If a function $f$ is measurable in the completion space then there is a function $g$ measurable in the...
$begingroup$
I have a question about this proof of a theorem from Yeh’s Real Analysis (3rd edition).
The theorem and proof are in the attached images:
My question pertains to the part that says:
${D : g < r_n} = {D setminus N : g < r_n} cup {N : g < r_n}$
How do we get that $D setminus N cup N = D$? This is only true if $N subseteq D$, but I’m not sure exactly how this came to be. I’m guessing it’s from the fact that D is a complete measure space? But it’s a bit vague to me.
Thanks!
real-analysis measure-theory
asked Nov 29 '18 at 18:23
Jane DoeJane Doe
335113
$endgroup$
|
show 3 more comments
$begingroup$
I have a question about this proof of a theorem from Yeh’s Real Analysis (3rd edition).
The theorem and proof are in the attached images:
My question pertains to the part that says:
${D : g < r_n} = {D setminus N : g < r_n} cup {N : g < r_n}$
How do we get that $D setminus N cup N = D$? This is only true if $N subseteq D$, but I’m not sure exactly how this came to be. I’m guessing it’s from the fact that D is a complete measure space? But it’s a bit vague to me.
Thanks!
real-analysis measure-theory
asked Nov 29 '18 at 18:23
Jane DoeJane Doe
335113
$endgroup$
$begingroup$
What is $D$ to begin with? And why it is not $X$?
$endgroup$
– Will M.
Nov 29 '18 at 18:26
$begingroup$
@WillM. I don’t think I understand your question. $D$ is defined in the statement of the theorem. It could be $X$, I suppose? But why would it necessarily be?
$endgroup$
– Jane Doe
Nov 29 '18 at 18:30
$begingroup$
So, the author is actually assuming the measure space to be $(D, mathfrak{A})$?
$endgroup$
– Will M.
Nov 29 '18 at 20:11
$begingroup$
@WillM. No, $mathfrak{A}$ is a $sigma$-algebra of subsets of $X$ and $D in mathfrak{A}$ so $D subseteq X$.
$endgroup$
– Jane Doe
Nov 29 '18 at 20:18
1
$begingroup$
@JaneDoe I see from your profile that you have not accepted answers to many of the questions you have asked. If any particular answer has helped you resolve the question, then consider upvoting and accepting an answer. This is considered good practice on this site, and helps remove your question from the unanswered list.
$endgroup$
– Brahadeesh
Nov 30 '18 at 10:46
|
show 3 more comments
$begingroup$
I have a question about this proof of a theorem from Yeh’s Real Analysis (3rd edition).
The theorem and proof are in the attached images:
My question pertains to the part that says:
${D : g < r_n} = {D setminus N : g < r_n} cup {N : g < r_n}$
How do we get that $D setminus N cup N = D$? This is only true if $N subseteq D$, but I’m not sure exactly how this came to be. I’m guessing it’s from the fact that D is a complete measure space? But it’s a bit vague to me.
Thanks!
real-analysis measure-theory
asked Nov 29 '18 at 18:23
Jane DoeJane Doe
335113
$endgroup$
I have a question about this proof of a theorem from Yeh’s Real Analysis (3rd edition).
The theorem and proof are in the attached images:
My question pertains to the part that says:
${D : g < r_n} = {D setminus N : g < r_n} cup {N : g < r_n}$
How do we get that $D setminus N cup N = D$? This is only true if $N subseteq D$, but I’m not sure exactly how this came to be. I’m guessing it’s from the fact that D is a complete measure space? But it’s a bit vague to me.
Thanks!
real-analysis measure-theory
real-analysis measure-theory
asked Nov 29 '18 at 18:23
Jane DoeJane Doe
335113
asked Nov 29 '18 at 18:23
Jane DoeJane Doe
335113
asked Nov 29 '18 at 18:23
Jane DoeJane Doe
335113
asked Nov 29 '18 at 18:23
Jane DoeJane Doe
335113
asked Nov 29 '18 at 18:23
Jane DoeJane Doe
335113
335113
$begingroup$
What is $D$ to begin with? And why it is not $X$?
$endgroup$
– Will M.
Nov 29 '18 at 18:26
$begingroup$
@WillM. I don’t think I understand your question. $D$ is defined in the statement of the theorem. It could be $X$, I suppose? But why would it necessarily be?
$endgroup$
– Jane Doe
Nov 29 '18 at 18:30
$begingroup$
So, the author is actually assuming the measure space to be $(D, mathfrak{A})$?
$endgroup$
– Will M.
Nov 29 '18 at 20:11
$begingroup$
@WillM. No, $mathfrak{A}$ is a $sigma$-algebra of subsets of $X$ and $D in mathfrak{A}$ so $D subseteq X$.
$endgroup$
– Jane Doe
Nov 29 '18 at 20:18
1
$begingroup$
@JaneDoe I see from your profile that you have not accepted answers to many of the questions you have asked. If any particular answer has helped you resolve the question, then consider upvoting and accepting an answer. This is considered good practice on this site, and helps remove your question from the unanswered list.
