Working with this 'almost everywhere' statement
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Consider a real function $f$ and take $mathbb R$ together with Lebesgue measure $m$. I want to check if I have the right reasoning with regards to working with the following "almost everywhere" statement; I have that $f(x)neq0$ for almost all $xinmathbb R$.
Now, I know that this means that the set of $xinmathbb R$ for which $f(x)=0$ is a null set with respect to Lebesgue measure. That is, $m({xinmathbb R:f(x)=0})=0$. But since the statement that $f(x)neq0$ holds for almost all $xinmathbb R$, does that mean that $m({xinmathbb R:f(x)neq0})=m(mathbb R)=+infty$? My thinking is that since the statement holds for all $mathbb R$ except possibly on a subset of $mathbb R$ with measure zero, then the measure of the set on which the statement holds must coincide with the measure of $mathbb R$; is this the correct reasoning?
real-analysis functional-analysis measure-theory lebesgue-measure
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up vote
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Consider a real function $f$ and take $mathbb R$ together with Lebesgue measure $m$. I want to check if I have the right reasoning with regards to working with the following "almost everywhere" statement; I have that $f(x)neq0$ for almost all $xinmathbb R$.
Now, I know that this means that the set of $xinmathbb R$ for which $f(x)=0$ is a null set with respect to Lebesgue measure. That is, $m({xinmathbb R:f(x)=0})=0$. But since the statement that $f(x)neq0$ holds for almost all $xinmathbb R$, does that mean that $m({xinmathbb R:f(x)neq0})=m(mathbb R)=+infty$? My thinking is that since the statement holds for all $mathbb R$ except possibly on a subset of $mathbb R$ with measure zero, then the measure of the set on which the statement holds must coincide with the measure of $mathbb R$; is this the correct reasoning?
real-analysis functional-analysis measure-theory lebesgue-measure
Yes, you are right.
– Kavi Rama Murthy
8 hours ago
@mathworker21, I don't think that it is too worrisome (unless you can indicate why it ought to be). I was unsure how to make use of the additivity property of the measure in a rigorous manner, but I see now how designating this set as $E$ makes that much clearer.
– Jeremy Jeffrey James
8 hours ago
@JeremyJeffreyJames my apologies for the comment. I was wrong
– mathworker21
8 hours ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Consider a real function $f$ and take $mathbb R$ together with Lebesgue measure $m$. I want to check if I have the right reasoning with regards to working with the following "almost everywhere" statement; I have that $f(x)neq0$ for almost all $xinmathbb R$.
Now, I know that this means that the set of $xinmathbb R$ for which $f(x)=0$ is a null set with respect to Lebesgue measure. That is, $m({xinmathbb R:f(x)=0})=0$. But since the statement that $f(x)neq0$ holds for almost all $xinmathbb R$, does that mean that $m({xinmathbb R:f(x)neq0})=m(mathbb R)=+infty$? My thinking is that since the statement holds for all $mathbb R$ except possibly on a subset of $mathbb R$ with measure zero, then the measure of the set on which the statement holds must coincide with the measure of $mathbb R$; is this the correct reasoning?
real-analysis functional-analysis measure-theory lebesgue-measure
Consider a real function $f$ and take $mathbb R$ together with Lebesgue measure $m$. I want to check if I have the right reasoning with regards to working with the following "almost everywhere" statement; I have that $f(x)neq0$ for almost all $xinmathbb R$.
Now, I know that this means that the set of $xinmathbb R$ for which $f(x)=0$ is a null set with respect to Lebesgue measure. That is, $m({xinmathbb R:f(x)=0})=0$. But since the statement that $f(x)neq0$ holds for almost all $xinmathbb R$, does that mean that $m({xinmathbb R:f(x)neq0})=m(mathbb R)=+infty$? My thinking is that since the statement holds for all $mathbb R$ except possibly on a subset of $mathbb R$ with measure zero, then the measure of the set on which the statement holds must coincide with the measure of $mathbb R$; is this the correct reasoning?
real-analysis functional-analysis measure-theory lebesgue-measure
real-analysis functional-analysis measure-theory lebesgue-measure
asked 8 hours ago
Jeremy Jeffrey James
912513
912513
Yes, you are right.
– Kavi Rama Murthy
8 hours ago
@mathworker21, I don't think that it is too worrisome (unless you can indicate why it ought to be). I was unsure how to make use of the additivity property of the measure in a rigorous manner, but I see now how designating this set as $E$ makes that much clearer.
