Mixing
Implications with the Lebesgue integral
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Let $(X,mathcalB,mu)$ a probability space and $f:X to mathbbR$ a non negative and integrable function. My question is, for $a>0$,
if $int_Xfdmugeq a$, then there exist a Borel set, $Bin mathcalB$, with $mu(B)>0$ such that $f(x)geq a$, for all $xin B$ ?
Please, help me! Thanks.
real-analysis probability measure-theory lebesgue-integral lebesgue-measure
asked Jul 21 at 0:00
Alex C.
32
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0
down vote
favorite
Let $(X,mathcalB,mu)$ a probability space and $f:X to mathbbR$ a non negative and integrable function. My question is, for $a>0$,
if $int_Xfdmugeq a$, then there exist a Borel set, $Bin mathcalB$, with $mu(B)>0$ such that $f(x)geq a$, for all $xin B$ ?
Please, help me! Thanks.
real-analysis probability measure-theory lebesgue-integral lebesgue-measure
asked Jul 21 at 0:00
Alex C.
32
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(X,mathcalB,mu)$ a probability space and $f:X to mathbbR$ a non negative and integrable function. My question is, for $a>0$,
if $int_Xfdmugeq a$, then there exist a Borel set, $Bin mathcalB$, with $mu(B)>0$ such that $f(x)geq a$, for all $xin B$ ?
Please, help me! Thanks.
real-analysis probability measure-theory lebesgue-integral lebesgue-measure
asked Jul 21 at 0:00
Alex C.
32
Let $(X,mathcalB,mu)$ a probability space and $f:X to mathbbR$ a non negative and integrable function. My question is, for $a>0$,
if $int_Xfdmugeq a$, then there exist a Borel set, $Bin mathcalB$, with $mu(B)>0$ such that $f(x)geq a$, for all $xin B$ ?
Please, help me! Thanks.
real-analysis probability measure-theory lebesgue-integral lebesgue-measure
asked Jul 21 at 0:00
Alex C.
32
edited Jul 21 at 0:06
asked Jul 21 at 0:00
Alex C.
32
asked Jul 21 at 0:00
Alex C.
32
asked Jul 21 at 0:00
Alex C.
32
32
add a comment |Â
add a comment |Â
1 Answer
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Yes. Take $B=x:f(x) geq a$. If $mu (B)=0$ the $f<a $ almost everywhere which implies $int f , dmu <a$.
answered Jul 21 at 0:15
Kavi Rama Murthy
20.6k2830
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes. Take $B=x:f(x) geq a$. If $mu (B)=0$ the $f<a $ almost everywhere which implies $int f , dmu <a$.
answered Jul 21 at 0:15
Kavi Rama Murthy
20.6k2830
add a comment |Â
up vote
1
down vote
accepted
Yes. Take $B=x:f(x) geq a$. If $mu (B)=0$ the $f<a $ almost everywhere which implies $int f , dmu <a$.
answered Jul 21 at 0:15
Kavi Rama Murthy
20.6k2830
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes. Take $B=x:f(x) geq a$. If $mu (B)=0$ the $f<a $ almost everywhere which implies $int f , dmu <a$.
answered Jul 21 at 0:15
Kavi Rama Murthy
20.6k2830
Yes. Take $B=x:f(x) geq a$. If $mu (B)=0$ the $f<a $ almost everywhere which implies $int f , dmu <a$.
answered Jul 21 at 0:15
Kavi Rama Murthy
20.6k2830
answered Jul 21 at 0:15
Kavi Rama Murthy
20.6k2830
answered Jul 21 at 0:15
Kavi Rama Murthy
20.6k2830
answered Jul 21 at 0:15
Kavi Rama Murthy
20.6k2830
20.6k2830
add a comment |Â
add a comment |Â
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