Mixing
Prove that if $f=g$ for almost all $xin B$ then $int_Bf=int_Bg$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let $BsubseteqmathbbR^n$ be a closed generalized rectangle
and let $f,gcolon BtomathbbR$ be two integrable functions. Prove
that if $B'=left xin Bmid fleft(xright)neq gleft(xright)right $
has measure zero then $int_Bf=int_Bg$.
My attempt:
$f,g$ are integrable then so is $f-g$ and $f-g=0$ for almost all
$xin B$. Moreover there are generalized rectangles $R_1,R_2,ldotssubseteq B$
where $R_i^intcap R_j^int=varnothing$ such that $B'subseteqbigcup R_i$
and $sum_iVleft(R_iright)<varepsilon$ for any $varepsilon$.
Now I thought if I could cover $B'$ by a finite number of rectangles
in $B$ I could then complete them also by a finite number of rectangles
to a partition $Pi$ of $B$. Then by splitting a Riemann
Sum over that partition I think I can finish.
So is there a way to cover $B'$ by a finite number of rectangles?
and/or Is there other way proving this?
EDIT:
Definitions for integrable function:
$fcolon BtomathbbR$ is integrable if for any $varepsilon>0$
there is a partition $Pi=left B_1,ldots,B_Nright $ of
$B$ such that $sum_i=1^NVleft(B_iright)undersetB_itextOSCf<varepsilon$.
OR
$fcolon BtomathbbR$ is integrable if for any $varepsilon>0$
there is $delta>0$ such that for any partition $Pi=left B_1,ldots,B_Nright $
of $B$ with $textdiamPi<delta$: $sum_i=1^NVleft(B_iright)undersetB_itextOSCf<varepsilon$
Definition for the integral of integrable function:
If $fcolon BtomathbbR$ is integrable then $int_Bf=sup_PiunderlineSleft(f,Piright)$
where $underlineS$ is the lower darboux sum
OR
If $fcolon BtomathbbR$ is integrable then $int_Bf=lim_textdiamPito0Sleft(f,Pi,Xright)$
where $Sleft(f,Pi,Xright)$ is a riemann sum
integration measure-theory
asked Aug 2 at 15:42
Jon
480412
add a comment |Â
up vote
1
down vote
favorite
Let $BsubseteqmathbbR^n$ be a closed generalized rectangle
and let $f,gcolon BtomathbbR$ be two integrable functions. Prove
that if $B'=left xin Bmid fleft(xright)neq gleft(xright)right $
has measure zero then $int_Bf=int_Bg$.
My attempt:
$f,g$ are integrable then so is $f-g$ and $f-g=0$ for almost all
$xin B$. Moreover there are generalized rectangles $R_1,R_2,ldotssubseteq B$
where $R_i^intcap R_j^int=varnothing$ such that $B'subseteqbigcup R_i$
and $sum_iVleft(R_iright)<varepsilon$ for any $varepsilon$.
Now I thought if I could cover $B'$ by a finite number of rectangles
in $B$ I could then complete them also by a finite number of rectangles
to a partition $Pi$ of $B$. Then by splitting a Riemann
Sum over that partition I think I can finish.
So is there a way to cover $B'$ by a finite number of rectangles?
and/or Is there other way proving this?
EDIT:
Definitions for integrable function:
$fcolon BtomathbbR$ is integrable if for any $varepsilon>0$
there is a partition $Pi=left B_1,ldots,B_Nright $ of
$B$ such that $sum_i=1^NVleft(B_iright)undersetB_itextOSCf<varepsilon$.
OR
$fcolon BtomathbbR$ is integrable if for any $varepsilon>0$
there is $delta>0$ such that for any partition $Pi=left B_1,ldots,B_Nright $
of $B$ with $textdiamPi<delta$: $sum_i=1^NVleft(B_iright)undersetB_itextOSCf<varepsilon$
Definition for the integral of integrable function:
If $fcolon BtomathbbR$ is integrable then $int_Bf=sup_PiunderlineSleft(f,Piright)$
where $underlineS$ is the lower darboux sum
OR
If $fcolon BtomathbbR$ is integrable then $int_Bf=lim_textdiamPito0Sleft(f,Pi,Xright)$
where $Sleft(f,Pi,Xright)$ is a riemann sum
integration measure-theory
asked Aug 2 at 15:42
Jon
480412
What definition of integration are you using? Riemann integration?
