zero times infinity in an integral
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Suppose $f$ and $g$ are measurable functions defined on a ball $B$ of $mathbb{R}^n$ ($ngeq1$), satisfying
$$0leqint_{B}f(x)g(x)dx=f(x_{0})int_{B}g(x)dx,$$
for some $x_{0}in B$. My question is: if $f(x_{0})=0$, can we say that $int_{B}g(x)dx$ is not $infty$?
real-analysis integration
asked Nov 29 at 5:26
M. Rahmat
309211
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0
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Suppose $f$ and $g$ are measurable functions defined on a ball $B$ of $mathbb{R}^n$ ($ngeq1$), satisfying
$$0leqint_{B}f(x)g(x)dx=f(x_{0})int_{B}g(x)dx,$$
for some $x_{0}in B$. My question is: if $f(x_{0})=0$, can we say that $int_{B}g(x)dx$ is not $infty$?
real-analysis integration
asked Nov 29 at 5:26
M. Rahmat
309211
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $f$ and $g$ are measurable functions defined on a ball $B$ of $mathbb{R}^n$ ($ngeq1$), satisfying
$$0leqint_{B}f(x)g(x)dx=f(x_{0})int_{B}g(x)dx,$$
for some $x_{0}in B$. My question is: if $f(x_{0})=0$, can we say that $int_{B}g(x)dx$ is not $infty$?
real-analysis integration
asked Nov 29 at 5:26
M. Rahmat
309211
Suppose $f$ and $g$ are measurable functions defined on a ball $B$ of $mathbb{R}^n$ ($ngeq1$), satisfying
$$0leqint_{B}f(x)g(x)dx=f(x_{0})int_{B}g(x)dx,$$
for some $x_{0}in B$. My question is: if $f(x_{0})=0$, can we say that $int_{B}g(x)dx$ is not $infty$?
real-analysis integration
real-analysis integration
asked Nov 29 at 5:26
M. Rahmat
309211
asked Nov 29 at 5:26
M. Rahmat
309211
asked Nov 29 at 5:26
M. Rahmat
309211
asked Nov 29 at 5:26
M. Rahmat
309211
asked Nov 29 at 5:26
M. Rahmat
309211
309211
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
You cannot say that. Take $fequiv 0$ and a $g$ whose integral is $infty$.
answered Nov 29 at 5:42
Kavi Rama Murthy
48.2k31854
add a comment |
up vote
1
down vote
This still works if $int_Bg = infty$.
The Lebesgue integral is defined for extended real-valued functions and the definition $0 cdot infty = 0$ is imposed. This is necessary to insure consistency.
The integral of a nonnegative function over a measurable E is defined as
$$int_E f = sup left{ int_E g ,,| ,,g,, text{measurable, of compact support, and } , 0 leqslant g leqslant fright} $$
and if $f = 0$ we have $int_E f = 0$ even if $m(E) = infty$
Hence,
$$0 = int_E 0 = 0 cdot m(E) = 0 cdot infty$$
Thus, if $f(x_0) = 0$ and $int_Bg = infty$ then
$$f(x_0) int_Bg = 0$$
answered Nov 29 at 6:13
RRL
48.5k42573
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You cannot say that. Take $fequiv 0$ and a $g$ whose integral is $infty$.
answered Nov 29 at 5:42
Kavi Rama Murthy
48.2k31854
add a comment |
up vote
2
down vote
accepted
You cannot say that. Take $fequiv 0$ and a $g$ whose integral is $infty$.
answered Nov 29 at 5:42
Kavi Rama Murthy
48.2k31854
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You cannot say that. Take $fequiv 0$ and a $g$ whose integral is $infty$.
answered Nov 29 at 5:42
Kavi Rama Murthy
48.2k31854
You cannot say that. Take $fequiv 0$ and a $g$ whose integral is $infty$.
answered Nov 29 at 5:42
Kavi Rama Murthy
48.2k31854
answered Nov 29 at 5:42
Kavi Rama Murthy
48.2k31854
answered Nov 29 at 5:42
Kavi Rama Murthy
48.2k31854
answered Nov 29 at 5:42
Kavi Rama Murthy
48.2k31854
48.2k31854
add a comment |
add a comment |
up vote
1
down vote
This still works if $int_Bg = infty$.
The Lebesgue integral is defined for extended real-valued functions and the definition $0 cdot infty = 0$ is imposed. This is necessary to insure consistency.
The integral of a nonnegative function over a measurable E is defined as
$$int_E f = sup left{ int_E g ,,| ,,g,, text{measurable, of compact support, and } , 0 leqslant g leqslant fright} $$
and if $f = 0$ we have $int_E f = 0$ even if $m(E) = infty$
Hence,
$$0 = int_E 0 = 0 cdot m(E) = 0 cdot infty$$
Thus, if $f(x_0) = 0$ and $int_Bg = infty$ then
$$f(x_0) int_Bg = 0$$
answered Nov 29 at 6:13
RRL
48.5k42573
add a comment |
up vote
1
down vote
This still works if $int_Bg = infty$.
The Lebesgue integral is defined for extended real-valued functions and the definition $0 cdot infty = 0$ is imposed. This is necessary to insure consistency.
The integral of a nonnegative function over a measurable E is defined as
$$int_E f = sup left{ int_E g ,,| ,,g,, text{measurable, of compact support, and } , 0 leqslant g leqslant fright} $$
and if $f = 0$ we have $int_E f = 0$ even if $m(E) = infty$
Hence,
$$0 = int_E 0 = 0 cdot m(E) = 0 cdot infty$$
Thus, if $f(x_0) = 0$ and $int_Bg = infty$ then
$$f(x_0) int_Bg = 0$$
answered Nov 29 at 6:13
RRL
48.5k42573
add a comment |
up vote
1
down vote
up vote
1
down vote
This still works if $int_Bg = infty$.
The Lebesgue integral is defined for extended real-valued functions and the definition $0 cdot infty = 0$ is imposed. This is necessary to insure consistency.
The integral of a nonnegative function over a measurable E is defined as
$$int_E f = sup left{ int_E g ,,| ,,g,, text{measurable, of compact support, and } , 0 leqslant g leqslant fright} $$
and if $f = 0$ we have $int_E f = 0$ even if $m(E) = infty$
Hence,
$$0 = int_E 0 = 0 cdot m(E) = 0 cdot infty$$
Thus, if $f(x_0) = 0$ and $int_Bg = infty$ then
$$f(x_0) int_Bg = 0$$
answered Nov 29 at 6:13
RRL
48.5k42573
This still works if $int_Bg = infty$.
The Lebesgue integral is defined for extended real-valued functions and the definition $0 cdot infty = 0$ is imposed. This is necessary to insure consistency.
The integral of a nonnegative function over a measurable E is defined as
$$int_E f = sup left{ int_E g ,,| ,,g,, text{measurable, of compact support, and } , 0 leqslant g leqslant fright} $$
and if $f = 0$ we have $int_E f = 0$ even if $m(E) = infty$
Hence,
$$0 = int_E 0 = 0 cdot m(E) = 0 cdot infty$$
Thus, if $f(x_0) = 0$ and $int_Bg = infty$ then
$$f(x_0) int_Bg = 0$$
answered Nov 29 at 6:13
RRL
48.5k42573
answered Nov 29 at 6:13
RRL
48.5k42573
answered Nov 29 at 6:13
RRL
48.5k42573
answered Nov 29 at 6:13
RRL
48.5k42573
48.5k42573
add a comment |
add a comment |
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