A counterexample for an integral inequality
$begingroup$
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
asked Dec 15 '18 at 6:53
Markus SteinerMarkus Steiner
785
$endgroup$
add a comment |
$begingroup$
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
asked Dec 15 '18 at 6:53
Markus SteinerMarkus Steiner
785
$endgroup$
$begingroup$
$f<g$ still implies $int fleqint g$. But I do suspect we can't have equality.
$endgroup$
– Arthur
Dec 15 '18 at 6:55
add a comment |
$begingroup$
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
asked Dec 15 '18 at 6:53
Markus SteinerMarkus Steiner
785
$endgroup$
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
asked Dec 15 '18 at 6:53
Markus SteinerMarkus Steiner
785
asked Dec 15 '18 at 6:53
Markus SteinerMarkus Steiner
785
asked Dec 15 '18 at 6:53
Markus SteinerMarkus Steiner
785
asked Dec 15 '18 at 6:53
Markus SteinerMarkus Steiner
785
asked Dec 15 '18 at 6:53
Markus SteinerMarkus Steiner
785
785
$begingroup$
$f<g$ still implies $int fleqint g$. But I do suspect we can't have equality.
$endgroup$
– Arthur
Dec 15 '18 at 6:55
add a comment |
$begingroup$
$f<g$ 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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $int f=int g$.
answered Dec 15 '18 at 7:02
SongSong
8,111624
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $int f=int g$.
answered Dec 15 '18 at 7:02
SongSong
8,111624
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $int f=int g$.
answered Dec 15 '18 at 7:02
SongSong
8,111624
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $int f=int g$.
answered Dec 15 '18 at 7:02
SongSong
8,111624
$endgroup$
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 $int f=int g$.
answered Dec 15 '18 at 7:02
SongSong
8,111624
answered Dec 15 '18 at 7:02
SongSong
8,111624
answered Dec 15 '18 at 7:02
SongSong
8,111624
answered Dec 15 '18 at 7:02
SongSong
8,111624
8,111624
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
$f<g$ still implies $int fleqint g$. But I do suspect we can't have equality.
$endgroup$
– Arthur
Dec 15 '18 at 6:55