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?










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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















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?










share|cite|improve this question






















  • 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













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?










share|cite|improve this question













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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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


















  • 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










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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.






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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.






    share|cite|improve this answer

























      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.






      share|cite|improve this answer























        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












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        answered 8 hours ago









        астон вілла олоф мэллбэрг

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