Mixing
Understanding $T(x)=3xpmod 1$ [closed]
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Let $T:[0,1]to[0,1]$ be such that $T(x)=3xpmod{1}$ which is measurable with respect to the $sigma$-algebra of Borel on $[0,1]$, which we denote by $mathscr{B}_{[0,1]}$.
Prove that Lebesgue measure on $[0,1]$ is T-invariant.
I know that if $[a,b]in T(x)$ then as the measure is invariant $T^{-1}([a,b])=I_1cup I_2cup I_3$. However I do not know how to determine the intervals.
Question:
Can someone explain me how should I work with the inverse of $T$?
measure-theory dynamical-systems ergodic-theory
edited Jan 30 at 21:52
Did
249k23227466
asked Jan 30 at 21:45
Pedro GomesPedro Gomes
1,9942721
$endgroup$
closed as unclear what you're asking by Did, John B, metamorphy, A. Pongrácz, José Carlos Santos Jan 31 at 10:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $T:[0,1]to[0,1]$ be such that $T(x)=3xpmod{1}$ which is measurable with respect to the $sigma$-algebra of Borel on $[0,1]$, which we denote by $mathscr{B}_{[0,1]}$.
Prove that Lebesgue measure on $[0,1]$ is T-invariant.
I know that if $[a,b]in T(x)$ then as the measure is invariant $T^{-1}([a,b])=I_1cup I_2cup I_3$. However I do not know how to determine the intervals.
Question:
Can someone explain me how should I work with the inverse of $T$?
measure-theory dynamical-systems ergodic-theory
edited Jan 30 at 21:52
Did
249k23227466
asked Jan 30 at 21:45
Pedro GomesPedro Gomes
1,9942721
$endgroup$
closed as unclear what you're asking by Did, John B, metamorphy, A. Pongrácz, José Carlos Santos Jan 31 at 10:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
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How come "as the measure is invariant" has anything to do with having a preimage the union of $3$ intervals?
$endgroup$
– John B
Jan 30 at 21:47
$begingroup$
Because it the Lebesgue measure of the union of those three intervals which equals the whole interval.
$endgroup$
– Pedro Gomes
Jan 30 at 21:49
1
$begingroup$
Well, for example, $T^{-1}([0,frac15])=[0,frac1{15}]cup[frac13,frac25]cup[frac23,frac{11}{15}]$. Where is the mystery?
$endgroup$
– Did
Jan 30 at 21:50
$begingroup$
@Did The mystery is the logical link between the two. There is no apparent (at least for me) link between the degree of a map on the circle and the fact the measure is invariant. It's easy to exhibit continuous maps of the circle, preserving (say) Lebesgue measure, such that the preimage of an interval is not the union of three intervals. Or maps of degree 3 that do not preserve Lebesgue measure.
$endgroup$
– user120527
Jan 31 at 21:38
add a comment |
$begingroup$
Let $T:[0,1]to[0,1]$ be such that $T(x)=3xpmod{1}$ which is measurable with respect to the $sigma$-algebra of Borel on $[0,1]$, which we denote by $mathscr{B}_{[0,1]}$.
Prove that Lebesgue measure on $[0,1]$ is T-invariant.
I know that if $[a,b]in T(x)$ then as the measure is invariant $T^{-1}([a,b])=I_1cup I_2cup I_3$. However I do not know how to determine the intervals.
Question:
Can someone explain me how should I work with the inverse of $T$?
measure-theory dynamical-systems ergodic-theory
edited Jan 30 at 21:52
Did
249k23227466
asked Jan 30 at 21:45
Pedro GomesPedro Gomes
1,9942721
$endgroup$
Let $T:[0,1]to[0,1]$ be such that $T(x)=3xpmod{1}$ which is measurable with respect to the $sigma$-algebra of Borel on $[0,1]$, which we denote by $mathscr{B}_{[0,1]}$.
Prove that Lebesgue measure on $[0,1]$ is T-invariant.
