A clarification about Fubini's theorem
Suppose $f(x,y)$ is a measurable function defined on an open set $Gsubsetmathbb{R}^{m}times mathbb{R}^{n}$ ($m,ngeq1$), and $Esubsetmathbb{R}^{m}$ and $Fsubsetmathbb{R}^{n}$ are two compact sets such that $Etimes Fsubset G$. Suppose
$$g(y)=int_{E}f(x,y)dxgeq0$$ and is integrable on $F$, i.e., $$int_{F}|g(y)|dy=int_{F}g(y)dy<infty.$$ Can we say that $f$ is integrable on $Etimes F$, i.e.,
$$int_{F}int_{E}|f(x,y)|dxdy<infty?$$
real-analysis measure-theory
asked Dec 26 at 0:02
M. Rahmat
327212
add a comment |
Suppose $f(x,y)$ is a measurable function defined on an open set $Gsubsetmathbb{R}^{m}times mathbb{R}^{n}$ ($m,ngeq1$), and $Esubsetmathbb{R}^{m}$ and $Fsubsetmathbb{R}^{n}$ are two compact sets such that $Etimes Fsubset G$. Suppose
$$g(y)=int_{E}f(x,y)dxgeq0$$ and is integrable on $F$, i.e., $$int_{F}|g(y)|dy=int_{F}g(y)dy<infty.$$ Can we say that $f$ is integrable on $Etimes F$, i.e.,
$$int_{F}int_{E}|f(x,y)|dxdy<infty?$$
real-analysis measure-theory
asked Dec 26 at 0:02
M. Rahmat
327212
add a comment |
Suppose $f(x,y)$ is a measurable function defined on an open set $Gsubsetmathbb{R}^{m}times mathbb{R}^{n}$ ($m,ngeq1$), and $Esubsetmathbb{R}^{m}$ and $Fsubsetmathbb{R}^{n}$ are two compact sets such that $Etimes Fsubset G$. Suppose
$$g(y)=int_{E}f(x,y)dxgeq0$$ and is integrable on $F$, i.e., $$int_{F}|g(y)|dy=int_{F}g(y)dy<infty.$$ Can we say that $f$ is integrable on $Etimes F$, i.e.,
$$int_{F}int_{E}|f(x,y)|dxdy<infty?$$
real-analysis measure-theory
asked Dec 26 at 0:02
M. Rahmat
327212
Suppose $f(x,y)$ is a measurable function defined on an open set $Gsubsetmathbb{R}^{m}times mathbb{R}^{n}$ ($m,ngeq1$), and $Esubsetmathbb{R}^{m}$ and $Fsubsetmathbb{R}^{n}$ are two compact sets such that $Etimes Fsubset G$. Suppose
$$g(y)=int_{E}f(x,y)dxgeq0$$ and is integrable on $F$, i.e., $$int_{F}|g(y)|dy=int_{F}g(y)dy<infty.$$ Can we say that $f$ is integrable on $Etimes F$, i.e.,
$$int_{F}int_{E}|f(x,y)|dxdy<infty?$$
real-analysis measure-theory
real-analysis measure-theory
asked Dec 26 at 0:02
M. Rahmat
327212
asked Dec 26 at 0:02
M. Rahmat
327212
asked Dec 26 at 0:02
M. Rahmat
327212
asked Dec 26 at 0:02
M. Rahmat
327212
asked Dec 26 at 0:02
M. Rahmat
327212
327212
add a comment |
add a comment |
1 Answer
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No, you cannot draw such a conclusion. Just take $f(x,y)=x h(y)$ where $h$ is not integrable. If $E$ is symmetric then $g(y)=0$ for all $y$ but $f$ is not integrable.
answered Dec 26 at 0:06
Kavi Rama Murthy
49.6k31854
add a comment |
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No, you cannot draw such a conclusion. Just take $f(x,y)=x h(y)$ where $h$ is not integrable. If $E$ is symmetric then $g(y)=0$ for all $y$ but $f$ is not integrable.
answered Dec 26 at 0:06
Kavi Rama Murthy
49.6k31854
add a comment |
No, you cannot draw such a conclusion. Just take $f(x,y)=x h(y)$ where $h$ is not integrable. If $E$ is symmetric then $g(y)=0$ for all $y$ but $f$ is not integrable.
answered Dec 26 at 0:06
Kavi Rama Murthy
49.6k31854
add a comment |
No, you cannot draw such a conclusion. Just take $f(x,y)=x h(y)$ where $h$ is not integrable. If $E$ is symmetric then $g(y)=0$ for all $y$ but $f$ is not integrable.
answered Dec 26 at 0:06
Kavi Rama Murthy
49.6k31854
No, you cannot draw such a conclusion. Just take $f(x,y)=x h(y)$ where $h$ is not integrable. If $E$ is symmetric then $g(y)=0$ for all $y$ but $f$ is not integrable.
answered Dec 26 at 0:06
Kavi Rama Murthy
49.6k31854
answered Dec 26 at 0:06
Kavi Rama Murthy
49.6k31854
answered Dec 26 at 0:06
Kavi Rama Murthy
49.6k31854
answered Dec 26 at 0:06
Kavi Rama Murthy
49.6k31854
49.6k31854
add a comment |
add a comment |
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