proof: $f+g$ is measurable

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I am struggling to understand the line "$xin X : (f+g)(x) > a = cup S_r$". How two sets can be equal? Could you elaborate on this?

Thank you in advance.

measure-theory

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asked Jul 30 at 11:22

Sihyun Kim

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I am struggling to understand the line "$xin X : (f+g)(x) > a = cup S_r$". How two sets can be equal? Could you elaborate on this?

Thank you in advance.

measure-theory

share|cite|improve this question

asked Jul 30 at 11:22

Sihyun Kim

701210

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I am struggling to understand the line "$xin X : (f+g)(x) > a = cup S_r$". How two sets can be equal? Could you elaborate on this?

Thank you in advance.

measure-theory

share|cite|improve this question

asked Jul 30 at 11:22

Sihyun Kim

701210

I am struggling to understand the line "$xin X : (f+g)(x) > a = cup S_r$". How two sets can be equal? Could you elaborate on this?

Thank you in advance.

measure-theory

share|cite|improve this question

asked Jul 30 at 11:22

Sihyun Kim

701210

share|cite|improve this question

share|cite|improve this question

asked Jul 30 at 11:22

Sihyun Kim

701210

asked Jul 30 at 11:22

Sihyun Kim

701210

asked Jul 30 at 11:22

Sihyun Kim

701210

701210

add a comment | 

add a comment | 

2 Answers
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accepted

1. If $x in bigcup S_r$, then there is a rational $r$ such that $f(x)>r$ and $g(x)> alpha -r$. It follows that $(f+g)(x)> alpha$.




2. Let $(f+g)(x)> alpha$ and suppose that $x notin bigcup S_r$. Now choose a sequence $(r_n)$ of rationals such that

$f(x) > r_n$ for all $n$ and $r_n to f(x)$. Since $x notin S_r_n$ for all $n$, we have

$g(x) le alpha -r_n$ for all $n$. With $n to infty$ we get $g(x) le alpha-f(x)$, a contradiction !

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answered Jul 30 at 11:43

Fred

37k1237

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It is evident that $S_r=f>rcapg>alpha-rsubseteqf+g>alpha$ for every $rinmathbb Q$.

If conversely $f(x)+g(x)>alpha$ or equivalently $f(x)>alpha-g(x)$ then, because $mathbb Q$ is dense in $mathbb R$, we can find some $rinmathbb Q$ with $r<f(x)$ (or equivalently $f(x)>r$) but still $r>alpha-g(x)$ (or equivalently $g(x)>alpha-r$). So we found an $rinmathbb Q$ with $xin S_r$.

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answered Jul 30 at 12:04

drhab

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

1. If $x in bigcup S_r$, then there is a rational $r$ such that $f(x)>r$ and $g(x)> alpha -r$. It follows that $(f+g)(x)> alpha$.




2. Let $(f+g)(x)> alpha$ and suppose that $x notin bigcup S_r$. Now choose a sequence $(r_n)$ of rationals such that

$f(x) > r_n$ for all $n$ and $r_n to f(x)$. Since $x notin S_r_n$ for all $n$, we have

$g(x) le alpha -r_n$ for all $n$. With $n to infty$ we get $g(x) le alpha-f(x)$, a contradiction !

share|cite|improve this answer

answered Jul 30 at 11:43

Fred

37k1237

add a comment | 

up vote
4
down vote



accepted

1. If $x in bigcup S_r$, then there is a rational $r$ such that $f(x)>r$ and $g(x)> alpha -r$. It follows that $(f+g)(x)> alpha$.




2. Let $(f+g)(x)> alpha$ and suppose that $x notin bigcup S_r$. Now choose a sequence $(r_n)$ of rationals such that

$f(x) > r_n$ for all $n$ and $r_n to f(x)$. Since $x notin S_r_n$ for all $n$, we have

$g(x) le alpha -r_n$ for all $n$. With $n to infty$ we get $g(x) le alpha-f(x)$, a contradiction !

share|cite|improve this answer

answered Jul 30 at 11:43

Fred

37k1237

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

1. If $x in bigcup S_r$, then there is a rational $r$ such that $f(x)>r$ and $g(x)> alpha -r$. It follows that $(f+g)(x)> alpha$.