– Eric Wofsey
Aug 2 at 15:45
@EricWofsey Edited in the question
– Jon
Aug 2 at 16:00
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $BsubseteqmathbbR^n$ be a closed generalized rectangle
and let $f,gcolon BtomathbbR$ be two integrable functions. Prove
that if $B'=left xin Bmid fleft(xright)neq gleft(xright)right $
has measure zero then $int_Bf=int_Bg$.
My attempt:
$f,g$ are integrable then so is $f-g$ and $f-g=0$ for almost all
$xin B$. Moreover there are generalized rectangles $R_1,R_2,ldotssubseteq B$
where $R_i^intcap R_j^int=varnothing$ such that $B'subseteqbigcup R_i$
and $sum_iVleft(R_iright)<varepsilon$ for any $varepsilon$.
Now I thought if I could cover $B'$ by a finite number of rectangles
in $B$ I could then complete them also by a finite number of rectangles
to a partition $Pi$ of $B$. Then by splitting a Riemann
Sum over that partition I think I can finish.
So is there a way to cover $B'$ by a finite number of rectangles?
and/or Is there other way proving this?
EDIT:
Definitions for integrable function:
$fcolon BtomathbbR$ is integrable if for any $varepsilon>0$
there is a partition $Pi=left B_1,ldots,B_Nright $ of
$B$ such that $sum_i=1^NVleft(B_iright)undersetB_itextOSCf<varepsilon$.
OR
$fcolon BtomathbbR$ is integrable if for any $varepsilon>0$
there is $delta>0$ such that for any partition $Pi=left B_1,ldots,B_Nright $
of $B$ with $textdiamPi<delta$: $sum_i=1^NVleft(B_iright)undersetB_itextOSCf<varepsilon$
Definition for the integral of integrable function:
If $fcolon BtomathbbR$ is integrable then $int_Bf=sup_PiunderlineSleft(f,Piright)$
where $underlineS$ is the lower darboux sum
OR
If $fcolon BtomathbbR$ is integrable then $int_Bf=lim_textdiamPito0Sleft(f,Pi,Xright)$
where $Sleft(f,Pi,Xright)$ is a riemann sum
integration measure-theory
asked Aug 2 at 15:42
Jon
480412
Let $BsubseteqmathbbR^n$ be a closed generalized rectangle
and let $f,gcolon BtomathbbR$ be two integrable functions. Prove
that if $B'=left xin Bmid fleft(xright)neq gleft(xright)right $
has measure zero then $int_Bf=int_Bg$.
My attempt:
$f,g$ are integrable then so is $f-g$ and $f-g=0$ for almost all
$xin B$. Moreover there are generalized rectangles $R_1,R_2,ldotssubseteq B$
where $R_i^intcap R_j^int=varnothing$ such that $B'subseteqbigcup R_i$
and $sum_iVleft(R_iright)<varepsilon$ for any $varepsilon$.
Now I thought if I could cover $B'$ by a finite number of rectangles
in $B$ I could then complete them also by a finite number of rectangles
to a partition $Pi$ of $B$. Then by splitting a Riemann
Sum over that partition I think I can finish.
So is there a way to cover $B'$ by a finite number of rectangles?
and/or Is there other way proving this?
EDIT:
Definitions for integrable function:
$fcolon BtomathbbR$ is integrable if for any $varepsilon>0$
there is a partition $Pi=left B_1,ldots,B_Nright $ of
$B$ such that $sum_i=1^NVleft(B_iright)undersetB_itextOSCf<varepsilon$.
