Need to show that sets belongs to Borel sigma algebra
I have two sets $$A_1={x=(x_1,x_2,...): sup x_n>1}$$ and $$A_2={x=(x_1,x_2,...): liminf_{n to infty} x_n>1}$$
I need to show that theese sets belongs to Borel sigma algebra in space $mathbb{R}^infty$.
Honestly, I have no Idea how to start and what to do.
probability probability-theory borel-sets
edited Nov 28 '18 at 11:42
DanielWainfleet
34.2k31647
asked Nov 28 '18 at 9:57
Atstovas
937
add a comment |
I have two sets $$A_1={x=(x_1,x_2,...): sup x_n>1}$$ and $$A_2={x=(x_1,x_2,...): liminf_{n to infty} x_n>1}$$
I need to show that theese sets belongs to Borel sigma algebra in space $mathbb{R}^infty$.
Honestly, I have no Idea how to start and what to do.
probability probability-theory borel-sets
edited Nov 28 '18 at 11:42
DanielWainfleet
34.2k31647
asked Nov 28 '18 at 9:57
Atstovas
937
My edit was for a trivial typo.
– DanielWainfleet
Nov 28 '18 at 11:43
add a comment |
I have two sets $$A_1={x=(x_1,x_2,...): sup x_n>1}$$ and $$A_2={x=(x_1,x_2,...): liminf_{n to infty} x_n>1}$$
I need to show that theese sets belongs to Borel sigma algebra in space $mathbb{R}^infty$.
Honestly, I have no Idea how to start and what to do.
probability probability-theory borel-sets
edited Nov 28 '18 at 11:42
DanielWainfleet
34.2k31647
asked Nov 28 '18 at 9:57
Atstovas
937
I have two sets $$A_1={x=(x_1,x_2,...): sup x_n>1}$$ and $$A_2={x=(x_1,x_2,...): liminf_{n to infty} x_n>1}$$
I need to show that theese sets belongs to Borel sigma algebra in space $mathbb{R}^infty$.
Honestly, I have no Idea how to start and what to do.
probability probability-theory borel-sets
probability probability-theory borel-sets
edited Nov 28 '18 at 11:42
DanielWainfleet
34.2k31647
asked Nov 28 '18 at 9:57
Atstovas
937
edited Nov 28 '18 at 11:42
DanielWainfleet
34.2k31647
asked Nov 28 '18 at 9:57
Atstovas
937
edited Nov 28 '18 at 11:42
DanielWainfleet
34.2k31647
edited Nov 28 '18 at 11:42
DanielWainfleet
34.2k31647
edited Nov 28 '18 at 11:42
DanielWainfleet
34.2k31647
34.2k31647
asked Nov 28 '18 at 9:57
Atstovas
937
asked Nov 28 '18 at 9:57
Atstovas
937
asked Nov 28 '18 at 9:57
Atstovas
937
937
My edit was for a trivial typo.
– DanielWainfleet
Nov 28 '18 at 11:43
add a comment |
My edit was for a trivial typo.
– DanielWainfleet
Nov 28 '18 at 11:43
My edit was for a trivial typo.
– DanielWainfleet
Nov 28 '18 at 11:43
My edit was for a trivial typo.
– DanielWainfleet
Nov 28 '18 at 11:43
add a comment |
2 Answers
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Although I like the Answer from Kavi Rama Murthy, here is a direct (naive) way:
(1).For $nin Bbb N$ let $A_{1,n}={(x_j)_{jin Bbb N}: x_n>1}.$ Each $A_{1,n}$ is open in the Tychonoff product topology. So $A_1=cup_{nin Bbb N}A_{1,n}$ is Borel.
(2). For $(k,n)in Bbb N^2$ let $A_{2,k,n}={(x_j)_{jin Bbb N}: forall jgeq n,(x_j> 1+frac {1}{k})}.$ Then $A_2 =cup_{(k,n)in Bbb N^2}A_{2,k,n}$ so it suffices to show that each $A_{2,k,n}$ is Borel.
For $k,n in Bbb N$ and $min {0}cup Bbb N$ let $A_{2,j,n,m}=$ $={(x_j)_{jin Bbb N}: (nleq jleq n+mimplies(x_j>1+frac {1}{k})}.$ Each $A_{2,k,n,m}$ is open so $A_{2,k,n}=cap_{m=0}^{infty}A_{2,k,n,m}$ is Borel.