2. Let $(f+g)(x)> alpha$ and suppose that $x notin bigcup S_r$. Now choose a sequence $(r_n)$ of rationals such that

$f(x) > r_n$ for all $n$ and $r_n to f(x)$. Since $x notin S_r_n$ for all $n$, we have

$g(x) le alpha -r_n$ for all $n$. With $n to infty$ we get $g(x) le alpha-f(x)$, a contradiction !

share|cite|improve this answer

answered Jul 30 at 11:43

Fred

37k1237

1. If $x in bigcup S_r$, then there is a rational $r$ such that $f(x)>r$ and $g(x)> alpha -r$. It follows that $(f+g)(x)> alpha$.




2. Let $(f+g)(x)> alpha$ and suppose that $x notin bigcup S_r$. Now choose a sequence $(r_n)$ of rationals such that

$f(x) > r_n$ for all $n$ and $r_n to f(x)$. Since $x notin S_r_n$ for all $n$, we have

$g(x) le alpha -r_n$ for all $n$. With $n to infty$ we get $g(x) le alpha-f(x)$, a contradiction !

share|cite|improve this answer

answered Jul 30 at 11:43

Fred

37k1237

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 11:43

Fred

37k1237

answered Jul 30 at 11:43

Fred

37k1237

answered Jul 30 at 11:43

Fred

37k1237

37k1237

add a comment | 

add a comment | 

up vote
1
down vote

It is evident that $S_r=f>rcapg>alpha-rsubseteqf+g>alpha$ for every $rinmathbb Q$.

If conversely $f(x)+g(x)>alpha$ or equivalently $f(x)>alpha-g(x)$ then, because $mathbb Q$ is dense in $mathbb R$, we can find some $rinmathbb Q$ with $r<f(x)$ (or equivalently $f(x)>r$) but still $r>alpha-g(x)$ (or equivalently $g(x)>alpha-r$). So we found an $rinmathbb Q$ with $xin S_r$.

share|cite|improve this answer

answered Jul 30 at 12:04

drhab

85.9k540118

add a comment | 

up vote
1
down vote

It is evident that $S_r=f>rcapg>alpha-rsubseteqf+g>alpha$ for every $rinmathbb Q$.

If conversely $f(x)+g(x)>alpha$ or equivalently $f(x)>alpha-g(x)$ then, because $mathbb Q$ is dense in $mathbb R$, we can find some $rinmathbb Q$ with $r<f(x)$ (or equivalently $f(x)>r$) but still $r>alpha-g(x)$ (or equivalently $g(x)>alpha-r$). So we found an $rinmathbb Q$ with $xin S_r$.

share|cite|improve this answer

answered Jul 30 at 12:04

drhab

85.9k540118

add a comment | 

up vote
1
down vote

up vote
1
down vote

It is evident that $S_r=f>rcapg>alpha-rsubseteqf+g>alpha$ for every $rinmathbb Q$.

If conversely $f(x)+g(x)>alpha$ or equivalently $f(x)>alpha-g(x)$ then, because $mathbb Q$ is dense in $mathbb R$, we can find some $rinmathbb Q$ with $r<f(x)$ (or equivalently $f(x)>r$) but still $r>alpha-g(x)$ (or equivalently $g(x)>alpha-r$). So we found an $rinmathbb Q$ with $xin S_r$.

share|cite|improve this answer

answered Jul 30 at 12:04

drhab

85.9k540118

It is evident that $S_r=f>rcapg>alpha-rsubseteqf+g>alpha$ for every $rinmathbb Q$.

If conversely $f(x)+g(x)>alpha$ or equivalently $f(x)>alpha-g(x)$ then, because $mathbb Q$ is dense in $mathbb R$, we can find some $rinmathbb Q$ with $r<f(x)$ (or equivalently $f(x)>r$) but still $r>alpha-g(x)$ (or equivalently $g(x)>alpha-r$). So we found an $rinmathbb Q$ with $xin S_r$.

share|cite|improve this answer

answered Jul 30 at 12:04

drhab

85.9k540118

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 12:04

drhab

85.9k540118

answered Jul 30 at 12:04

drhab

85.9k540118

answered Jul 30 at 12:04

drhab

85.9k540118

85.9k540118

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