$endgroup$
– Brahadeesh
Nov 30 '18 at 10:46
|
show 3 more comments
$begingroup$
What is $D$ to begin with? And why it is not $X$?
$endgroup$
– Will M.
Nov 29 '18 at 18:26
$begingroup$
@WillM. I don’t think I understand your question. $D$ is defined in the statement of the theorem. It could be $X$, I suppose? But why would it necessarily be?
$endgroup$
– Jane Doe
Nov 29 '18 at 18:30
$begingroup$
So, the author is actually assuming the measure space to be $(D, mathfrak{A})$?
$endgroup$
– Will M.
Nov 29 '18 at 20:11
$begingroup$
@WillM. No, $mathfrak{A}$ is a $sigma$-algebra of subsets of $X$ and $D in mathfrak{A}$ so $D subseteq X$.
$endgroup$
– Jane Doe
Nov 29 '18 at 20:18
1
$begingroup$
@JaneDoe I see from your profile that you have not accepted answers to many of the questions you have asked. If any particular answer has helped you resolve the question, then consider upvoting and accepting an answer. This is considered good practice on this site, and helps remove your question from the unanswered list.
$endgroup$
– Brahadeesh
Nov 30 '18 at 10:46
$begingroup$
What is $D$ to begin with? And why it is not $X$?
$endgroup$
– Will M.
Nov 29 '18 at 18:26
$begingroup$
What is $D$ to begin with? And why it is not $X$?
$endgroup$
– Will M.
Nov 29 '18 at 18:26
$begingroup$
@WillM. I don’t think I understand your question. $D$ is defined in the statement of the theorem. It could be $X$, I suppose? But why would it necessarily be?
$endgroup$
– Jane Doe
Nov 29 '18 at 18:30
$begingroup$
@WillM. I don’t think I understand your question. $D$ is defined in the statement of the theorem. It could be $X$, I suppose? But why would it necessarily be?
$endgroup$
– Jane Doe
Nov 29 '18 at 18:30
$begingroup$
So, the author is actually assuming the measure space to be $(D, mathfrak{A})$?
$endgroup$
– Will M.
Nov 29 '18 at 20:11
$begingroup$
So, the author is actually assuming the measure space to be $(D, mathfrak{A})$?
$endgroup$
– Will M.
Nov 29 '18 at 20:11
$begingroup$
@WillM. No, $mathfrak{A}$ is a $sigma$-algebra of subsets of $X$ and $D in mathfrak{A}$ so $D subseteq X$.
$endgroup$
– Jane Doe
Nov 29 '18 at 20:18
$begingroup$
@WillM. No, $mathfrak{A}$ is a $sigma$-algebra of subsets of $X$ and $D in mathfrak{A}$ so $D subseteq X$.
$endgroup$
– Jane Doe
Nov 29 '18 at 20:18
1
1
$begingroup$
@JaneDoe I see from your profile that you have not accepted answers to many of the questions you have asked. If any particular answer has helped you resolve the question, then consider upvoting and accepting an answer. This is considered good practice on this site, and helps remove your question from the unanswered list.
$endgroup$
– Brahadeesh
Nov 30 '18 at 10:46
$begingroup$
@JaneDoe I see from your profile that you have not accepted answers to many of the questions you have asked. If any particular answer has helped you resolve the question, then consider upvoting and accepting an answer. This is considered good practice on this site, and helps remove your question from the unanswered list.
$endgroup$
– Brahadeesh
Nov 30 '18 at 10:46
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Although your question pertains to equation $(1)$, observe that the same assumption, namely that $D = (D setminus N) cup N$, is made earlier when defining $g$ on $D$ by defining it separately on $D setminus N$ and $N$.
You are right that the author is assuming that $N subset D$, and that this is not true from the way $N$ is defined. However, if we let $N' := N cap D$, then the same proof goes through with $N'$ in place of $N$. For instance, it is clear that $N' in mathfrak{A}$, $N' subset D$, $N'$ is a null set and $C_n subset B_n subset N'$. Note that you are not using the fact that $D$ is a complete measure space, you only need that $D, N in mathfrak{A}$ to conclude that $D cap N in mathfrak{A}$, etc.
So, we can take without loss of generality that $N$ is a subset of $D$. But the author should have mentioned this for the sake of clarity, in my opinion. Good catch.
answered Nov 30 '18 at 9:47
BrahadeeshBrahadeesh
6,32942362
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Although your question pertains to equation $(1)$, observe that the same assumption, namely that $D = (D setminus N) cup N$, is made earlier when defining $g$ on $D$ by defining it separately on $D setminus N$ and $N$.
