A measure preserving transformation Let $(\Bbb{R},\mathcal{B},\mu)$ a probability space,where $\mathcal{B}$ is the Borel sigma algebra,and $d\mu=\frac{1}{\pi} \frac{1}{x^2+1}dx$
Let $T: \Bbb{R} \to \Bbb{R}$ such that $Tx=\frac{1}{2}(x-\frac{1}{x})$ for $x \neq 0.$
Prove that $T$ is $\mu-$invariant,i.e: $\mu(T^{-1}(A))=\mu(A),\forall A \in \mathcal{B}$
The strategy i'm following here is to prove that $T$ preserves the measure of all of the half open intervals $(a,b]$ where $-\infty \leq a <b \leq +\infty$ 
The collection  of these intervals forms a semi-algebra(or semi-ring) that generates the Borel sigma algebra,so it is enough to prove the statement for these sets.(the result will follow from a theorem we proved in class)
I managed to prove that $T$ preserves the measure of bounded intervals of this form.But i have a difficulty proving it for unbounded intervals,especially for intevals of the form $(-\infty,a]$
Can someone help me to finish my proof or guide me to a different solution if it exists?
Thank you in advance.
 A: For any $A=(a, b]$ with $-\infty \leq a <b \leq +\infty$
we can write the measure of set $A$ for $a, b$ finite as
$$
\mu (A) =\int _A d\mu =\int ^{b}_{a} \frac{1}{\pi} \frac{1}{x^2+1}dx
=\dfrac {1}{\pi} (\arctan b -\arctan a)
$$
and we have
$$
\lim _{a\rightarrow -\infty} \arctan a = -\dfrac {\pi}{2}, 
\lim _{b\rightarrow \infty} \arctan b = \dfrac {\pi}{2}
$$
so the above equation holds for all set $A$.
then 
$$
\mu (T^{-1} (A))=\int _{\{x\in T^{-1}  \ (A)\}} \frac{1}{\pi} \frac{1}{x^2+1}dx
$$
$$
=\int ^{b-\sqrt {b^{2}+1}} _{a-\sqrt {a^{2}+1}}\frac{1}{\pi} \frac{1}{x^2+1}dx+\int ^{b+\sqrt {b^{2}+1}} _{a+\sqrt {a^{2}+1}}\frac{1}{\pi} \frac{1}{x^2+1}dx
$$
$$
=\dfrac {1}{\pi} (\arctan (b-\sqrt {b^{2}+1}) -\arctan (a-\sqrt {a^{2}+1}))+
 \dfrac {1}{\pi} (\arctan (b+\sqrt {b^{2}+1}) -\arctan (a+\sqrt {a^{2}+1}))
$$
since
$$
\arctan (b-\sqrt {b^{2}+1}) + \arctan (b+\sqrt {b^{2}+1})= \arctan \dfrac {b-\sqrt {b^{2}+1} + b+\sqrt {b^{2}+1}} {1-(b-\sqrt {b^{2}+1})(b+\sqrt {b^{2}+1})}
$$
$$
=\arctan b
$$
and same for $a$, we are done.
with regard to the problem of $a\rightarrow-\infty$
we have
$$
a-\sqrt {a^{2}+1}\rightarrow -\infty
$$
$$
a+\sqrt {a^{2}+1}\rightarrow 0
$$
Then 
$$
\mu (T^{-1} ((-\infty, b]))=\lim _{a\rightarrow -\infty} \{\int ^{b-\sqrt {b^{2}+1}} _{a-\sqrt {a^{2}+1}}\frac{1}{\pi} \frac{1}{x^2+1}dx+\int ^{b+\sqrt {b^{2}+1}} _{a+\sqrt {a^{2}+1}}\frac{1}{\pi} \frac{1}{x^2+1}dx \}
$$
$$
= \int ^{b-\sqrt {b^{2}+1}} _{-\infty}\frac{1}{\pi} \frac{1}{x^2+1}dx+\int ^{b+\sqrt {b^{2}+1}} _{0}\frac{1}{\pi} \frac{1}{x^2+1}dx \
$$
$$
=  \dfrac {1}{\pi} [\arctan (b-\sqrt {b^{2}+1}) - (- \dfrac {\pi }{2}) + \arctan (b+\sqrt {b^{2}+1}) - \arctan 0 ]
$$
$$
=  \dfrac {1}{\pi} [\arctan b + \dfrac {\pi }{2}]=\lim _{a\rightarrow -\infty} \mu ((a, b])=\mu ((-\infty, b])
$$
same for $b\rightarrow\infty$.