– Jeremy Jeffrey James
8 hours ago
@JeremyJeffreyJames my apologies for the comment. I was wrong
– mathworker21
8 hours ago
add a comment |
Yes, you are right.
– Kavi Rama Murthy
8 hours ago
@mathworker21, I don't think that it is too worrisome (unless you can indicate why it ought to be). I was unsure how to make use of the additivity property of the measure in a rigorous manner, but I see now how designating this set as $E$ makes that much clearer.
– Jeremy Jeffrey James
8 hours ago
@JeremyJeffreyJames my apologies for the comment. I was wrong
– mathworker21
8 hours ago
Yes, you are right.
– Kavi Rama Murthy
8 hours ago
Yes, you are right.
– Kavi Rama Murthy
8 hours ago
@mathworker21, I don't think that it is too worrisome (unless you can indicate why it ought to be). I was unsure how to make use of the additivity property of the measure in a rigorous manner, but I see now how designating this set as $E$ makes that much clearer.
– Jeremy Jeffrey James
8 hours ago
@mathworker21, I don't think that it is too worrisome (unless you can indicate why it ought to be). I was unsure how to make use of the additivity property of the measure in a rigorous manner, but I see now how designating this set as $E$ makes that much clearer.
– Jeremy Jeffrey James
8 hours ago
@JeremyJeffreyJames my apologies for the comment. I was wrong
– mathworker21
8 hours ago
@JeremyJeffreyJames my apologies for the comment. I was wrong
– mathworker21
8 hours ago
add a comment |
1 Answer
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2
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No issues with what you have said : $f(x) neq 0$ almost everywhere, means that the set ${x : f(x) = 0}$ has Lebesgue measure zero, which implies that the set ${x : f(x) neq 0}$ has the same Lebesgue measure as $mathbb R$, which is $+infty$.
However, note that the converse is not true : if I tell you that the set ${x : f(x) neq 0}$ has measure $+infty$, then this does not imply that $f(x) neq 0$ almost everywhere. For example, take $f(x) = 0$ if $[x]$ is a multiple of $2$, and $1$ otherwise. Then ${x : f(x) neq 0}$ and ${x : f(x) = 0}$ have measure $+infty$.
In short, I want to tell you that when you say "$f(x) neq 0$ almost everywhere means ${x : f(x) neq 0}$ has measure infinite", we say it with the idea that the left side of the "means" is like a definition and the right side is an equivalent statement. However, the "means" for you, is an implication : the statements are not equivalent, as I showed. You should therefore be careful about a definition and an implication : that ${f(x) =0}$ has Lebesgue measure zero is the definition of $f(x) neq 0$ almmost everywhere. On the other hand, that ${f(x) neq 0}$ has full measure is an implication of $f(x) neq 0$ everywhere, and not equivalent to it. As long as you are aware of this, you can be relatively be relaxed about what you are writing.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
No issues with what you have said : $f(x) neq 0$ almost everywhere, means that the set ${x : f(x) = 0}$ has Lebesgue measure zero, which implies that the set ${x : f(x) neq 0}$ has the same Lebesgue measure as $mathbb R$, which is $+infty$.
However, note that the converse is not true : if I tell you that the set ${x : f(x) neq 0}$ has measure $+infty$, then this does not imply that $f(x) neq 0$ almost everywhere. For example, take $f(x) = 0$ if $[x]$ is a multiple of $2$, and $1$ otherwise. Then ${x : f(x) neq 0}$ and ${x : f(x) = 0}$ have measure $+infty$.
In short, I want to tell you that when you say "$f(x) neq 0$ almost everywhere means ${x : f(x) neq 0}$ has measure infinite", we say it with the idea that the left side of the "means" is like a definition and the right side is an equivalent statement. However, the "means" for you, is an implication : the statements are not equivalent, as I showed. You should therefore be careful about a definition and an implication : that ${f(x) =0}$ has Lebesgue measure zero is the definition of $f(x) neq 0$ almmost everywhere. On the other hand, that ${f(x) neq 0}$ has full measure is an implication of $f(x) neq 0$ everywhere, and not equivalent to it. As long as you are aware of this, you can be relatively be relaxed about what you are writing.
add a comment |
up vote
2
down vote
accepted
No issues with what you have said : $f(x) neq 0$ almost everywhere, means that the set ${x : f(x) = 0}$ has Lebesgue measure zero, which implies that the set ${x : f(x) neq 0}$ has the same Lebesgue measure as $mathbb R$, which is $+infty$.