OR
$fcolon BtomathbbR$ is integrable if for any $varepsilon>0$
there is $delta>0$ such that for any partition $Pi=left B_1,ldots,B_Nright $
of $B$ with $textdiamPi<delta$: $sum_i=1^NVleft(B_iright)undersetB_itextOSCf<varepsilon$
Definition for the integral of integrable function:
If $fcolon BtomathbbR$ is integrable then $int_Bf=sup_PiunderlineSleft(f,Piright)$
where $underlineS$ is the lower darboux sum
OR
If $fcolon BtomathbbR$ is integrable then $int_Bf=lim_textdiamPito0Sleft(f,Pi,Xright)$
where $Sleft(f,Pi,Xright)$ is a riemann sum
integration measure-theory
asked Aug 2 at 15:42
Jon
480412
edited Aug 2 at 15:59
asked Aug 2 at 15:42
Jon
480412
asked Aug 2 at 15:42
Jon
480412
asked Aug 2 at 15:42
Jon
480412
480412
What definition of integration are you using? Riemann integration?
– Eric Wofsey
Aug 2 at 15:45
@EricWofsey Edited in the question
– Jon
Aug 2 at 16:00
add a comment |Â
What definition of integration are you using? Riemann integration?
– Eric Wofsey
Aug 2 at 15:45
@EricWofsey Edited in the question
– Jon
Aug 2 at 16:00
What definition of integration are you using? Riemann integration?
– Eric Wofsey
Aug 2 at 15:45
What definition of integration are you using? Riemann integration?
– Eric Wofsey
Aug 2 at 15:45
@EricWofsey Edited in the question
– Jon
Aug 2 at 16:00
@EricWofsey Edited in the question
– Jon
Aug 2 at 16:00
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
For the Lebesgue integral this follows almost directly from the definition. For the Riemann integral over a rectangle in $mathbbR^n$, the proof is more subtle in order to avoid measure theory -- other than the basic notion of zero measure.
Note that $h = f-g$ is Riemann integrable on $B$ (according to your definitions) and the set
$B' = x in B: h(x) = f(x)-g(x) neq 0$ has measure zero.
Take any partition $P$ of the rectangle $B$. Any subrectangle $R$ of $P$ has non-zero content and measure by definition of a partition. Hence $R$ is not a subset of $B'$ and must contain at least one point where $h(x) = 0$. This implies that $inf_R h(x) leqslant 0 leqslant sup_R h(x)$.
Forming upper and lower Riemann sums, we have for any partition $P$
$$underlineS(h ,P) leqslant 0 leqslant overlineS(h ,P)$$
Hence,
$$underlineint_B h = sup_P underlineS(h,P) leqslant 0 leqslant inf_P overlineS(h, P) = overlineint_B h $$
Since $h$ is Riemann integrable on $B$, the upper and lower integrals must be equal and
$$int_Bf - int_Bg =int_Bh = underlineint_B h = overlineint_B h = 0$$
answered Aug 3 at 5:20
RRL
43.4k42260
add a comment |Â
up vote
0
down vote
$f-g=0$, except on a set of measure $0$. Therefore $int_B(f-g)=int_B0=0$, or $int_Bf=int_Bg$. Why isn't it obvious?
If you need to be pedantic let $B=B_1cup B_2$, where $B_1$ is the set where $f=g$ and $B_2$ is the set of measure $0$ where $fne g$. Then $int_B(f-g)=int_B_1(f-g)+int_B_2(f-g)$. The first term=$0$, since the integrand is $0$, while the second term=$0$, since the domain of integration has measure $0$.
answered Aug 2 at 19:05
herb steinberg
93529
Can you sketch a proof of the fact that the Riemann integral on a set of measure zero (not content zero) is equal to zero ?
– Tony Piccolo
Aug 2 at 23:20
I was using measure theory (Lebesgue integral). It is a standard theorem.
– herb steinberg
Aug 3 at 0:05
My aim was to know if there exists a proof that doesn't use Lebesgue theory.
– Tony Piccolo
Aug 3 at 6:06
Once you use the term "measure zero", what else could you use?