Remark. In (2) we use the fact that $(x_j)_{jin Bbb N}in A_2$ iff there exists $kin Bbb N$ such that ${j:x_jleq 1+frac {1}{k}}$ is finite.
answered Nov 28 '18 at 12:37
DanielWainfleet
34.2k31647
add a comment |
The coordinate maps $p_n:mathbb R^{infty} to mathbb R$ defined by $p_n(x)=x_n$ are continuous, hence Borel measurable. Supremum and $lim inf$ of a seqeuence of measurable functions is measurable. Hence ${x: sup p_n ^{-1} (1,infty)}$ is measurable. Similarly ${x: lim sup p_n ^{-1} (1,infty)}$ is measurable.
answered Nov 28 '18 at 10:02
Kavi Rama Murthy
50.7k31854
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Although I like the Answer from Kavi Rama Murthy, here is a direct (naive) way:
(1).For $nin Bbb N$ let $A_{1,n}={(x_j)_{jin Bbb N}: x_n>1}.$ Each $A_{1,n}$ is open in the Tychonoff product topology. So $A_1=cup_{nin Bbb N}A_{1,n}$ is Borel.
(2). For $(k,n)in Bbb N^2$ let $A_{2,k,n}={(x_j)_{jin Bbb N}: forall jgeq n,(x_j> 1+frac {1}{k})}.$ Then $A_2 =cup_{(k,n)in Bbb N^2}A_{2,k,n}$ so it suffices to show that each $A_{2,k,n}$ is Borel.
For $k,n in Bbb N$ and $min {0}cup Bbb N$ let $A_{2,j,n,m}=$ $={(x_j)_{jin Bbb N}: (nleq jleq n+mimplies(x_j>1+frac {1}{k})}.$ Each $A_{2,k,n,m}$ is open so $A_{2,k,n}=cap_{m=0}^{infty}A_{2,k,n,m}$ is Borel.
Remark. In (2) we use the fact that $(x_j)_{jin Bbb N}in A_2$ iff there exists $kin Bbb N$ such that ${j:x_jleq 1+frac {1}{k}}$ is finite.
answered Nov 28 '18 at 12:37
DanielWainfleet
34.2k31647
add a comment |
Although I like the Answer from Kavi Rama Murthy, here is a direct (naive) way:
(1).For $nin Bbb N$ let $A_{1,n}={(x_j)_{jin Bbb N}: x_n>1}.$ Each $A_{1,n}$ is open in the Tychonoff product topology. So $A_1=cup_{nin Bbb N}A_{1,n}$ is Borel.
(2). For $(k,n)in Bbb N^2$ let $A_{2,k,n}={(x_j)_{jin Bbb N}: forall jgeq n,(x_j> 1+frac {1}{k})}.$ Then $A_2 =cup_{(k,n)in Bbb N^2}A_{2,k,n}$ so it suffices to show that each $A_{2,k,n}$ is Borel.
For $k,n in Bbb N$ and $min {0}cup Bbb N$ let $A_{2,j,n,m}=$ $={(x_j)_{jin Bbb N}: (nleq jleq n+mimplies(x_j>1+frac {1}{k})}.$ Each $A_{2,k,n,m}$ is open so $A_{2,k,n}=cap_{m=0}^{infty}A_{2,k,n,m}$ is Borel.
Remark. In (2) we use the fact that $(x_j)_{jin Bbb N}in A_2$ iff there exists $kin Bbb N$ such that ${j:x_jleq 1+frac {1}{k}}$ is finite.
answered Nov 28 '18 at 12:37
DanielWainfleet
34.2k31647
add a comment |
Although I like the Answer from Kavi Rama Murthy, here is a direct (naive) way:
(1).For $nin Bbb N$ let $A_{1,n}={(x_j)_{jin Bbb N}: x_n>1}.$ Each $A_{1,n}$ is open in the Tychonoff product topology. So $A_1=cup_{nin Bbb N}A_{1,n}$ is Borel.
(2). For $(k,n)in Bbb N^2$ let $A_{2,k,n}={(x_j)_{jin Bbb N}: forall jgeq n,(x_j> 1+frac {1}{k})}.$ Then $A_2 =cup_{(k,n)in Bbb N^2}A_{2,k,n}$ so it suffices to show that each $A_{2,k,n}$ is Borel.
For $k,n in Bbb N$ and $min {0}cup Bbb N$ let $A_{2,j,n,m}=$ $={(x_j)_{jin Bbb N}: (nleq jleq n+mimplies(x_j>1+frac {1}{k})}.$ Each $A_{2,k,n,m}$ is open so $A_{2,k,n}=cap_{m=0}^{infty}A_{2,k,n,m}$ is Borel.
