Measure-preserving transformations

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In the following image the last line why does the measure of the set $G$ equal $1$ ?enter image description here




general-topology measure-theory proof-explanation intuition ergodic-theory




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  • Continuity of measure.
    – fourierwho
    May 19 at 23:28










  • @fourierwho how please ?
    – Neil hawking
    May 19 at 23:32














up vote
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down vote

favorite
1












In the following image the last line why does the measure of the set $G$ equal $1$ ?enter image description here




general-topology measure-theory proof-explanation intuition ergodic-theory




share|cite|improve this question






















  • Continuity of measure.
    – fourierwho
    May 19 at 23:28










  • @fourierwho how please ?
    – Neil hawking
    May 19 at 23:32












up vote
2
down vote

favorite
1









up vote
2
down vote

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In the following image the last line why does the measure of the set $G$ equal $1$ ?enter image description here




general-topology measure-theory proof-explanation intuition ergodic-theory




share|cite|improve this question














In the following image the last line why does the measure of the set $G$ equal $1$ ?enter image description here





general-topology measure-theory proof-explanation intuition ergodic-theory





share|cite|improve this question













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edited Aug 20 at 10:27

























asked May 19 at 23:23









Neil hawking

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  • Continuity of measure.
    – fourierwho
    May 19 at 23:28










  • @fourierwho how please ?
    – Neil hawking
    May 19 at 23:32
















  • Continuity of measure.
    – fourierwho
    May 19 at 23:28










  • @fourierwho how please ?
    – Neil hawking
    May 19 at 23:32















Continuity of measure.
– fourierwho
May 19 at 23:28




Continuity of measure.
– fourierwho
May 19 at 23:28












@fourierwho how please ?
– Neil hawking
May 19 at 23:32




@fourierwho how please ?
– Neil hawking
May 19 at 23:32










1 Answer
1






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The text defines $G$ as $G = bigcap_n = 1^infty G_n$, where $G_n = bigcup_k = 0^infty xi^-k(V_n)$. It is therein proved that $mu(G_n) = 1$. Recall that if $A$ and $B$ are events/measurable sets and $mu(A) = mu(B) = 1$, then $mu(A cap B) = 1$. Thus, for each $N in mathbbN$, $mu(G_1 cap dots cap G_N) = 1$ (by induction). Therefore, by continuity of measure,
beginequation*
mu(G) = lim_N to infty mu(G_1 cap dots cap G_N) = 1.
endequation*






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  • Thanks merci danke
    – Neil hawking
    May 19 at 23:45










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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










The text defines $G$ as $G = bigcap_n = 1^infty G_n$, where $G_n = bigcup_k = 0^infty xi^-k(V_n)$. It is therein proved that $mu(G_n) = 1$. Recall that if $A$ and $B$ are events/measurable sets and $mu(A) = mu(B) = 1$, then $mu(A cap B) = 1$. Thus, for each $N in mathbbN$, $mu(G_1 cap dots cap G_N) = 1$ (by induction). Therefore, by continuity of measure,
beginequation*
mu(G) = lim_N to infty mu(G_1 cap dots cap G_N) = 1.
endequation*






share|cite|improve this answer




















  • Thanks merci danke
    – Neil hawking
    May 19 at 23:45














up vote
2
down vote



accepted










The text defines $G$ as $G = bigcap_n = 1^infty G_n$, where $G_n = bigcup_k = 0^infty xi^-k(V_n)$. It is therein proved that $mu(G_n) = 1$. Recall that if $A$ and $B$ are events/measurable sets and $mu(A) = mu(B) = 1$, then $mu(A cap B) = 1$. Thus, for each $N in mathbbN$, $mu(G_1 cap dots cap G_N) = 1$ (by induction). Therefore, by continuity of measure,
beginequation*
mu(G) = lim_N to infty mu(G_1 cap dots cap G_N) = 1.
endequation*






share|cite|improve this answer




















  • Thanks merci danke
    – Neil hawking
    May 19 at 23:45












up vote
2
down vote



accepted







up vote
2
down vote



accepted






The text defines $G$ as $G = bigcap_n = 1^infty G_n$, where $G_n = bigcup_k = 0^infty xi^-k(V_n)$. It is therein proved that $mu(G_n) = 1$. Recall that if $A$ and $B$ are events/measurable sets and $mu(A) = mu(B) = 1$, then $mu(A cap B) = 1$. Thus, for each $N in mathbbN$, $mu(G_1 cap dots cap G_N) = 1$ (by induction). Therefore, by continuity of measure,
beginequation*
mu(G) = lim_N to infty mu(G_1 cap dots cap G_N) = 1.
endequation*






share|cite|improve this answer












The text defines $G$ as $G = bigcap_n = 1^infty G_n$, where $G_n = bigcup_k = 0^infty xi^-k(V_n)$. It is therein proved that $mu(G_n) = 1$. Recall that if $A$ and $B$ are events/measurable sets and $mu(A) = mu(B) = 1$, then $mu(A cap B) = 1$. Thus, for each $N in mathbbN$, $mu(G_1 cap dots cap G_N) = 1$ (by induction). Therefore, by continuity of measure,
beginequation*
mu(G) = lim_N to infty mu(G_1 cap dots cap G_N) = 1.
endequation*







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered May 19 at 23:35









fourierwho

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  • Thanks merci danke
    – Neil hawking
    May 19 at 23:45
















  • Thanks merci danke
    – Neil hawking
    May 19 at 23:45















Thanks merci danke
– Neil hawking
May 19 at 23:45




Thanks merci danke
– Neil hawking
May 19 at 23:45












 

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