I know that if $[a,b]in T(x)$ then as the measure is invariant $T^{-1}([a,b])=I_1cup I_2cup I_3$. However I do not know how to determine the intervals.
Question:
Can someone explain me how should I work with the inverse of $T$?
measure-theory dynamical-systems ergodic-theory
measure-theory dynamical-systems ergodic-theory
edited Jan 30 at 21:52
Did
249k23227466
asked Jan 30 at 21:45
Pedro GomesPedro Gomes
1,9942721
edited Jan 30 at 21:52
Did
249k23227466
asked Jan 30 at 21:45
Pedro GomesPedro Gomes
1,9942721
edited Jan 30 at 21:52
Did
249k23227466
edited Jan 30 at 21:52
Did
249k23227466
edited Jan 30 at 21:52
Did
249k23227466
249k23227466
asked Jan 30 at 21:45
Pedro GomesPedro Gomes
1,9942721
asked Jan 30 at 21:45
Pedro GomesPedro Gomes
1,9942721
asked Jan 30 at 21:45
Pedro GomesPedro Gomes
1,9942721
1,9942721
closed as unclear what you're asking by Did, John B, metamorphy, A. Pongrácz, José Carlos Santos Jan 31 at 10:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Did, John B, metamorphy, A. Pongrácz, José Carlos Santos Jan 31 at 10:54
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
How come "as the measure is invariant" has anything to do with having a preimage the union of $3$ intervals?
$endgroup$
– John B
Jan 30 at 21:47
$begingroup$
Because it the Lebesgue measure of the union of those three intervals which equals the whole interval.
$endgroup$
– Pedro Gomes
Jan 30 at 21:49
1
$begingroup$
Well, for example, $T^{-1}([0,frac15])=[0,frac1{15}]cup[frac13,frac25]cup[frac23,frac{11}{15}]$. Where is the mystery?
$endgroup$
– Did
Jan 30 at 21:50
$begingroup$
@Did The mystery is the logical link between the two. There is no apparent (at least for me) link between the degree of a map on the circle and the fact the measure is invariant. It's easy to exhibit continuous maps of the circle, preserving (say) Lebesgue measure, such that the preimage of an interval is not the union of three intervals. Or maps of degree 3 that do not preserve Lebesgue measure.
$endgroup$
– user120527
Jan 31 at 21:38
add a comment |
2
$begingroup$
How come "as the measure is invariant" has anything to do with having a preimage the union of $3$ intervals?
$endgroup$
– John B
Jan 30 at 21:47
$begingroup$
Because it the Lebesgue measure of the union of those three intervals which equals the whole interval.
$endgroup$
– Pedro Gomes
Jan 30 at 21:49
1
$begingroup$
Well, for example, $T^{-1}([0,frac15])=[0,frac1{15}]cup[frac13,frac25]cup[frac23,frac{11}{15}]$. Where is the mystery?
$endgroup$
– Did
Jan 30 at 21:50
$begingroup$
@Did The mystery is the logical link between the two. There is no apparent (at least for me) link between the degree of a map on the circle and the fact the measure is invariant. It's easy to exhibit continuous maps of the circle, preserving (say) Lebesgue measure, such that the preimage of an interval is not the union of three intervals. Or maps of degree 3 that do not preserve Lebesgue measure.
$endgroup$
– user120527
Jan 31 at 21:38
2
2
$begingroup$
How come "as the measure is invariant" has anything to do with having a preimage the union of $3$ intervals?
$endgroup$
– John B
Jan 30 at 21:47
$begingroup$
How come "as the measure is invariant" has anything to do with having a preimage the union of $3$ intervals?
$endgroup$
– John B
Jan 30 at 21:47
$begingroup$
Because it the Lebesgue measure of the union of those three intervals which equals the whole interval.
$endgroup$
– Pedro Gomes
Jan 30 at 21:49
$begingroup$
Because it the Lebesgue measure of the union of those three intervals which equals the whole interval.