– herb steinberg
Aug 3 at 15:16
See the accepted answer ..... I mean that.
– Tony Piccolo
Aug 3 at 19:02
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
For the Lebesgue integral this follows almost directly from the definition. For the Riemann integral over a rectangle in $mathbbR^n$, the proof is more subtle in order to avoid measure theory -- other than the basic notion of zero measure.
Note that $h = f-g$ is Riemann integrable on $B$ (according to your definitions) and the set
$B' = x in B: h(x) = f(x)-g(x) neq 0$ has measure zero.
Take any partition $P$ of the rectangle $B$. Any subrectangle $R$ of $P$ has non-zero content and measure by definition of a partition. Hence $R$ is not a subset of $B'$ and must contain at least one point where $h(x) = 0$. This implies that $inf_R h(x) leqslant 0 leqslant sup_R h(x)$.
Forming upper and lower Riemann sums, we have for any partition $P$
$$underlineS(h ,P) leqslant 0 leqslant overlineS(h ,P)$$
Hence,
$$underlineint_B h = sup_P underlineS(h,P) leqslant 0 leqslant inf_P overlineS(h, P) = overlineint_B h $$
Since $h$ is Riemann integrable on $B$, the upper and lower integrals must be equal and
$$int_Bf - int_Bg =int_Bh = underlineint_B h = overlineint_B h = 0$$
answered Aug 3 at 5:20
RRL
43.4k42260
add a comment |Â
up vote
2
down vote
accepted
For the Lebesgue integral this follows almost directly from the definition. For the Riemann integral over a rectangle in $mathbbR^n$, the proof is more subtle in order to avoid measure theory -- other than the basic notion of zero measure.
Note that $h = f-g$ is Riemann integrable on $B$ (according to your definitions) and the set
$B' = x in B: h(x) = f(x)-g(x) neq 0$ has measure zero.
Take any partition $P$ of the rectangle $B$. Any subrectangle $R$ of $P$ has non-zero content and measure by definition of a partition. Hence $R$ is not a subset of $B'$ and must contain at least one point where $h(x) = 0$. This implies that $inf_R h(x) leqslant 0 leqslant sup_R h(x)$.
Forming upper and lower Riemann sums, we have for any partition $P$
$$underlineS(h ,P) leqslant 0 leqslant overlineS(h ,P)$$
Hence,
$$underlineint_B h = sup_P underlineS(h,P) leqslant 0 leqslant inf_P overlineS(h, P) = overlineint_B h $$
Since $h$ is Riemann integrable on $B$, the upper and lower integrals must be equal and
$$int_Bf - int_Bg =int_Bh = underlineint_B h = overlineint_B h = 0$$
answered Aug 3 at 5:20
RRL
43.4k42260
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
For the Lebesgue integral this follows almost directly from the definition. For the Riemann integral over a rectangle in $mathbbR^n$, the proof is more subtle in order to avoid measure theory -- other than the basic notion of zero measure.
Note that $h = f-g$ is Riemann integrable on $B$ (according to your definitions) and the set
$B' = x in B: h(x) = f(x)-g(x) neq 0$ has measure zero.
Take any partition $P$ of the rectangle $B$. Any subrectangle $R$ of $P$ has non-zero content and measure by definition of a partition. Hence $R$ is not a subset of $B'$ and must contain at least one point where $h(x) = 0$. This implies that $inf_R h(x) leqslant 0 leqslant sup_R h(x)$.
Forming upper and lower Riemann sums, we have for any partition $P$
$$underlineS(h ,P) leqslant 0 leqslant overlineS(h ,P)$$
Hence,
$$underlineint_B h = sup_P underlineS(h,P) leqslant 0 leqslant inf_P overlineS(h, P) = overlineint_B h $$
Since $h$ is Riemann integrable on $B$, the upper and lower integrals must be equal and
$$int_Bf - int_Bg =int_Bh = underlineint_B h = overlineint_B h = 0$$
answered Aug 3 at 5:20
RRL
43.4k42260
For the Lebesgue integral this follows almost directly from the definition. For the Riemann integral over a rectangle in $mathbbR^n$, the proof is more subtle in order to avoid measure theory -- other than the basic notion of zero measure.
