$forall alpha in [0,1] exists B_alpha in mathcal B(Bbb R): Bbb P(B_alpha) = alpha$ [duplicate]

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  continuity of a probability measure if $mu (x)=0$
  
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Let $Bbb P$ be a probability measure on $Bbb R$ with $Bbb P(x) = 0$ for all $x in Bbb R$. I want to show that for all $alpha in [0,1]$ there exists a $B_alpha in mathcal B(Bbb R)$ such that $Bbb P(B_alpha) = alpha$.

My approach is to take an interval $I_m = [-m,m]$ (let's assume $Bbb P([-m,m]) > alpha$) and construct a sequence of subintervals $[-m + q_k, m - q'_k]$ with $(q_k,q'_k) in Bbb Q_+^2$ such that $[-m + q_k, m - q'_k] supseteq [-m + q_k+1, m - q'_k+1]$ and $Bbb P ([-m + q_k, m - q'_k]) ge alpha $ for all $k in Bbb N$. Then my idea was to use the continuity from above of the measure but I'm not sure whether this sequence indeed converges to a set with measure $alpha$.

probability-theory measure-theory

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asked Jul 18 at 16:04

Pazu

359213

marked as duplicate by Eric Wofsey, Community♦ Jul 18 at 16:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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up vote
1
down vote

favorite

1

This question already has an answer here:

• 
  continuity of a probability measure if $mu (x)=0$
  
  2 answers
  
  

Let $Bbb P$ be a probability measure on $Bbb R$ with $Bbb P(x) = 0$ for all $x in Bbb R$. I want to show that for all $alpha in [0,1]$ there exists a $B_alpha in mathcal B(Bbb R)$ such that $Bbb P(B_alpha) = alpha$.

My approach is to take an interval $I_m = [-m,m]$ (let's assume $Bbb P([-m,m]) > alpha$) and construct a sequence of subintervals $[-m + q_k, m - q'_k]$ with $(q_k,q'_k) in Bbb Q_+^2$ such that $[-m + q_k, m - q'_k] supseteq [-m + q_k+1, m - q'_k+1]$ and $Bbb P ([-m + q_k, m - q'_k]) ge alpha $ for all $k in Bbb N$. Then my idea was to use the continuity from above of the measure but I'm not sure whether this sequence indeed converges to a set with measure $alpha$.

probability-theory measure-theory

share|cite|improve this question

asked Jul 18 at 16:04

Pazu

359213

marked as duplicate by Eric Wofsey, Community♦ Jul 18 at 16:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

This question already has an answer here:

• 
  continuity of a probability measure if $mu (x)=0$
  
  2 answers
  
  

Let $Bbb P$ be a probability measure on $Bbb R$ with $Bbb P(x) = 0$ for all $x in Bbb R$. I want to show that for all $alpha in [0,1]$ there exists a $B_alpha in mathcal B(Bbb R)$ such that $Bbb P(B_alpha) = alpha$.

My approach is to take an interval $I_m = [-m,m]$ (let's assume $Bbb P([-m,m]) > alpha$) and construct a sequence of subintervals $[-m + q_k, m - q'_k]$ with $(q_k,q'_k) in Bbb Q_+^2$ such that $[-m + q_k, m - q'_k] supseteq [-m + q_k+1, m - q'_k+1]$ and $Bbb P ([-m + q_k, m - q'_k]) ge alpha $ for all $k in Bbb N$. Then my idea was to use the continuity from above of the measure but I'm not sure whether this sequence indeed converges to a set with measure $alpha$.

probability-theory measure-theory

share|cite|improve this question

asked Jul 18 at 16:04

Pazu

359213

This question already has an answer here:

• 
  continuity of a probability measure if $mu (x)=0$
  
  2 answers
  
  

Let $Bbb P$ be a probability measure on $Bbb R$ with $Bbb P(x) = 0$ for all $x in Bbb R$. I want to show that for all $alpha in [0,1]$ there exists a $B_alpha in mathcal B(Bbb R)$ such that $Bbb P(B_alpha) = alpha$.

My approach is to take an interval $I_m = [-m,m]$ (let's assume $Bbb P([-m,m]) > alpha$) and construct a sequence of subintervals $[-m + q_k, m - q'_k]$ with $(q_k,q'_k) in Bbb Q_+^2$ such that $[-m + q_k, m - q'_k] supseteq [-m + q_k+1, m - q'_k+1]$ and $Bbb P ([-m + q_k, m - q'_k]) ge alpha $ for all $k in Bbb N$. Then my idea was to use the continuity from above of the measure but I'm not sure whether this sequence indeed converges to a set with measure $alpha$.

