A counterexample for an integral inequality












1












$begingroup$


I'm looking for functions $f,g in L^1$ such that $f < g$ and $int f = int g$.



I know that $f le g Rightarrow int f le int g$, but I suppose the implication with the strict inequality doesn't hold (am I correct?).






integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    $f<g$ still implies $int fleqint g$. But I do suspect we can't have equality.
    $endgroup$
    – Arthur
    Dec 15 '18 at 6:55


















1












$begingroup$


I'm looking for functions $f,g in L^1$ such that $f < g$ and $int f = int g$.



I know that $f le g Rightarrow int f le int g$, but I suppose the implication with the strict inequality doesn't hold (am I correct?).






integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    $f<g$ still implies $int fleqint g$. But I do suspect we can't have equality.
    $endgroup$
    – Arthur
    Dec 15 '18 at 6:55
















1












1








1


1



$begingroup$


I'm looking for functions $f,g in L^1$ such that $f < g$ and $int f = int g$.



I know that $f le g Rightarrow int f le int g$, but I suppose the implication with the strict inequality doesn't hold (am I correct?).






integration







share|cite|improve this question









$endgroup$




I'm looking for functions $f,g in L^1$ such that $f < g$ and $int f = int g$.



I know that $f le g Rightarrow int f le int g$, but I suppose the implication with the strict inequality doesn't hold (am I correct?).






integration




integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 '18 at 6:53









Markus SteinerMarkus Steiner

1036




1036












  • $begingroup$
    $f<g$ still implies $int fleqint g$. But I do suspect we can't have equality.
    $endgroup$
    – Arthur
    Dec 15 '18 at 6:55




















  • $begingroup$
    $f<g$ still implies $int fleqint g$. But I do suspect we can't have equality.
    $endgroup$
    – Arthur
    Dec 15 '18 at 6:55


















$begingroup$
$f<g$ still implies $int fleqint g$. But I do suspect we can't have equality.
$endgroup$
– Arthur
Dec 15 '18 at 6:55






$begingroup$
$f<g$ still implies $int fleqint g$. But I do suspect we can't have equality.
$endgroup$
– Arthur
Dec 15 '18 at 6:55












1 Answer
1






active

oldest

votes


















3












$begingroup$

We can show that if $fgeq 0$ satisfies $int f= 0$, then $f$ vanishes almost everywhere. Thus, there is no $f,gin L^1$ with $fleq g$ such that
$
f< g
$ on a set of positive measure and $int f=int g$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I was wondering if this is true also for Riemann's integral. I mean if f is bounded (not necessarily continuous) on [a,b] is it true that f > 0 ⇒ ∫f > 0?
    $endgroup$
    – Markus Steiner
    Dec 15 '18 at 8:48










  • $begingroup$
    Yes. You can see it by noting that every Riemann integrable function is Lebesgue integrable.
    $endgroup$
    – Song
    Dec 15 '18 at 9:22











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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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3












$begingroup$

We can show that if $fgeq 0$ satisfies $int f= 0$, then $f$ vanishes almost everywhere. Thus, there is no $f,gin L^1$ with $fleq g$ such that
$
f< g
$ on a set of positive measure and $int f=int g$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I was wondering if this is true also for Riemann's integral. I mean if f is bounded (not necessarily continuous) on [a,b] is it true that f > 0 ⇒ ∫f > 0?
    $endgroup$
    – Markus Steiner
    Dec 15 '18 at 8:48










  • $begingroup$
    Yes. You can see it by noting that every Riemann integrable function is Lebesgue integrable.
    $endgroup$
    – Song
    Dec 15 '18 at 9:22
















3












$begingroup$

We can show that if $fgeq 0$ satisfies $int f= 0$, then $f$ vanishes almost everywhere. Thus, there is no $f,gin L^1$ with $fleq g$ such that
$
f< g
$ on a set of positive measure and $int f=int g$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I was wondering if this is true also for Riemann's integral. I mean if f is bounded (not necessarily continuous) on [a,b] is it true that f > 0 ⇒ ∫f > 0?
    $endgroup$
    – Markus Steiner
    Dec 15 '18 at 8:48










  • $begingroup$
    Yes. You can see it by noting that every Riemann integrable function is Lebesgue integrable.
    $endgroup$
    – Song
    Dec 15 '18 at 9:22














3












3








3





$begingroup$

We can show that if $fgeq 0$ satisfies $int f= 0$, then $f$ vanishes almost everywhere. Thus, there is no $f,gin L^1$ with $fleq g$ such that
$
f< g
$ on a set of positive measure and $int f=int g$.






share|cite|improve this answer









$endgroup$



We can show that if $fgeq 0$ satisfies $int f= 0$, then $f$ vanishes almost everywhere. Thus, there is no $f,gin L^1$ with $fleq g$ such that
$
f< g
$ on a set of positive measure and $int f=int g$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 15 '18 at 7:02









SongSong

12.9k631




12.9k631












  • $begingroup$
    I was wondering if this is true also for Riemann's integral. I mean if f is bounded (not necessarily continuous) on [a,b] is it true that f > 0 ⇒ ∫f > 0?
    $endgroup$
    – Markus Steiner
    Dec 15 '18 at 8:48










  • $begingroup$
    Yes. You can see it by noting that every Riemann integrable function is Lebesgue integrable.
    $endgroup$
    – Song
    Dec 15 '18 at 9:22


















  • $begingroup$
    I was wondering if this is true also for Riemann's integral. I mean if f is bounded (not necessarily continuous) on [a,b] is it true that f > 0 ⇒ ∫f > 0?
    $endgroup$
    – Markus Steiner
    Dec 15 '18 at 8:48










  • $begingroup$
    Yes. You can see it by noting that every Riemann integrable function is Lebesgue integrable.
    $endgroup$
    – Song
    Dec 15 '18 at 9:22
















$begingroup$
I was wondering if this is true also for Riemann's integral. I mean if f is bounded (not necessarily continuous) on [a,b] is it true that f > 0 ⇒ ∫f > 0?
$endgroup$
– Markus Steiner
Dec 15 '18 at 8:48




$begingroup$
I was wondering if this is true also for Riemann's integral. I mean if f is bounded (not necessarily continuous) on [a,b] is it true that f > 0 ⇒ ∫f > 0?
$endgroup$
– Markus Steiner
Dec 15 '18 at 8:48












$begingroup$
Yes. You can see it by noting that every Riemann integrable function is Lebesgue integrable.
$endgroup$
– Song
Dec 15 '18 at 9:22




$begingroup$
Yes. You can see it by noting that every Riemann integrable function is Lebesgue integrable.
$endgroup$
– Song
Dec 15 '18 at 9:22


















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