Note that $h = f-g$ is Riemann integrable on $B$ (according to your definitions) and the set
$B' = x in B: h(x) = f(x)-g(x) neq 0$ has measure zero.
Take any partition $P$ of the rectangle $B$. Any subrectangle $R$ of $P$ has non-zero content and measure by definition of a partition. Hence $R$ is not a subset of $B'$ and must contain at least one point where $h(x) = 0$. This implies that $inf_R h(x) leqslant 0 leqslant sup_R h(x)$.
Forming upper and lower Riemann sums, we have for any partition $P$
$$underlineS(h ,P) leqslant 0 leqslant overlineS(h ,P)$$
Hence,
$$underlineint_B h = sup_P underlineS(h,P) leqslant 0 leqslant inf_P overlineS(h, P) = overlineint_B h $$
Since $h$ is Riemann integrable on $B$, the upper and lower integrals must be equal and
$$int_Bf - int_Bg =int_Bh = underlineint_B h = overlineint_B h = 0$$
answered Aug 3 at 5:20
RRL
43.4k42260
edited Aug 3 at 6:11
answered Aug 3 at 5:20
RRL
43.4k42260
answered Aug 3 at 5:20
RRL
43.4k42260
answered Aug 3 at 5:20
RRL
43.4k42260
43.4k42260
add a comment |Â
add a comment |Â
up vote
0
down vote
$f-g=0$, except on a set of measure $0$. Therefore $int_B(f-g)=int_B0=0$, or $int_Bf=int_Bg$. Why isn't it obvious?
If you need to be pedantic let $B=B_1cup B_2$, where $B_1$ is the set where $f=g$ and $B_2$ is the set of measure $0$ where $fne g$. Then $int_B(f-g)=int_B_1(f-g)+int_B_2(f-g)$. The first term=$0$, since the integrand is $0$, while the second term=$0$, since the domain of integration has measure $0$.
answered Aug 2 at 19:05
herb steinberg
93529
Can you sketch a proof of the fact that the Riemann integral on a set of measure zero (not content zero) is equal to zero ?
– Tony Piccolo
Aug 2 at 23:20
I was using measure theory (Lebesgue integral). It is a standard theorem.
– herb steinberg
Aug 3 at 0:05
My aim was to know if there exists a proof that doesn't use Lebesgue theory.
– Tony Piccolo
Aug 3 at 6:06
Once you use the term "measure zero", what else could you use?
– herb steinberg
Aug 3 at 15:16
See the accepted answer ..... I mean that.
– Tony Piccolo
Aug 3 at 19:02
add a comment |Â
up vote
0
down vote
$f-g=0$, except on a set of measure $0$. Therefore $int_B(f-g)=int_B0=0$, or $int_Bf=int_Bg$. Why isn't it obvious?
If you need to be pedantic let $B=B_1cup B_2$, where $B_1$ is the set where $f=g$ and $B_2$ is the set of measure $0$ where $fne g$. Then $int_B(f-g)=int_B_1(f-g)+int_B_2(f-g)$. The first term=$0$, since the integrand is $0$, while the second term=$0$, since the domain of integration has measure $0$.
answered Aug 2 at 19:05
herb steinberg
93529
Can you sketch a proof of the fact that the Riemann integral on a set of measure zero (not content zero) is equal to zero ?
– Tony Piccolo
Aug 2 at 23:20
I was using measure theory (Lebesgue integral). It is a standard theorem.
– herb steinberg
Aug 3 at 0:05
My aim was to know if there exists a proof that doesn't use Lebesgue theory.
– Tony Piccolo
Aug 3 at 6:06
Once you use the term "measure zero", what else could you use?
– herb steinberg
Aug 3 at 15:16
See the accepted answer ..... I mean that.
