2}} < +\infty \end{align*}
(e) Similarly, one has that \begin{align*} T^{-1}(A_{i}) = \left(-i - \frac{1}{i},-i\right)\cup\left(i,i + \frac{1}{i}\right) \Rightarrow mT^{-1}(A_{i}) = \frac{2}{i} \Rightarrow mT^{-1}(A) = \sum_{i=1}^{\infty}\frac{2}{i} = +\infty \end{align*}
Is there any theoretical flaw in the solutions I have proposed? Please let me know.