What does $f = g$ almost everywhere mean? The Next CEO of Stack OverflowMeasurability of almost everywhere continuous functionsShow that a function almost everywhere continuous is measurableProve that every Lebesgue measurable function is equal almost everywhere to a Borel measurable functionA function constant almost everywhereWhat does “almost everywhere” mean in convergence almost everywhere?Sequences of Borel measurable functions and limit almost everywhereIf $f$ is finite almost everywhere and $E$ is of finite Lebesgue measure, then f is “almost” boundedProving for a Lebesgue measure the convergence in measure is equivalent to the convergence almost everywhere.A confusion on the definition of *almost everywhere*measurable and finite almost everywhere
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What does $f = g$ almost everywhere mean?
The Next CEO of Stack OverflowMeasurability of almost everywhere continuous functionsShow that a function almost everywhere continuous is measurableProve that every Lebesgue measurable function is equal almost everywhere to a Borel measurable functionA function constant almost everywhereWhat does “almost everywhere” mean in convergence almost everywhere?Sequences of Borel measurable functions and limit almost everywhereIf $f$ is finite almost everywhere and $E$ is of finite Lebesgue measure, then f is “almost” boundedProving for a Lebesgue measure the convergence in measure is equivalent to the convergence almost everywhere.A confusion on the definition of *almost everywhere*measurable and finite almost everywhere
$begingroup$
I need to prove the following proposition.
If $f,g: [a,b] rightarrow mathbbR, $ $f$ is Lebesgue measurable, and $ f = g $ almost everywhere, then $g$ is Lebesgue measurable. But I do not know what exactly is almost everywhere.
real-analysis lebesgue-measure
edited Mar 27 at 18:01
zhw.
74.8k43175
asked Mar 22 at 19:41
José MarínJosé Marín
195210
$endgroup$
add a comment |
$begingroup$
I need to prove the following proposition.
If $f,g: [a,b] rightarrow mathbbR, $ $f$ is Lebesgue measurable, and $ f = g $ almost everywhere, then $g$ is Lebesgue measurable. But I do not know what exactly is almost everywhere.
real-analysis lebesgue-measure
edited Mar 27 at 18:01
zhw.
74.8k43175
asked Mar 22 at 19:41
José MarínJosé Marín
195210
$endgroup$
1
$begingroup$
It means the set of points where $fne g$ has measure $0$
$endgroup$
– J. W. Tanner
Mar 22 at 19:44
1
$begingroup$
Exactly what Tanner said. The set $A:=xin[a,b] : f(x)not = g(x)$ has measure zero. In this case it means that for all $varepsilon>0$ there exists intervals $I_1,...,I_n$ of total length $<varepsilon$ such that $Asubseteq bigcup_i=1^n I_i$.
$endgroup$
– Yanko
Mar 22 at 19:46
$begingroup$
We usually specify what measure we are talking about to avoid confusion. If no measure is specified, we usually mean Lebesgue measure.
$endgroup$
– Jakobian
Mar 22 at 19:50
$begingroup$
You should restate the problem, as some hypotheses are missing
$endgroup$
– zhw.
Mar 22 at 19:54
$begingroup$
I just edited so it should be clear.
$endgroup$
– José Marín
Mar 22 at 19:56
add a comment |
$begingroup$
I need to prove the following proposition.
If $f,g: [a,b] rightarrow mathbbR, $ $f$ is Lebesgue measurable, and $ f = g $ almost everywhere, then $g$ is Lebesgue measurable. But I do not know what exactly is almost everywhere.
real-analysis lebesgue-measure
edited Mar 27 at 18:01
zhw.
74.8k43175
asked Mar 22 at 19:41
José MarínJosé Marín
195210
$endgroup$
I need to prove the following proposition.
If $f,g: [a,b] rightarrow mathbbR, $ $f$ is Lebesgue measurable, and $ f = g $ almost everywhere, then $g$ is Lebesgue measurable. But I do not know what exactly is almost everywhere.
real-analysis lebesgue-measure
real-analysis lebesgue-measure
edited Mar 27 at 18:01
zhw.
74.8k43175
asked Mar 22 at 19:41
José MarínJosé Marín
195210
edited Mar 27 at 18:01
zhw.
74.8k43175
asked Mar 22 at 19:41
José MarínJosé Marín
195210
edited Mar 27 at 18:01
zhw.
74.8k43175
edited Mar 27 at 18:01
zhw.
74.8k43175
edited Mar 27 at 18:01
zhw.
74.8k43175
74.8k43175
asked Mar 22 at 19:41
José MarínJosé Marín
195210
asked Mar 22 at 19:41
José MarínJosé Marín
195210
asked Mar 22 at 19:41
José MarínJosé Marín
195210
195210
1
$begingroup$
It means the set of points where $fne g$ has measure $0$
$endgroup$
– J. W. Tanner
Mar 22 at 19:44
1
$begingroup$
Exactly what Tanner said. The set $A:=xin[a,b] : f(x)not = g(x)$ has measure zero. In this case it means that for all $varepsilon>0$ there exists intervals $I_1,...,I_n$ of total length $<varepsilon$ such that $Asubseteq bigcup_i=1^n I_i$.
