Show that $int_{0}^{1}sup{f_n} dmu=infty$ under this certain condition.
Let $([0,1],mu)$ be a measure space.
Let $f_n :[0,1] to [0,infty)$ be integrable functions.
If $int_{0}^{1} f_n dmu = 1$ and $int_{0}^{1/n} f_n dmu > 1-1/n$ then show
$int_{0}^{1}sup{f_n} dmu=infty$.
This seems to be true because $sup f_n$ has a arbitrarily large value near $0$.
But I have no idea how to handle such value has a enough width so that $int_{0}^{1}sup{f_n} dmu=infty$.
lebesgue-integral
asked Dec 11 '18 at 14:42
Maddy
958421
add a comment |
Let $([0,1],mu)$ be a measure space.
Let $f_n :[0,1] to [0,infty)$ be integrable functions.
If $int_{0}^{1} f_n dmu = 1$ and $int_{0}^{1/n} f_n dmu > 1-1/n$ then show
$int_{0}^{1}sup{f_n} dmu=infty$.
This seems to be true because $sup f_n$ has a arbitrarily large value near $0$.
But I have no idea how to handle such value has a enough width so that $int_{0}^{1}sup{f_n} dmu=infty$.
lebesgue-integral
asked Dec 11 '18 at 14:42
Maddy
958421
add a comment |
Let $([0,1],mu)$ be a measure space.
Let $f_n :[0,1] to [0,infty)$ be integrable functions.
If $int_{0}^{1} f_n dmu = 1$ and $int_{0}^{1/n} f_n dmu > 1-1/n$ then show
$int_{0}^{1}sup{f_n} dmu=infty$.
This seems to be true because $sup f_n$ has a arbitrarily large value near $0$.
But I have no idea how to handle such value has a enough width so that $int_{0}^{1}sup{f_n} dmu=infty$.
lebesgue-integral
asked Dec 11 '18 at 14:42
Maddy
958421
Let $([0,1],mu)$ be a measure space.
Let $f_n :[0,1] to [0,infty)$ be integrable functions.
If $int_{0}^{1} f_n dmu = 1$ and $int_{0}^{1/n} f_n dmu > 1-1/n$ then show
$int_{0}^{1}sup{f_n} dmu=infty$.
This seems to be true because $sup f_n$ has a arbitrarily large value near $0$.
But I have no idea how to handle such value has a enough width so that $int_{0}^{1}sup{f_n} dmu=infty$.
lebesgue-integral
lebesgue-integral
asked Dec 11 '18 at 14:42
Maddy
958421
asked Dec 11 '18 at 14:42
Maddy
958421
asked Dec 11 '18 at 14:42
Maddy
958421
asked Dec 11 '18 at 14:42
Maddy
958421
asked Dec 11 '18 at 14:42
Maddy
958421
958421
add a comment |
add a comment |
1 Answer
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Is this a valid solution?
Note $g=sup f_n$ dominates $f_n$.
We have $int_{1/n}^{1}f_n dmu < 1/n$.
Fix $N$ so on $[1/N,1]$ we have $int_{1/N}^{1}f_n dmu < 1/n$. Let $nto infty$.
Then $f_n to 0$ on $[1/N,1]$. Claiming this for all $N in mathbb{N}$ gets $f_n to 0$ a.e.
Now suppose $int g dmu <infty$. Then applying DCT on $f_n$ gets contradiction.
Because $int f_n =1$ for all $n$ while $int lim f_n =0$.
answered Dec 11 '18 at 14:49
Maddy
958421
add a comment |
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1 Answer
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1 Answer
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Is this a valid solution?
Note $g=sup f_n$ dominates $f_n$.
We have $int_{1/n}^{1}f_n dmu < 1/n$.
Fix $N$ so on $[1/N,1]$ we have $int_{1/N}^{1}f_n dmu < 1/n$. Let $nto infty$.
Then $f_n to 0$ on $[1/N,1]$. Claiming this for all $N in mathbb{N}$ gets $f_n to 0$ a.e.
Now suppose $int g dmu <infty$. Then applying DCT on $f_n$ gets contradiction.
Because $int f_n =1$ for all $n$ while $int lim f_n =0$.
answered Dec 11 '18 at 14:49
Maddy
958421
add a comment |
Is this a valid solution?
Note $g=sup f_n$ dominates $f_n$.
We have $int_{1/n}^{1}f_n dmu < 1/n$.
Fix $N$ so on $[1/N,1]$ we have $int_{1/N}^{1}f_n dmu < 1/n$. Let $nto infty$.
Then $f_n to 0$ on $[1/N,1]$. Claiming this for all $N in mathbb{N}$ gets $f_n to 0$ a.e.
Now suppose $int g dmu <infty$. Then applying DCT on $f_n$ gets contradiction.
Because $int f_n =1$ for all $n$ while $int lim f_n =0$.
answered Dec 11 '18 at 14:49
Maddy
958421
add a comment |
Is this a valid solution?
Note $g=sup f_n$ dominates $f_n$.
We have $int_{1/n}^{1}f_n dmu < 1/n$.
Fix $N$ so on $[1/N,1]$ we have $int_{1/N}^{1}f_n dmu < 1/n$. Let $nto infty$.
Then $f_n to 0$ on $[1/N,1]$. Claiming this for all $N in mathbb{N}$ gets $f_n to 0$ a.e.
Now suppose $int g dmu <infty$. Then applying DCT on $f_n$ gets contradiction.
Because $int f_n =1$ for all $n$ while $int lim f_n =0$.
answered Dec 11 '18 at 14:49
Maddy
958421
Is this a valid solution?
Note $g=sup f_n$ dominates $f_n$.
We have $int_{1/n}^{1}f_n dmu < 1/n$.
Fix $N$ so on $[1/N,1]$ we have $int_{1/N}^{1}f_n dmu < 1/n$. Let $nto infty$.
Then $f_n to 0$ on $[1/N,1]$. Claiming this for all $N in mathbb{N}$ gets $f_n to 0$ a.e.
Now suppose $int g dmu <infty$. Then applying DCT on $f_n$ gets contradiction.
Because $int f_n =1$ for all $n$ while $int lim f_n =0$.
answered Dec 11 '18 at 14:49
Maddy
958421
answered Dec 11 '18 at 14:49
Maddy
958421
answered Dec 11 '18 at 14:49
Maddy
958421
answered Dec 11 '18 at 14:49
Maddy
958421
958421
add a comment |
add a comment |
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