Lebesgue-measurable almost everywhere equals Borel measurable function [closed]
$begingroup$
Let $, f: mathbb{R}^n rightarrow [-infty, +infty]$ be a Lebesgue measurable function.
I want to show that f is $lambda_n$- alomost everywhere equal to to a Borel measurable function $, f'$.
I would very much appreciate your help.
Best, KingDingeling
measure-theory lebesgue-measure measurable-functions borel-measures
asked Jan 7 at 12:05
KingDingelingKingDingeling
1126
$endgroup$
closed as off-topic by Nosrati, Adrian Keister, amWhy, José Carlos Santos, egreg Jan 7 at 23:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Adrian Keister, amWhy, José Carlos Santos, egreg
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $, f: mathbb{R}^n rightarrow [-infty, +infty]$ be a Lebesgue measurable function.
I want to show that f is $lambda_n$- alomost everywhere equal to to a Borel measurable function $, f'$.
I would very much appreciate your help.
Best, KingDingeling
measure-theory lebesgue-measure measurable-functions borel-measures
asked Jan 7 at 12:05
KingDingelingKingDingeling
1126
$endgroup$
closed as off-topic by Nosrati, Adrian Keister, amWhy, José Carlos Santos, egreg Jan 7 at 23:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Adrian Keister, amWhy, José Carlos Santos, egreg
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $, f: mathbb{R}^n rightarrow [-infty, +infty]$ be a Lebesgue measurable function.
I want to show that f is $lambda_n$- alomost everywhere equal to to a Borel measurable function $, f'$.
I would very much appreciate your help.
Best, KingDingeling
measure-theory lebesgue-measure measurable-functions borel-measures
asked Jan 7 at 12:05
KingDingelingKingDingeling
1126
$endgroup$
Let $, f: mathbb{R}^n rightarrow [-infty, +infty]$ be a Lebesgue measurable function.
I want to show that f is $lambda_n$- alomost everywhere equal to to a Borel measurable function $, f'$.
I would very much appreciate your help.
Best, KingDingeling
measure-theory lebesgue-measure measurable-functions borel-measures
measure-theory lebesgue-measure measurable-functions borel-measures
asked Jan 7 at 12:05
KingDingelingKingDingeling
1126
asked Jan 7 at 12:05
KingDingelingKingDingeling
1126
asked Jan 7 at 12:05
KingDingelingKingDingeling
1126
asked Jan 7 at 12:05
KingDingelingKingDingeling
1126
asked Jan 7 at 12:05
KingDingelingKingDingeling
1126
1126
closed as off-topic by Nosrati, Adrian Keister, amWhy, José Carlos Santos, egreg Jan 7 at 23:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Adrian Keister, amWhy, José Carlos Santos, egreg
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Nosrati, Adrian Keister, amWhy, José Carlos Santos, egreg Jan 7 at 23:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Adrian Keister, amWhy, José Carlos Santos, egreg
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I suppose by "Lebesgue mesurable' you mean that inverse image of any Borel set as well as those of ${infty}$, ${-infty}$ are Lebesgue measurable sets. If $f$ is a simple function this follows easily since any Lebesgue measurable set is almost everywhere equal to a Borel set. Now take limits.
answered Jan 7 at 12:12
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
$begingroup$
I am sorry, but I don't know how to do the algebraic induction here.
$endgroup$
– KingDingeling
Jan 7 at 12:41
add a comment |
$begingroup$
Sketch (just an elaboration of the other answer):
$f$ is the pointwise limit of simple functions $f_n.$
- Each $f_n$ is a sum of terms $cchi_A$ where $cin mathbb R$ and $A$ is Lebesgue measurable.
- Each $A=Ecup N$ where $E$ is a Borel set and $lambda (N)=0.$
- Therefore, if $g_n$ is defined in the "obvious way", you can show that $g_n$ is Borel measurable and then show
$f$ is the pointwise limit of $g_n$.
answered Jan 7 at 16:52
MatematletaMatematleta
10.2k2918
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I suppose by "Lebesgue mesurable' you mean that inverse image of any Borel set as well as those of ${infty}$, ${-infty}$ are Lebesgue measurable sets. If $f$ is a simple function this follows easily since any Lebesgue measurable set is almost everywhere equal to a Borel set. Now take limits.
answered Jan 7 at 12:12
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
$begingroup$
I am sorry, but I don't know how to do the algebraic induction here.