$endgroup$
– Yanko
Mar 22 at 19:46
$begingroup$
We usually specify what measure we are talking about to avoid confusion. If no measure is specified, we usually mean Lebesgue measure.
$endgroup$
– Jakobian
Mar 22 at 19:50
$begingroup$
You should restate the problem, as some hypotheses are missing
$endgroup$
– zhw.
Mar 22 at 19:54
$begingroup$
I just edited so it should be clear.
$endgroup$
– José Marín
Mar 22 at 19:56
add a comment |
1
$begingroup$
It means the set of points where $fne g$ has measure $0$
$endgroup$
– J. W. Tanner
Mar 22 at 19:44
1
$begingroup$
Exactly what Tanner said. The set $A:=xin[a,b] : f(x)not = g(x)$ has measure zero. In this case it means that for all $varepsilon>0$ there exists intervals $I_1,...,I_n$ of total length $<varepsilon$ such that $Asubseteq bigcup_i=1^n I_i$.
$endgroup$
– Yanko
Mar 22 at 19:46
$begingroup$
We usually specify what measure we are talking about to avoid confusion. If no measure is specified, we usually mean Lebesgue measure.
$endgroup$
– Jakobian
Mar 22 at 19:50
$begingroup$
You should restate the problem, as some hypotheses are missing
$endgroup$
– zhw.
Mar 22 at 19:54
$begingroup$
I just edited so it should be clear.
$endgroup$
– José Marín
Mar 22 at 19:56
1
1
$begingroup$
It means the set of points where $fne g$ has measure $0$
$endgroup$
– J. W. Tanner
Mar 22 at 19:44
$begingroup$
It means the set of points where $fne g$ has measure $0$
$endgroup$
– J. W. Tanner
Mar 22 at 19:44
1
1
$begingroup$
Exactly what Tanner said. The set $A:=xin[a,b] : f(x)not = g(x)$ has measure zero. In this case it means that for all $varepsilon>0$ there exists intervals $I_1,...,I_n$ of total length $<varepsilon$ such that $Asubseteq bigcup_i=1^n I_i$.
$endgroup$
– Yanko
Mar 22 at 19:46
$begingroup$
Exactly what Tanner said. The set $A:=xin[a,b] : f(x)not = g(x)$ has measure zero. In this case it means that for all $varepsilon>0$ there exists intervals $I_1,...,I_n$ of total length $<varepsilon$ such that $Asubseteq bigcup_i=1^n I_i$.
$endgroup$
– Yanko
Mar 22 at 19:46
$begingroup$
We usually specify what measure we are talking about to avoid confusion. If no measure is specified, we usually mean Lebesgue measure.
$endgroup$
– Jakobian
Mar 22 at 19:50
$begingroup$
We usually specify what measure we are talking about to avoid confusion. If no measure is specified, we usually mean Lebesgue measure.
$endgroup$
– Jakobian
Mar 22 at 19:50
$begingroup$
You should restate the problem, as some hypotheses are missing
$endgroup$
– zhw.
Mar 22 at 19:54
$begingroup$
You should restate the problem, as some hypotheses are missing
$endgroup$
– zhw.
Mar 22 at 19:54
$begingroup$
I just edited so it should be clear.
$endgroup$
– José Marín
Mar 22 at 19:56
$begingroup$
I just edited so it should be clear.
$endgroup$
– José Marín
Mar 22 at 19:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This terminology is from measure theory. Two functions are said to be equal almost everywhere on $[a,b]$ if there exists a set $Esubseteq [a,b]$ of measure $0$ such that $f(x)=g(x)$ for all $xin [a,b]setminus E$.
A subset $Esubseteq [a,b]$ is said to have measure $0$ if for every $epsilon>0$ there exist countably many intervals $((a_k, b_k))_k$ such that $Esubseteq bigcup_k (a_k, b_k)$ and $sum_k (b_k-a_k) < epsilon$.
How you can go about showing that if $f$ is measurable then so is $g$ depends on the definition you are using. Either way, I hope this helps!
edited Mar 24 at 2:06
J. W. Tanner
4,0611320
answered Mar 22 at 19:50
QuokaQuoka
1,568316
$endgroup$
$begingroup$
Yes, it helps thanks!
$endgroup$
– José Marín
Mar 22 at 20:03
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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$begingroup$
This terminology is from measure theory. Two functions are said to be equal almost everywhere on $[a,b]$ if there exists a set $Esubseteq [a,b]$ of measure $0$ such that $f(x)=g(x)$ for all $xin [a,b]setminus E$.
A subset $Esubseteq [a,b]$ is said to have measure $0$ if for every $epsilon>0$ there exist countably many intervals $((a_k, b_k))_k$ such that $Esubseteq bigcup_k (a_k, b_k)$ and $sum_k (b_k-a_k) < epsilon$.