You are right that the author is assuming that $N subset D$, and that this is not true from the way $N$ is defined. However, if we let $N' := N cap D$, then the same proof goes through with $N'$ in place of $N$. For instance, it is clear that $N' in mathfrak{A}$, $N' subset D$, $N'$ is a null set and $C_n subset B_n subset N'$. Note that you are not using the fact that $D$ is a complete measure space, you only need that $D, N in mathfrak{A}$ to conclude that $D cap N in mathfrak{A}$, etc.
So, we can take without loss of generality that $N$ is a subset of $D$. But the author should have mentioned this for the sake of clarity, in my opinion. Good catch.
answered Nov 30 '18 at 9:47
BrahadeeshBrahadeesh
6,32942362
$endgroup$
add a comment |
$begingroup$
Although your question pertains to equation $(1)$, observe that the same assumption, namely that $D = (D setminus N) cup N$, is made earlier when defining $g$ on $D$ by defining it separately on $D setminus N$ and $N$.
You are right that the author is assuming that $N subset D$, and that this is not true from the way $N$ is defined. However, if we let $N' := N cap D$, then the same proof goes through with $N'$ in place of $N$. For instance, it is clear that $N' in mathfrak{A}$, $N' subset D$, $N'$ is a null set and $C_n subset B_n subset N'$. Note that you are not using the fact that $D$ is a complete measure space, you only need that $D, N in mathfrak{A}$ to conclude that $D cap N in mathfrak{A}$, etc.
So, we can take without loss of generality that $N$ is a subset of $D$. But the author should have mentioned this for the sake of clarity, in my opinion. Good catch.
answered Nov 30 '18 at 9:47
BrahadeeshBrahadeesh
6,32942362
$endgroup$
add a comment |
$begingroup$
Although your question pertains to equation $(1)$, observe that the same assumption, namely that $D = (D setminus N) cup N$, is made earlier when defining $g$ on $D$ by defining it separately on $D setminus N$ and $N$.
You are right that the author is assuming that $N subset D$, and that this is not true from the way $N$ is defined. However, if we let $N' := N cap D$, then the same proof goes through with $N'$ in place of $N$. For instance, it is clear that $N' in mathfrak{A}$, $N' subset D$, $N'$ is a null set and $C_n subset B_n subset N'$. Note that you are not using the fact that $D$ is a complete measure space, you only need that $D, N in mathfrak{A}$ to conclude that $D cap N in mathfrak{A}$, etc.
So, we can take without loss of generality that $N$ is a subset of $D$. But the author should have mentioned this for the sake of clarity, in my opinion. Good catch.
answered Nov 30 '18 at 9:47
BrahadeeshBrahadeesh
6,32942362
$endgroup$
Although your question pertains to equation $(1)$, observe that the same assumption, namely that $D = (D setminus N) cup N$, is made earlier when defining $g$ on $D$ by defining it separately on $D setminus N$ and $N$.
You are right that the author is assuming that $N subset D$, and that this is not true from the way $N$ is defined. However, if we let $N' := N cap D$, then the same proof goes through with $N'$ in place of $N$. For instance, it is clear that $N' in mathfrak{A}$, $N' subset D$, $N'$ is a null set and $C_n subset B_n subset N'$. Note that you are not using the fact that $D$ is a complete measure space, you only need that $D, N in mathfrak{A}$ to conclude that $D cap N in mathfrak{A}$, etc.
So, we can take without loss of generality that $N$ is a subset of $D$. But the author should have mentioned this for the sake of clarity, in my opinion. Good catch.
answered Nov 30 '18 at 9:47
BrahadeeshBrahadeesh
6,32942362
answered Nov 30 '18 at 9:47
BrahadeeshBrahadeesh
6,32942362
answered Nov 30 '18 at 9:47
BrahadeeshBrahadeesh
6,32942362
answered Nov 30 '18 at 9:47
BrahadeeshBrahadeesh
6,32942362
6,32942362
add a comment |
add a comment |
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$begingroup$
What is $D$ to begin with? And why it is not $X$?
$endgroup$
– Will M.
Nov 29 '18 at 18:26
$begingroup$
@WillM. I don’t think I understand your question. $D$ is defined in the statement of the theorem. It could be $X$, I suppose? But why would it necessarily be?
$endgroup$
– Jane Doe
Nov 29 '18 at 18:30
$begingroup$
So, the author is actually assuming the measure space to be $(D, mathfrak{A})$?
$endgroup$
– Will M.
Nov 29 '18 at 20:11
$begingroup$
@WillM. No, $mathfrak{A}$ is a $sigma$-algebra of subsets of $X$ and $D in mathfrak{A}$ so $D subseteq X$.
$endgroup$
– Jane Doe
Nov 29 '18 at 20:18
1
$begingroup$
@JaneDoe I see from your profile that you have not accepted answers to many of the questions you have asked. If any particular answer has helped you resolve the question, then consider upvoting and accepting an answer. This is considered good practice on this site, and helps remove your question from the unanswered list.
$endgroup$
– Brahadeesh
Nov 30 '18 at 10:46