$endgroup$
– KingDingeling
Jan 7 at 12:41
add a comment |
$begingroup$
I suppose by "Lebesgue mesurable' you mean that inverse image of any Borel set as well as those of ${infty}$, ${-infty}$ are Lebesgue measurable sets. If $f$ is a simple function this follows easily since any Lebesgue measurable set is almost everywhere equal to a Borel set. Now take limits.
answered Jan 7 at 12:12
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
$begingroup$
I am sorry, but I don't know how to do the algebraic induction here.
$endgroup$
– KingDingeling
Jan 7 at 12:41
add a comment |
$begingroup$
I suppose by "Lebesgue mesurable' you mean that inverse image of any Borel set as well as those of ${infty}$, ${-infty}$ are Lebesgue measurable sets. If $f$ is a simple function this follows easily since any Lebesgue measurable set is almost everywhere equal to a Borel set. Now take limits.
answered Jan 7 at 12:12
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
$endgroup$
I suppose by "Lebesgue mesurable' you mean that inverse image of any Borel set as well as those of ${infty}$, ${-infty}$ are Lebesgue measurable sets. If $f$ is a simple function this follows easily since any Lebesgue measurable set is almost everywhere equal to a Borel set. Now take limits.
answered Jan 7 at 12:12
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Jan 7 at 12:12
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Jan 7 at 12:12
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
answered Jan 7 at 12:12
Kavi Rama MurthyKavi Rama Murthy
52.8k32055
52.8k32055
$begingroup$
I am sorry, but I don't know how to do the algebraic induction here.
$endgroup$
– KingDingeling
Jan 7 at 12:41
add a comment |
$begingroup$
I am sorry, but I don't know how to do the algebraic induction here.
$endgroup$
– KingDingeling
Jan 7 at 12:41
$begingroup$
I am sorry, but I don't know how to do the algebraic induction here.
$endgroup$
– KingDingeling
Jan 7 at 12:41
$begingroup$
I am sorry, but I don't know how to do the algebraic induction here.
$endgroup$
– KingDingeling
Jan 7 at 12:41
add a comment |
$begingroup$
Sketch (just an elaboration of the other answer):
$f$ is the pointwise limit of simple functions $f_n.$
- Each $f_n$ is a sum of terms $cchi_A$ where $cin mathbb R$ and $A$ is Lebesgue measurable.
- Each $A=Ecup N$ where $E$ is a Borel set and $lambda (N)=0.$
- Therefore, if $g_n$ is defined in the "obvious way", you can show that $g_n$ is Borel measurable and then show
$f$ is the pointwise limit of $g_n$.
answered Jan 7 at 16:52
MatematletaMatematleta
10.2k2918
$endgroup$
add a comment |
$begingroup$
Sketch (just an elaboration of the other answer):
$f$ is the pointwise limit of simple functions $f_n.$
- Each $f_n$ is a sum of terms $cchi_A$ where $cin mathbb R$ and $A$ is Lebesgue measurable.
- Each $A=Ecup N$ where $E$ is a Borel set and $lambda (N)=0.$
- Therefore, if $g_n$ is defined in the "obvious way", you can show that $g_n$ is Borel measurable and then show
$f$ is the pointwise limit of $g_n$.
answered Jan 7 at 16:52
MatematletaMatematleta
10.2k2918
$endgroup$
add a comment |
$begingroup$
Sketch (just an elaboration of the other answer):
$f$ is the pointwise limit of simple functions $f_n.$
- Each $f_n$ is a sum of terms $cchi_A$ where $cin mathbb R$ and $A$ is Lebesgue measurable.
- Each $A=Ecup N$ where $E$ is a Borel set and $lambda (N)=0.$
- Therefore, if $g_n$ is defined in the "obvious way", you can show that $g_n$ is Borel measurable and then show
$f$ is the pointwise limit of $g_n$.
answered Jan 7 at 16:52
MatematletaMatematleta
10.2k2918
$endgroup$
Sketch (just an elaboration of the other answer):
$f$ is the pointwise limit of simple functions $f_n.$
- Each $f_n$ is a sum of terms $cchi_A$ where $cin mathbb R$ and $A$ is Lebesgue measurable.
- Each $A=Ecup N$ where $E$ is a Borel set and $lambda (N)=0.$
- Therefore, if $g_n$ is defined in the "obvious way", you can show that $g_n$ is Borel measurable and then show
$f$ is the pointwise limit of $g_n$.
answered Jan 7 at 16:52
MatematletaMatematleta
10.2k2918
answered Jan 7 at 16:52
MatematletaMatematleta
10.2k2918
answered Jan 7 at 16:52
MatematletaMatematleta
10.2k2918
answered Jan 7 at 16:52
MatematletaMatematleta
10.2k2918
10.2k2918
add a comment |
add a comment |