– Tony Piccolo
Aug 3 at 19:02
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$f-g=0$, except on a set of measure $0$. Therefore $int_B(f-g)=int_B0=0$, or $int_Bf=int_Bg$. Why isn't it obvious?
If you need to be pedantic let $B=B_1cup B_2$, where $B_1$ is the set where $f=g$ and $B_2$ is the set of measure $0$ where $fne g$. Then $int_B(f-g)=int_B_1(f-g)+int_B_2(f-g)$. The first term=$0$, since the integrand is $0$, while the second term=$0$, since the domain of integration has measure $0$.
answered Aug 2 at 19:05
herb steinberg
93529
$f-g=0$, except on a set of measure $0$. Therefore $int_B(f-g)=int_B0=0$, or $int_Bf=int_Bg$. Why isn't it obvious?
If you need to be pedantic let $B=B_1cup B_2$, where $B_1$ is the set where $f=g$ and $B_2$ is the set of measure $0$ where $fne g$. Then $int_B(f-g)=int_B_1(f-g)+int_B_2(f-g)$. The first term=$0$, since the integrand is $0$, while the second term=$0$, since the domain of integration has measure $0$.
answered Aug 2 at 19:05
herb steinberg
93529
edited Aug 2 at 19:19
answered Aug 2 at 19:05
herb steinberg
93529
answered Aug 2 at 19:05
herb steinberg
93529
answered Aug 2 at 19:05
herb steinberg
93529
93529
Can you sketch a proof of the fact that the Riemann integral on a set of measure zero (not content zero) is equal to zero ?
– Tony Piccolo
Aug 2 at 23:20
I was using measure theory (Lebesgue integral). It is a standard theorem.
– herb steinberg
Aug 3 at 0:05
My aim was to know if there exists a proof that doesn't use Lebesgue theory.
– Tony Piccolo
Aug 3 at 6:06
Once you use the term "measure zero", what else could you use?
– herb steinberg
Aug 3 at 15:16
See the accepted answer ..... I mean that.
– Tony Piccolo
Aug 3 at 19:02
add a comment |Â
Can you sketch a proof of the fact that the Riemann integral on a set of measure zero (not content zero) is equal to zero ?
– Tony Piccolo
Aug 2 at 23:20
I was using measure theory (Lebesgue integral). It is a standard theorem.
– herb steinberg
Aug 3 at 0:05
My aim was to know if there exists a proof that doesn't use Lebesgue theory.
– Tony Piccolo
Aug 3 at 6:06
Once you use the term "measure zero", what else could you use?
– herb steinberg
Aug 3 at 15:16
See the accepted answer ..... I mean that.
– Tony Piccolo
Aug 3 at 19:02
Can you sketch a proof of the fact that the Riemann integral on a set of measure zero (not content zero) is equal to zero ?
– Tony Piccolo
Aug 2 at 23:20
Can you sketch a proof of the fact that the Riemann integral on a set of measure zero (not content zero) is equal to zero ?
– Tony Piccolo
Aug 2 at 23:20
I was using measure theory (Lebesgue integral). It is a standard theorem.
– herb steinberg
Aug 3 at 0:05
I was using measure theory (Lebesgue integral). It is a standard theorem.
– herb steinberg
Aug 3 at 0:05
My aim was to know if there exists a proof that doesn't use Lebesgue theory.
– Tony Piccolo
Aug 3 at 6:06
My aim was to know if there exists a proof that doesn't use Lebesgue theory.
– Tony Piccolo
Aug 3 at 6:06
Once you use the term "measure zero", what else could you use?
– herb steinberg
Aug 3 at 15:16
Once you use the term "measure zero", what else could you use?
– herb steinberg
Aug 3 at 15:16
See the accepted answer ..... I mean that.
– Tony Piccolo
Aug 3 at 19:02
See the accepted answer ..... I mean that.
– Tony Piccolo
Aug 3 at 19:02
add a comment |Â
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What definition of integration are you using? Riemann integration?
– Eric Wofsey
Aug 2 at 15:45
@EricWofsey Edited in the question
– Jon
Aug 2 at 16:00