Showing that a transformation is measure preserving Note: This question is related to this question about the construction of stochastic processes. Specifically, it relates to the transformation $\mathbb S: \Omega \rightarrow \Omega$ that is mentioned. The following is an example to help understand such transformations. If such transformation are measure preserving, then the distribution function of $X_t$ is identical for all $t \geq 0$.

If we suppose that $\Omega = [0,1)$ and that $Pr$ is the uniform measure and that 
$$
\mathbb S(\omega) =
\begin{cases}
2 \omega &amp; \omega \in [0, 1/2) \\
2 \omega - 1 &amp; \omega \in [1/2, 1),
\end{cases}
$$
how would we show that $\mathbb S$ is measure-preserving?

It definitely suffices to verify this property for just the open intervals in $\Omega$. But trying to so explicitly for an arbitrary interval $(a,b) \subset \Omega$ is complicated by the fact that it's hard to say more than $\mathbb S((a,b)) \subset [0,1)$ when $(a,b) \subset [0,1/2)$ and that again
$\mathbb S((a,b)) \subset [0,1)$ when $(a,b) \subset [1/2)$. Given this, what is an easy way to go about arguing that
$\mathbb S$ is measure-preserving?
To show that transformation is measure preserving you need to show that full preimage $S^{-1}(A)$ of any set A in the Borel $\sigma$-field on [0,1) is again in the same $\sigma$-field, i.e. it is measurable, and that $Pr\{S^{-1}(A)\} = Pr\{A\}.$

As you have correctly pointed out, it will suffice to show it for any base of usual topology, namely, all open intervals.

It is evident that preimage of $(a,b)$ is $(\frac{a}{2},\frac{b}{2})\cup (\frac{1}{2}+\frac{a}{2},\frac{1}{2}+\frac{b}{2})$. It is, evidently, in Borel $\sigma$-field, as it is a union of two open intervals, and $Pr\{(\frac{a}{2},\frac{b}{2})\cup (\frac{1}{2}+\frac{a}{2},\frac{1}{2}+\frac{b}{2})\} = Pr\{(a,b)\} = b-a$

We proved that preimage of any open interval is measurable and has the same measure as the interval. As open intervals form a base of the usual topology we can extend this result to any measurable set and it completes the proof.