However, note that the converse is not true : if I tell you that the set ${x : f(x) neq 0}$ has measure $+infty$, then this does not imply that $f(x) neq 0$ almost everywhere. For example, take $f(x) = 0$ if $[x]$ is a multiple of $2$, and $1$ otherwise. Then ${x : f(x) neq 0}$ and ${x : f(x) = 0}$ have measure $+infty$.
In short, I want to tell you that when you say "$f(x) neq 0$ almost everywhere means ${x : f(x) neq 0}$ has measure infinite", we say it with the idea that the left side of the "means" is like a definition and the right side is an equivalent statement. However, the "means" for you, is an implication : the statements are not equivalent, as I showed. You should therefore be careful about a definition and an implication : that ${f(x) =0}$ has Lebesgue measure zero is the definition of $f(x) neq 0$ almmost everywhere. On the other hand, that ${f(x) neq 0}$ has full measure is an implication of $f(x) neq 0$ everywhere, and not equivalent to it. As long as you are aware of this, you can be relatively be relaxed about what you are writing.
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
No issues with what you have said : $f(x) neq 0$ almost everywhere, means that the set ${x : f(x) = 0}$ has Lebesgue measure zero, which implies that the set ${x : f(x) neq 0}$ has the same Lebesgue measure as $mathbb R$, which is $+infty$.
However, note that the converse is not true : if I tell you that the set ${x : f(x) neq 0}$ has measure $+infty$, then this does not imply that $f(x) neq 0$ almost everywhere. For example, take $f(x) = 0$ if $[x]$ is a multiple of $2$, and $1$ otherwise. Then ${x : f(x) neq 0}$ and ${x : f(x) = 0}$ have measure $+infty$.
In short, I want to tell you that when you say "$f(x) neq 0$ almost everywhere means ${x : f(x) neq 0}$ has measure infinite", we say it with the idea that the left side of the "means" is like a definition and the right side is an equivalent statement. However, the "means" for you, is an implication : the statements are not equivalent, as I showed. You should therefore be careful about a definition and an implication : that ${f(x) =0}$ has Lebesgue measure zero is the definition of $f(x) neq 0$ almmost everywhere. On the other hand, that ${f(x) neq 0}$ has full measure is an implication of $f(x) neq 0$ everywhere, and not equivalent to it. As long as you are aware of this, you can be relatively be relaxed about what you are writing.
No issues with what you have said : $f(x) neq 0$ almost everywhere, means that the set ${x : f(x) = 0}$ has Lebesgue measure zero, which implies that the set ${x : f(x) neq 0}$ has the same Lebesgue measure as $mathbb R$, which is $+infty$.
However, note that the converse is not true : if I tell you that the set ${x : f(x) neq 0}$ has measure $+infty$, then this does not imply that $f(x) neq 0$ almost everywhere. For example, take $f(x) = 0$ if $[x]$ is a multiple of $2$, and $1$ otherwise. Then ${x : f(x) neq 0}$ and ${x : f(x) = 0}$ have measure $+infty$.
In short, I want to tell you that when you say "$f(x) neq 0$ almost everywhere means ${x : f(x) neq 0}$ has measure infinite", we say it with the idea that the left side of the "means" is like a definition and the right side is an equivalent statement. However, the "means" for you, is an implication : the statements are not equivalent, as I showed. You should therefore be careful about a definition and an implication : that ${f(x) =0}$ has Lebesgue measure zero is the definition of $f(x) neq 0$ almmost everywhere. On the other hand, that ${f(x) neq 0}$ has full measure is an implication of $f(x) neq 0$ everywhere, and not equivalent to it. As long as you are aware of this, you can be relatively be relaxed about what you are writing.
answered 8 hours ago
астон вілла олоф мэллбэрг
35.7k33274
35.7k33274
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Yes, you are right.
– Kavi Rama Murthy
8 hours ago
@mathworker21, I don't think that it is too worrisome (unless you can indicate why it ought to be). I was unsure how to make use of the additivity property of the measure in a rigorous manner, but I see now how designating this set as $E$ makes that much clearer.
– Jeremy Jeffrey James
8 hours ago
@JeremyJeffreyJames my apologies for the comment. I was wrong
– mathworker21
8 hours ago