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$T: [0,1)^{2}\rightarrow[0,1)^{2}$ by

$T(x,y) = (2x,\frac{y}{2})$, with $0 \leq x < \frac{1}{2}$

and

$T(x,y) = (2x-1, \frac{y+1}{2})$, with $\frac{1}{2} \leq x < 1$

In class we said this $T$ is

a) invertivle

b) measurable

c) measure preserving.

My last Analysis and Stochasitc class is quite a time ago and so I do not see that easy how $T$ fulfills a)-c)....

I hope someone can tell my why this is true, or give me some hints how I can find that out myseld :)

Best, Luca

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To check it's invertible, you could simply find the inverse. It's a piecewise linear map so it shouldn't be too bad. You know it's measurable because it's piece-wise continuous. You should check the measure-preserving property by drawing what happens to points under $T$. – Christopher A. Wong Oct 9 '12 at 23:37

1 Answer

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youre taking the square, smooshing it down and stretching it out. then cut off the excess and put it on top. (draw a picture). clearly invertible (just undo what you did). preserves measure (doesnt change areas, should be clear, what happens to the area of a little rectangle under $T^{-1}$?).

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Hey! First thanks for the Answers! My $T^{-1}(x,y) = (\frac{x}{2},2y)$ for $0 \leq y < \frac{1}{2}$ And $= (\frac{x+1}{2},2y-1)$ for $\frac{1}{2} \leq y < 1$. – Luca Oct 10 '12 at 0:40
I pushed enter too early ;) Hey! First thanks for the Answers! The baker trafo describes taking a rectangle, strech it, fold it and turn it a quater. Hence if I name the two halfs $A$ and $B$, then $A$ has $\frac{1}{2}$ of the area and $T(A)$ sends it again to $\frac{1}{2}$ of the area. And then $T(A)$ intersects $A$ and $B$ in a square of area $\frac{1}{2}$. Well so informally, if we take the measure of $A$, and go the streching/folding.. back, than we still have the same area for $A$? How do I write this more formally? – Luca Oct 10 '12 at 0:47

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