How you can go about showing that if $f$ is measurable then so is $g$ depends on the definition you are using. Either way, I hope this helps!
edited Mar 24 at 2:06
J. W. Tanner
4,0611320
answered Mar 22 at 19:50
QuokaQuoka
1,568316
$endgroup$
$begingroup$
Yes, it helps thanks!
$endgroup$
– José Marín
Mar 22 at 20:03
add a comment |
$begingroup$
This terminology is from measure theory. Two functions are said to be equal almost everywhere on $[a,b]$ if there exists a set $Esubseteq [a,b]$ of measure $0$ such that $f(x)=g(x)$ for all $xin [a,b]setminus E$.
A subset $Esubseteq [a,b]$ is said to have measure $0$ if for every $epsilon>0$ there exist countably many intervals $((a_k, b_k))_k$ such that $Esubseteq bigcup_k (a_k, b_k)$ and $sum_k (b_k-a_k) < epsilon$.
How you can go about showing that if $f$ is measurable then so is $g$ depends on the definition you are using. Either way, I hope this helps!
edited Mar 24 at 2:06
J. W. Tanner
4,0611320
answered Mar 22 at 19:50
QuokaQuoka
1,568316
$endgroup$
$begingroup$
Yes, it helps thanks!
$endgroup$
– José Marín
Mar 22 at 20:03
add a comment |
$begingroup$
This terminology is from measure theory. Two functions are said to be equal almost everywhere on $[a,b]$ if there exists a set $Esubseteq [a,b]$ of measure $0$ such that $f(x)=g(x)$ for all $xin [a,b]setminus E$.
A subset $Esubseteq [a,b]$ is said to have measure $0$ if for every $epsilon>0$ there exist countably many intervals $((a_k, b_k))_k$ such that $Esubseteq bigcup_k (a_k, b_k)$ and $sum_k (b_k-a_k) < epsilon$.
How you can go about showing that if $f$ is measurable then so is $g$ depends on the definition you are using. Either way, I hope this helps!
edited Mar 24 at 2:06
J. W. Tanner
4,0611320
answered Mar 22 at 19:50
QuokaQuoka
1,568316
$endgroup$
This terminology is from measure theory. Two functions are said to be equal almost everywhere on $[a,b]$ if there exists a set $Esubseteq [a,b]$ of measure $0$ such that $f(x)=g(x)$ for all $xin [a,b]setminus E$.
A subset $Esubseteq [a,b]$ is said to have measure $0$ if for every $epsilon>0$ there exist countably many intervals $((a_k, b_k))_k$ such that $Esubseteq bigcup_k (a_k, b_k)$ and $sum_k (b_k-a_k) < epsilon$.
How you can go about showing that if $f$ is measurable then so is $g$ depends on the definition you are using. Either way, I hope this helps!
edited Mar 24 at 2:06
J. W. Tanner
4,0611320
answered Mar 22 at 19:50
QuokaQuoka
1,568316
edited Mar 24 at 2:06
J. W. Tanner
4,0611320
edited Mar 24 at 2:06
J. W. Tanner
4,0611320
edited Mar 24 at 2:06
J. W. Tanner
4,0611320
4,0611320
answered Mar 22 at 19:50
QuokaQuoka
1,568316
answered Mar 22 at 19:50
QuokaQuoka
1,568316
answered Mar 22 at 19:50
QuokaQuoka
1,568316
1,568316
$begingroup$
Yes, it helps thanks!
$endgroup$
– José Marín
Mar 22 at 20:03
add a comment |
$begingroup$
Yes, it helps thanks!
$endgroup$
– José Marín
Mar 22 at 20:03
$begingroup$
Yes, it helps thanks!
$endgroup$
– José Marín
Mar 22 at 20:03
$begingroup$
Yes, it helps thanks!
$endgroup$
– José Marín
Mar 22 at 20:03
add a comment |
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1
$begingroup$
It means the set of points where $fne g$ has measure $0$
$endgroup$
– J. W. Tanner
Mar 22 at 19:44
1
$begingroup$
Exactly what Tanner said. The set $A:=xin[a,b] : f(x)not = g(x)$ has measure zero. In this case it means that for all $varepsilon>0$ there exists intervals $I_1,...,I_n$ of total length $<varepsilon$ such that $Asubseteq bigcup_i=1^n I_i$.
$endgroup$
– Yanko
Mar 22 at 19:46
$begingroup$
We usually specify what measure we are talking about to avoid confusion. If no measure is specified, we usually mean Lebesgue measure.
$endgroup$
– Jakobian
Mar 22 at 19:50
$begingroup$
You should restate the problem, as some hypotheses are missing
$endgroup$
– zhw.
Mar 22 at 19:54
$begingroup$
I just edited so it should be clear.
$endgroup$
– José Marín
Mar 22 at 19:56