$endgroup$
– Pedro Gomes
Jan 30 at 21:49
1
1
$begingroup$
Well, for example, $T^{-1}([0,frac15])=[0,frac1{15}]cup[frac13,frac25]cup[frac23,frac{11}{15}]$. Where is the mystery?
$endgroup$
– Did
Jan 30 at 21:50
$begingroup$
Well, for example, $T^{-1}([0,frac15])=[0,frac1{15}]cup[frac13,frac25]cup[frac23,frac{11}{15}]$. Where is the mystery?
$endgroup$
– Did
Jan 30 at 21:50
$begingroup$
@Did The mystery is the logical link between the two. There is no apparent (at least for me) link between the degree of a map on the circle and the fact the measure is invariant. It's easy to exhibit continuous maps of the circle, preserving (say) Lebesgue measure, such that the preimage of an interval is not the union of three intervals. Or maps of degree 3 that do not preserve Lebesgue measure.
$endgroup$
– user120527
Jan 31 at 21:38
$begingroup$
@Did The mystery is the logical link between the two. There is no apparent (at least for me) link between the degree of a map on the circle and the fact the measure is invariant. It's easy to exhibit continuous maps of the circle, preserving (say) Lebesgue measure, such that the preimage of an interval is not the union of three intervals. Or maps of degree 3 that do not preserve Lebesgue measure.
$endgroup$
– user120527
Jan 31 at 21:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A broad hint rather than a solution:
What things map to, say, $x = .30$ under $T$? Well, there's $u_1 = 0.1$, and then there's $u_2 = u_1 + frac{1}{3}$, and there's $u_3 = u_1 + frac{2}{3}$
More generally,
$$
T^{-1} ([a, b]) = [frac{a}{3}, frac{b}{3}] cup [frac{a+1}{3}, frac{b+1}{3}] cup [frac{a+2}{3}, frac{b+2}{3}].
$$
With that in mind, perhaps you can make some progress.
answered Jan 30 at 21:51
John HughesJohn Hughes
65.2k24293
$endgroup$
$begingroup$
Is it not $0.1$ instead of $0.0.1$?
$endgroup$
– Pedro Gomes
Jan 30 at 21:58
$begingroup$
So much for my ability to put a decimal in the right place! Thanks for noticing. Now fixed.
$endgroup$
– John Hughes
Jan 30 at 22:28
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A broad hint rather than a solution:
What things map to, say, $x = .30$ under $T$? Well, there's $u_1 = 0.1$, and then there's $u_2 = u_1 + frac{1}{3}$, and there's $u_3 = u_1 + frac{2}{3}$
More generally,
$$
T^{-1} ([a, b]) = [frac{a}{3}, frac{b}{3}] cup [frac{a+1}{3}, frac{b+1}{3}] cup [frac{a+2}{3}, frac{b+2}{3}].
$$
With that in mind, perhaps you can make some progress.
answered Jan 30 at 21:51
John HughesJohn Hughes
65.2k24293
$endgroup$
$begingroup$
Is it not $0.1$ instead of $0.0.1$?
$endgroup$
– Pedro Gomes
Jan 30 at 21:58
$begingroup$
So much for my ability to put a decimal in the right place! Thanks for noticing. Now fixed.
$endgroup$
– John Hughes
Jan 30 at 22:28
add a comment |
$begingroup$
A broad hint rather than a solution:
What things map to, say, $x = .30$ under $T$? Well, there's $u_1 = 0.1$, and then there's $u_2 = u_1 + frac{1}{3}$, and there's $u_3 = u_1 + frac{2}{3}$
More generally,
$$
T^{-1} ([a, b]) = [frac{a}{3}, frac{b}{3}] cup [frac{a+1}{3}, frac{b+1}{3}] cup [frac{a+2}{3}, frac{b+2}{3}].
$$
With that in mind, perhaps you can make some progress.
answered Jan 30 at 21:51
John HughesJohn Hughes
65.2k24293
$endgroup$
$begingroup$
Is it not $0.1$ instead of $0.0.1$?
$endgroup$
– Pedro Gomes
Jan 30 at 21:58
$begingroup$
So much for my ability to put a decimal in the right place! Thanks for noticing. Now fixed.
