<p>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 $Pr\{S^{-1}(A)\} = Pr\{A\}.$</p>

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

<p>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 $Pr\{(\frac{a}{2},\frac{b}{2})\cup (\frac{1}{2}+\frac{a}{2},\frac{1}{2}+\frac{b}{2})\} = Pr\{(a,b)\} = b-a$</p>

<p>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.</p>

CC BY-SA 3.0
2015-01-07T17:15:25.167Z
2015-01-07T17:15:25.167Z
Nikita Toropov
<p><em>Note: This question is related to this <a href="https://economics.stackexchange.com/questions/1880/understanding-the-construction-of-stochastic-processes">question about the construction of stochastic processes.</a> 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 $t \geq 0$.</em></p>

<p>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?</p>

<p>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?</p>

CC BY-SA 3.0
2015-01-04T06:14:25.480Z
In my opinion, this looks like it belongs to math.SE or Cross Validated. The question may be related to tools we use in Economics, but the question is devoid of any economic connection, and also, it is too abstract related to some other "mathematical /econometric" questions posted in here.
Alecos Papadopoulos
CC BY-SA 3.0
2015-01-04T11:02:45.240Z
Yeah, it may well be over the line. On the other hand, there is definitely some precedent for it on this site. In my opinion, I would treat it as one of those questions that would be appropriate on multiple sites---a question where it's up to the OP to decide where to ask it. Also, for what it's worth, it is literally an exercise from an econometrics textbook.
jmbejara
CC BY-SA 3.0
2015-01-04T20:28:22.663Z
It's definitely technical, but is relevant to PhD level economists.  Some economics questions are for everybody, and some are more mechanical and for the profession.
Bryce
<econometrics><time-series><probability><stochastic-processes>
Showing that a transformation is measure preserving
jmbejara
# Showing that a transformation is measure preserving

(Asked by: username_0 on 2015-01-04)

*Note: This question is related to this [question about the construction of stochastic processes.](https://economics.stackexchange.com/questions/1880/understanding-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
$$
\mathbb S(\omega) =
\begin{cases}
2 \omega & \omega \in [0, 1/2) \\
2 \omega - 1 & \omega \in [1/2, 1),
\end{cases}
$$
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?

**Tags:** econometrics, time-series, probability, stochastic-processes
> **Comments:**
> **username_1**: In my opinion, this looks like it belongs to math.SE or Cross Validated. The question may be related to tools we use in Economics, but the question is devoid of any economic connection, and also, it is too abstract related to some other "mathematical /econometric" questions posted in here.  
> **username_0**: Yeah, it may well be over the line. On the other hand, there is definitely some precedent for it on this site. In my opinion, I would treat it as one of those questions that would be appropriate on multiple sites---a question where it's up to the OP to decide where to ask it. Also, for what it's worth, it is literally an exercise from an econometrics textbook.  
> **username_2**: It's definitely technical, but is relevant to PhD level economists.  Some economics questions are for everybody, and some are more mechanical and for the profession.  

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## 1 answers

### Answer by username_3 on 2015-01-07. Score: 5 | Accepted answer

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.


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