This question already has an answer here:

• 
  continuity of a probability measure if $mu (x)=0$
  
  2 answers
  
  

probability-theory measure-theory

share|cite|improve this question

asked Jul 18 at 16:04

Pazu

359213

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 16:04

Pazu

359213

asked Jul 18 at 16:04

Pazu

359213

asked Jul 18 at 16:04

Pazu

359213

359213

marked as duplicate by Eric Wofsey, Community♦ Jul 18 at 16:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Eric Wofsey, Community♦ Jul 18 at 16:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add a comment | 

add a comment | 

2 Answers
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2
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accepted

Let $F(x)=mathbbP((-infty,x])$ be the c.d.f. associated with
$mathbbP$. The condition $mathbbP(x)=0$ for each $xinmathbbR$
implies that $F$ is a continuous function. Note that $lim_xrightarrow-inftyF(x)=0$
and $lim_xrightarrowinftyF(x)=1$. By intermediate value theorem,
the result follows.

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answered Jul 18 at 16:12

Danny Pak-Keung Chan

1,87628

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The cumulative density function is continuous iff $mathbbP(x)=0$ for all $x$. As this applies, we have a continuous $F$ with limit 0 at $-infty$ and with limit 1 at $infty$. So the range of $F$ is $[0,1]$, making your assertion true.

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answered Jul 18 at 16:22

A. Pongrácz

2,574321

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Let $F(x)=mathbbP((-infty,x])$ be the c.d.f. associated with
$mathbbP$. The condition $mathbbP(x)=0$ for each $xinmathbbR$
implies that $F$ is a continuous function. Note that $lim_xrightarrow-inftyF(x)=0$
and $lim_xrightarrowinftyF(x)=1$. By intermediate value theorem,
the result follows.

share|cite|improve this answer

answered Jul 18 at 16:12

Danny Pak-Keung Chan

1,87628

add a comment | 

up vote
2
down vote



accepted

Let $F(x)=mathbbP((-infty,x])$ be the c.d.f. associated with
$mathbbP$. The condition $mathbbP(x)=0$ for each $xinmathbbR$
implies that $F$ is a continuous function. Note that $lim_xrightarrow-inftyF(x)=0$
and $lim_xrightarrowinftyF(x)=1$. By intermediate value theorem,
the result follows.

share|cite|improve this answer

answered Jul 18 at 16:12

Danny Pak-Keung Chan

1,87628

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Let $F(x)=mathbbP((-infty,x])$ be the c.d.f. associated with
$mathbbP$. The condition $mathbbP(x)=0$ for each $xinmathbbR$
implies that $F$ is a continuous function. Note that $lim_xrightarrow-inftyF(x)=0$
and $lim_xrightarrowinftyF(x)=1$. By intermediate value theorem,
the result follows.

share|cite|improve this answer

answered Jul 18 at 16:12

Danny Pak-Keung Chan

1,87628

Let $F(x)=mathbbP((-infty,x])$ be the c.d.f. associated with
$mathbbP$. The condition $mathbbP(x)=0$ for each $xinmathbbR$
implies that $F$ is a continuous function. Note that $lim_xrightarrow-inftyF(x)=0$
and $lim_xrightarrowinftyF(x)=1$. By intermediate value theorem,
the result follows.

share|cite|improve this answer

answered Jul 18 at 16:12

Danny Pak-Keung Chan

1,87628

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 16:12

Danny Pak-Keung Chan

1,87628

answered Jul 18 at 16:12

Danny Pak-Keung Chan

1,87628

answered Jul 18 at 16:12

Danny Pak-Keung Chan

1,87628

1,87628

add a comment | 

add a comment | 

up vote
0
down vote

The cumulative density function is continuous iff $mathbbP(x)=0$ for all $x$. As this applies, we have a continuous $F$ with limit 0 at $-infty$ and with limit 1 at $infty$. So the range of $F$ is $[0,1]$, making your assertion true.

share|cite|improve this answer

answered Jul 18 at 16:22

A. Pongrácz

2,574321

add a comment | 

up vote
0
down vote

The cumulative density function is continuous iff $mathbbP(x)=0$ for all $x$. As this applies, we have a continuous $F$ with limit 0 at $-infty$ and with limit 1 at $infty$. So the range of $F$ is $[0,1]$, making your assertion true.

share|cite|improve this answer

answered Jul 18 at 16:22

A. Pongrácz

2,574321

add a comment | 

up vote
0
down vote

up vote
0
down vote

The cumulative density function is continuous iff $mathbbP(x)=0$ for all $x$. As this applies, we have a continuous $F$ with limit 0 at $-infty$ and with limit 1 at $infty$. So the range of $F$ is $[0,1]$, making your assertion true.

share|cite|improve this answer

answered Jul 18 at 16:22

A. Pongrácz

2,574321

The cumulative density function is continuous iff $mathbbP(x)=0$ for all $x$. As this applies, we have a continuous $F$ with limit 0 at $-infty$ and with limit 1 at $infty$. So the range of $F$ is $[0,1]$, making your assertion true.

share|cite|improve this answer

answered Jul 18 at 16:22

A. Pongrácz

2,574321

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 16:22

A. Pongrácz

2,574321

answered Jul 18 at 16:22

A. Pongrácz

2,574321

answered Jul 18 at 16:22

A. Pongrácz

2,574321

2,574321

add a comment | 

add a comment |