$endgroup$
– John Hughes
Jan 30 at 22:28
add a comment |
$begingroup$
A broad hint rather than a solution:
What things map to, say, $x = .30$ under $T$? Well, there's $u_1 = 0.1$, and then there's $u_2 = u_1 + frac{1}{3}$, and there's $u_3 = u_1 + frac{2}{3}$
More generally,
$$
T^{-1} ([a, b]) = [frac{a}{3}, frac{b}{3}] cup [frac{a+1}{3}, frac{b+1}{3}] cup [frac{a+2}{3}, frac{b+2}{3}].
$$
With that in mind, perhaps you can make some progress.
answered Jan 30 at 21:51
John HughesJohn Hughes
65.2k24293
$endgroup$
A broad hint rather than a solution:
What things map to, say, $x = .30$ under $T$? Well, there's $u_1 = 0.1$, and then there's $u_2 = u_1 + frac{1}{3}$, and there's $u_3 = u_1 + frac{2}{3}$
More generally,
$$
T^{-1} ([a, b]) = [frac{a}{3}, frac{b}{3}] cup [frac{a+1}{3}, frac{b+1}{3}] cup [frac{a+2}{3}, frac{b+2}{3}].
$$
With that in mind, perhaps you can make some progress.
answered Jan 30 at 21:51
John HughesJohn Hughes
65.2k24293
edited Jan 30 at 22:27
answered Jan 30 at 21:51
John HughesJohn Hughes
65.2k24293
answered Jan 30 at 21:51
John HughesJohn Hughes
65.2k24293
answered Jan 30 at 21:51
John HughesJohn Hughes
65.2k24293
65.2k24293
$begingroup$
Is it not $0.1$ instead of $0.0.1$?
$endgroup$
– Pedro Gomes
Jan 30 at 21:58
$begingroup$
So much for my ability to put a decimal in the right place! Thanks for noticing. Now fixed.
$endgroup$
– John Hughes
Jan 30 at 22:28
add a comment |
$begingroup$
Is it not $0.1$ instead of $0.0.1$?
$endgroup$
– Pedro Gomes
Jan 30 at 21:58
$begingroup$
So much for my ability to put a decimal in the right place! Thanks for noticing. Now fixed.
$endgroup$
– John Hughes
Jan 30 at 22:28
$begingroup$
Is it not $0.1$ instead of $0.0.1$?
$endgroup$
– Pedro Gomes
Jan 30 at 21:58
$begingroup$
Is it not $0.1$ instead of $0.0.1$?
$endgroup$
– Pedro Gomes
Jan 30 at 21:58
$begingroup$
So much for my ability to put a decimal in the right place! Thanks for noticing. Now fixed.
$endgroup$
– John Hughes
Jan 30 at 22:28
$begingroup$
So much for my ability to put a decimal in the right place! Thanks for noticing. Now fixed.
$endgroup$
– John Hughes
Jan 30 at 22:28
add a comment |
2
$begingroup$
How come "as the measure is invariant" has anything to do with having a preimage the union of $3$ intervals?
$endgroup$
– John B
Jan 30 at 21:47
$begingroup$
Because it the Lebesgue measure of the union of those three intervals which equals the whole interval.
$endgroup$
– Pedro Gomes
Jan 30 at 21:49
1
$begingroup$
Well, for example, $T^{-1}([0,frac15])=[0,frac1{15}]cup[frac13,frac25]cup[frac23,frac{11}{15}]$. Where is the mystery?
$endgroup$
– Did
Jan 30 at 21:50
$begingroup$
@Did The mystery is the logical link between the two. There is no apparent (at least for me) link between the degree of a map on the circle and the fact the measure is invariant. It's easy to exhibit continuous maps of the circle, preserving (say) Lebesgue measure, such that the preimage of an interval is not the union of three intervals. Or maps of degree 3 that do not preserve Lebesgue measure.
$endgroup$
– user120527
Jan 31 at 21:38