I've encountered the following example in "Ergodic Theory, with a view towards Number Theory", by Einsiedler and Ward. The writers introduce the system $(\mathbb{R}, \mu ,T)$ where $\mu$ and $T$ are defined as follows:
$$\begin{aligned} \mu([a,b]) &= \intop_a^b \frac{dx}{\pi(1+x^2)} \\ T(x) &= \begin{cases} \frac{1}{2}(x-\frac{1}{x}) &: x\neq0 \\ 0\ &: x=0 \end{cases} \end{aligned} $$ The relevant sigma-algebra is not explicitly named, so I guess it is Borel's.
This system is claimed to be measure preserving, and the argument for that claim is that for every $f\in L^1$, the substitution $y=Tx$ proves the equality:
$$ \intop_{-\infty}^\infty f \circ T(x) \frac{dx}{\pi(1+x^2)} =\intop_{-\infty}^\infty f(y) \frac{dy}{\pi(1+y^2)} $$
I can see why the claim follows from the equality, but I can't convince myself this equality is true - I'm pretty sure we have inequality for $f=id$, which is $L^1$ with respect to the measure we deal with.
Am I missing anything obvious?
The equality is correct.
\begin{align} \intop_{-\infty}^\infty f(y) \frac{dy}{\pi(1+y^2)} &= \intop_0^\infty f(T(x)) \frac{d(T(x))}{\pi(1+(T(x))^2)} = \\ &= \intop_0^\infty f \circ T(x) \frac{d(\frac{1}{2}(x-\frac{1}{x}))}{\pi(1+(\frac{1}{2}(x-\frac{1}{x}))^2)} = \\ &= \intop_0^\infty f \circ T(x) \frac{\frac{1}{2}(1+\frac{1}{x^2})dx}{\frac{1}{4}\pi(4+(x-\frac{1}{x})^2)} = \\ &= \intop_0^\infty f \circ T(x) \frac{2(1+\frac{1}{x^2})dx}{\pi(4+x^2-\frac{1}{x^2}-2)} = \\ &= \intop_0^\infty f \circ T(x) \frac{2(1+\frac{1}{x^2})dx}{\pi(2+x^2-\frac{1}{x^2})} = \\ &= \intop_0^\infty f \circ T(x) \frac{2(1+\frac{1}{x^2})dx}{\pi(1+\frac{1}{x^2})(1+x^2)} = \\ &= 2\intop_0^\infty f \circ T(x) \frac{dx}{\pi(1+x^2)} = \\ &= \intop_0^\infty f \circ T(x) \frac{dx}{\pi(1+x^2)} + \intop_0^\infty f \circ T(x) \frac{dx}{\pi(1+x^2)} = \\ &= \intop_0^\infty f \circ T(x) \frac{dx}{\pi(1+x^2)} + \intop_{-\infty}^0 f \circ T(x) \frac{dx}{\pi(1+x^2)} = \\ &= \intop_{-\infty}^\infty f \circ T(x) \frac{dx}{\pi(1+x^2)} \end{align}