Absolue continuity of the integral
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Let $f in L^1(X,mu)$ Show that for every $epsilon gt 0$ it exists $sigma gt 0$ such that $int_{A}f lt epsilon$ if $mu(A) lt sigma$
real-analysis measure-theory
asked Dec 6 '18 at 14:38
Jonathan BaramJonathan Baram
140113
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add a comment |
$begingroup$
Let $f in L^1(X,mu)$ Show that for every $epsilon gt 0$ it exists $sigma gt 0$ such that $int_{A}f lt epsilon$ if $mu(A) lt sigma$
real-analysis measure-theory
asked Dec 6 '18 at 14:38
Jonathan BaramJonathan Baram
140113
$endgroup$
$begingroup$
You want $|f|$ in the integral
$endgroup$
– zhw.
Dec 6 '18 at 20:15
add a comment |
$begingroup$
Let $f in L^1(X,mu)$ Show that for every $epsilon gt 0$ it exists $sigma gt 0$ such that $int_{A}f lt epsilon$ if $mu(A) lt sigma$
real-analysis measure-theory
asked Dec 6 '18 at 14:38
Jonathan BaramJonathan Baram
140113
$endgroup$
Let $f in L^1(X,mu)$ Show that for every $epsilon gt 0$ it exists $sigma gt 0$ such that $int_{A}f lt epsilon$ if $mu(A) lt sigma$
real-analysis measure-theory
real-analysis measure-theory
asked Dec 6 '18 at 14:38
Jonathan BaramJonathan Baram
140113
asked Dec 6 '18 at 14:38
Jonathan BaramJonathan Baram
140113
asked Dec 6 '18 at 14:38
Jonathan BaramJonathan Baram
140113
asked Dec 6 '18 at 14:38
Jonathan BaramJonathan Baram
140113
asked Dec 6 '18 at 14:38
Jonathan BaramJonathan Baram
140113
140113
$begingroup$
You want $|f|$ in the integral
$endgroup$
– zhw.
Dec 6 '18 at 20:15
add a comment |
$begingroup$
You want $|f|$ in the integral
$endgroup$
– zhw.
Dec 6 '18 at 20:15
$begingroup$
You want $|f|$ in the integral
$endgroup$
– zhw.
Dec 6 '18 at 20:15
$begingroup$
You want $|f|$ in the integral
$endgroup$
– zhw.
Dec 6 '18 at 20:15
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If the thesis were not true, it would exist an $varepsilon > 0$ such that for all $sigma$ then: if $mu(A) < sigma$, then $|int_{A} f| geq varepsilon$. If this happens, you can make the integral arbitrarily large.
answered Dec 6 '18 at 19:39
GlazunovGlazunov
213
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$begingroup$
That is my thought too but I was hoping someone could formulate it in a formal way
$endgroup$
– Jonathan Baram
Dec 6 '18 at 19:41
add a comment |
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Hint: Given $epsilon>0,$ you can find a bounded $b$ such that $int_X|f-b|<epsilon.$ The result is clearly true for $b.$
answered Dec 6 '18 at 20:19
zhw.zhw.
72.3k43175
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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
$begingroup$
If the thesis were not true, it would exist an $varepsilon > 0$ such that for all $sigma$ then: if $mu(A) < sigma$, then $|int_{A} f| geq varepsilon$. If this happens, you can make the integral arbitrarily large.
answered Dec 6 '18 at 19:39
GlazunovGlazunov
213
$endgroup$
$begingroup$
That is my thought too but I was hoping someone could formulate it in a formal way
$endgroup$
– Jonathan Baram
Dec 6 '18 at 19:41
add a comment |
$begingroup$
If the thesis were not true, it would exist an $varepsilon > 0$ such that for all $sigma$ then: if $mu(A) < sigma$, then $|int_{A} f| geq varepsilon$. If this happens, you can make the integral arbitrarily large.
answered Dec 6 '18 at 19:39
GlazunovGlazunov
213
$endgroup$
$begingroup$
That is my thought too but I was hoping someone could formulate it in a formal way
$endgroup$
– Jonathan Baram
Dec 6 '18 at 19:41
add a comment |
$begingroup$
If the thesis were not true, it would exist an $varepsilon > 0$ such that for all $sigma$ then: if $mu(A) < sigma$, then $|int_{A} f| geq varepsilon$. If this happens, you can make the integral arbitrarily large.
answered Dec 6 '18 at 19:39
GlazunovGlazunov
213
$endgroup$
If the thesis were not true, it would exist an $varepsilon > 0$ such that for all $sigma$ then: if $mu(A) < sigma$, then $|int_{A} f| geq varepsilon$. If this happens, you can make the integral arbitrarily large.
answered Dec 6 '18 at 19:39
GlazunovGlazunov
213
answered Dec 6 '18 at 19:39
GlazunovGlazunov
213
answered Dec 6 '18 at 19:39
GlazunovGlazunov
213
answered Dec 6 '18 at 19:39
GlazunovGlazunov
213
213
$begingroup$
That is my thought too but I was hoping someone could formulate it in a formal way
$endgroup$
– Jonathan Baram
Dec 6 '18 at 19:41
add a comment |
$begingroup$
That is my thought too but I was hoping someone could formulate it in a formal way
$endgroup$
– Jonathan Baram
Dec 6 '18 at 19:41
$begingroup$
That is my thought too but I was hoping someone could formulate it in a formal way
$endgroup$
– Jonathan Baram
Dec 6 '18 at 19:41
$begingroup$
That is my thought too but I was hoping someone could formulate it in a formal way
$endgroup$
– Jonathan Baram
Dec 6 '18 at 19:41
add a comment |
$begingroup$
Hint: Given $epsilon>0,$ you can find a bounded $b$ such that $int_X|f-b|<epsilon.$ The result is clearly true for $b.$
answered Dec 6 '18 at 20:19
zhw.zhw.
72.3k43175
$endgroup$
add a comment |
$begingroup$
Hint: Given $epsilon>0,$ you can find a bounded $b$ such that $int_X|f-b|<epsilon.$ The result is clearly true for $b.$
answered Dec 6 '18 at 20:19
zhw.zhw.
72.3k43175
$endgroup$
add a comment |
$begingroup$
Hint: Given $epsilon>0,$ you can find a bounded $b$ such that $int_X|f-b|<epsilon.$ The result is clearly true for $b.$
answered Dec 6 '18 at 20:19
zhw.zhw.
72.3k43175
$endgroup$
Hint: Given $epsilon>0,$ you can find a bounded $b$ such that $int_X|f-b|<epsilon.$ The result is clearly true for $b.$
answered Dec 6 '18 at 20:19
zhw.zhw.
72.3k43175
answered Dec 6 '18 at 20:19
zhw.zhw.
72.3k43175
answered Dec 6 '18 at 20:19
zhw.zhw.
72.3k43175
answered Dec 6 '18 at 20:19
zhw.zhw.
72.3k43175
72.3k43175
add a comment |
add a comment |
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$begingroup$
You want $|f|$ in the integral
$endgroup$
– zhw.
Dec 6 '18 at 20:15