Remark. In (2) we use the fact that $(x_j)_{jin Bbb N}in A_2$ iff there exists $kin Bbb N$ such that ${j:x_jleq 1+frac {1}{k}}$ is finite.
answered Nov 28 '18 at 12:37
DanielWainfleet
34.2k31647
Although I like the Answer from Kavi Rama Murthy, here is a direct (naive) way:
(1).For $nin Bbb N$ let $A_{1,n}={(x_j)_{jin Bbb N}: x_n>1}.$ Each $A_{1,n}$ is open in the Tychonoff product topology. So $A_1=cup_{nin Bbb N}A_{1,n}$ is Borel.
(2). For $(k,n)in Bbb N^2$ let $A_{2,k,n}={(x_j)_{jin Bbb N}: forall jgeq n,(x_j> 1+frac {1}{k})}.$ Then $A_2 =cup_{(k,n)in Bbb N^2}A_{2,k,n}$ so it suffices to show that each $A_{2,k,n}$ is Borel.
For $k,n in Bbb N$ and $min {0}cup Bbb N$ let $A_{2,j,n,m}=$ $={(x_j)_{jin Bbb N}: (nleq jleq n+mimplies(x_j>1+frac {1}{k})}.$ Each $A_{2,k,n,m}$ is open so $A_{2,k,n}=cap_{m=0}^{infty}A_{2,k,n,m}$ is Borel.
Remark. In (2) we use the fact that $(x_j)_{jin Bbb N}in A_2$ iff there exists $kin Bbb N$ such that ${j:x_jleq 1+frac {1}{k}}$ is finite.
answered Nov 28 '18 at 12:37
DanielWainfleet
34.2k31647
answered Nov 28 '18 at 12:37
DanielWainfleet
34.2k31647
answered Nov 28 '18 at 12:37
DanielWainfleet
34.2k31647
answered Nov 28 '18 at 12:37
DanielWainfleet
34.2k31647
34.2k31647
add a comment |
add a comment |
The coordinate maps $p_n:mathbb R^{infty} to mathbb R$ defined by $p_n(x)=x_n$ are continuous, hence Borel measurable. Supremum and $lim inf$ of a seqeuence of measurable functions is measurable. Hence ${x: sup p_n ^{-1} (1,infty)}$ is measurable. Similarly ${x: lim sup p_n ^{-1} (1,infty)}$ is measurable.
answered Nov 28 '18 at 10:02
Kavi Rama Murthy
50.7k31854
add a comment |
The coordinate maps $p_n:mathbb R^{infty} to mathbb R$ defined by $p_n(x)=x_n$ are continuous, hence Borel measurable. Supremum and $lim inf$ of a seqeuence of measurable functions is measurable. Hence ${x: sup p_n ^{-1} (1,infty)}$ is measurable. Similarly ${x: lim sup p_n ^{-1} (1,infty)}$ is measurable.
answered Nov 28 '18 at 10:02
Kavi Rama Murthy
50.7k31854
add a comment |
The coordinate maps $p_n:mathbb R^{infty} to mathbb R$ defined by $p_n(x)=x_n$ are continuous, hence Borel measurable. Supremum and $lim inf$ of a seqeuence of measurable functions is measurable. Hence ${x: sup p_n ^{-1} (1,infty)}$ is measurable. Similarly ${x: lim sup p_n ^{-1} (1,infty)}$ is measurable.
answered Nov 28 '18 at 10:02
Kavi Rama Murthy
50.7k31854
The coordinate maps $p_n:mathbb R^{infty} to mathbb R$ defined by $p_n(x)=x_n$ are continuous, hence Borel measurable. Supremum and $lim inf$ of a seqeuence of measurable functions is measurable. Hence ${x: sup p_n ^{-1} (1,infty)}$ is measurable. Similarly ${x: lim sup p_n ^{-1} (1,infty)}$ is measurable.
answered Nov 28 '18 at 10:02
Kavi Rama Murthy
50.7k31854
answered Nov 28 '18 at 10:02
Kavi Rama Murthy
50.7k31854
answered Nov 28 '18 at 10:02
Kavi Rama Murthy
50.7k31854
answered Nov 28 '18 at 10:02
Kavi Rama Murthy
50.7k31854
50.7k31854
add a comment |
add a comment |
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My edit was for a trivial typo.
– DanielWainfleet
Nov 28 '18 at 11:43