$f,g: [0,1] to mathbb{R}$ with equal Lebesgue integrals
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I am given two functions $f, g: [0,1] rightarrow mathbb{R}$, with $$int_{0}^{1} f(x) dx = int_{0}^{1} g(x) dx = 1,$$ (where the previous integrals are Lebesgue integrals), and I am asked to prove that for every $alpha in (0,1)$, there exists a measurable $E subset [0,1]$, such that $$int_{E} f(x) dx = int_{E} g(x) dx = alpha.$$ I have already proved this, in the special case that $f$, $g$ are simple functions, but I don't know how to generalize it. Could anyone give me a hint?
lebesgue-integral
edited Nov 21 at 15:54
Xander Henderson
14.1k103553
asked Nov 21 at 15:51
A.M.
11
add a comment |
up vote
0
down vote
favorite
I am given two functions $f, g: [0,1] rightarrow mathbb{R}$, with $$int_{0}^{1} f(x) dx = int_{0}^{1} g(x) dx = 1,$$ (where the previous integrals are Lebesgue integrals), and I am asked to prove that for every $alpha in (0,1)$, there exists a measurable $E subset [0,1]$, such that $$int_{E} f(x) dx = int_{E} g(x) dx = alpha.$$ I have already proved this, in the special case that $f$, $g$ are simple functions, but I don't know how to generalize it. Could anyone give me a hint?
lebesgue-integral
edited Nov 21 at 15:54
Xander Henderson
14.1k103553
asked Nov 21 at 15:51
A.M.
11
If you have already shown that it works for simple functions, then the next step of the bootstrapping argument is to show that it works for nonnegative functions, which can be done by considering monotone limits of simple functions. Once that is done, you can generalize to measurable functions by considering the positive and negative parts as nonnegative functions.
– Xander Henderson
Nov 21 at 15:56
Thank you very much.
– A.M.
Nov 21 at 16:51
I have already tried to show it following these steps but it fails. Could you please be more specific?
– A.M.
Nov 22 at 12:00
@XanderHenderson I don't see how it's clear that the property is preserved under monotone limits.
– David C. Ullrich
Nov 22 at 17:12
@DavidC.Ullrich To be honest, I didn't give this problem much thought---my first instinct would be to try that approach. Since the question doesn't indicate what has been tried (if anything), that seemed like a good place to start.
– Xander Henderson
Nov 22 at 17:23
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am given two functions $f, g: [0,1] rightarrow mathbb{R}$, with $$int_{0}^{1} f(x) dx = int_{0}^{1} g(x) dx = 1,$$ (where the previous integrals are Lebesgue integrals), and I am asked to prove that for every $alpha in (0,1)$, there exists a measurable $E subset [0,1]$, such that $$int_{E} f(x) dx = int_{E} g(x) dx = alpha.$$ I have already proved this, in the special case that $f$, $g$ are simple functions, but I don't know how to generalize it. Could anyone give me a hint?
lebesgue-integral
edited Nov 21 at 15:54
Xander Henderson
14.1k103553
asked Nov 21 at 15:51
A.M.
11
I am given two functions $f, g: [0,1] rightarrow mathbb{R}$, with $$int_{0}^{1} f(x) dx = int_{0}^{1} g(x) dx = 1,$$ (where the previous integrals are Lebesgue integrals), and I am asked to prove that for every $alpha in (0,1)$, there exists a measurable $E subset [0,1]$, such that $$int_{E} f(x) dx = int_{E} g(x) dx = alpha.$$ I have already proved this, in the special case that $f$, $g$ are simple functions, but I don't know how to generalize it. Could anyone give me a hint?
lebesgue-integral
lebesgue-integral
edited Nov 21 at 15:54
Xander Henderson
14.1k103553
asked Nov 21 at 15:51
A.M.
11
edited Nov 21 at 15:54
Xander Henderson
14.1k103553
asked Nov 21 at 15:51
A.M.
11
edited Nov 21 at 15:54
Xander Henderson
14.1k103553
edited Nov 21 at 15:54
Xander Henderson
14.1k103553
edited Nov 21 at 15:54
Xander Henderson
14.1k103553
14.1k103553
asked Nov 21 at 15:51
A.M.
11
asked Nov 21 at 15:51
A.M.
11
asked Nov 21 at 15:51
A.M.
11
11
If you have already shown that it works for simple functions, then the next step of the bootstrapping argument is to show that it works for nonnegative functions, which can be done by considering monotone limits of simple functions. Once that is done, you can generalize to measurable functions by considering the positive and negative parts as nonnegative functions.
– Xander Henderson
Nov 21 at 15:56
Thank you very much.
– A.M.
Nov 21 at 16:51
I have already tried to show it following these steps but it fails. Could you please be more specific?
– A.M.
Nov 22 at 12:00
@XanderHenderson I don't see how it's clear that the property is preserved under monotone limits.
– David C. Ullrich
Nov 22 at 17:12
@DavidC.Ullrich To be honest, I didn't give this problem much thought---my first instinct would be to try that approach. Since the question doesn't indicate what has been tried (if anything), that seemed like a good place to start.
– Xander Henderson
Nov 22 at 17:23
add a comment |
If you have already shown that it works for simple functions, then the next step of the bootstrapping argument is to show that it works for nonnegative functions, which can be done by considering monotone limits of simple functions. Once that is done, you can generalize to measurable functions by considering the positive and negative parts as nonnegative functions.
– Xander Henderson
Nov 21 at 15:56
Thank you very much.
– A.M.
Nov 21 at 16:51
I have already tried to show it following these steps but it fails. Could you please be more specific?
– A.M.
Nov 22 at 12:00
@XanderHenderson I don't see how it's clear that the property is preserved under monotone limits.
– David C. Ullrich
Nov 22 at 17:12
@DavidC.Ullrich To be honest, I didn't give this problem much thought---my first instinct would be to try that approach. Since the question doesn't indicate what has been tried (if anything), that seemed like a good place to start.
– Xander Henderson
Nov 22 at 17:23
If you have already shown that it works for simple functions, then the next step of the bootstrapping argument is to show that it works for nonnegative functions, which can be done by considering monotone limits of simple functions. Once that is done, you can generalize to measurable functions by considering the positive and negative parts as nonnegative functions.
– Xander Henderson
Nov 21 at 15:56
If you have already shown that it works for simple functions, then the next step of the bootstrapping argument is to show that it works for nonnegative functions, which can be done by considering monotone limits of simple functions. Once that is done, you can generalize to measurable functions by considering the positive and negative parts as nonnegative functions.
– Xander Henderson
Nov 21 at 15:56
Thank you very much.
– A.M.
Nov 21 at 16:51
Thank you very much.
– A.M.
Nov 21 at 16:51
I have already tried to show it following these steps but it fails. Could you please be more specific?
– A.M.
Nov 22 at 12:00
I have already tried to show it following these steps but it fails. Could you please be more specific?
– A.M.
Nov 22 at 12:00
@XanderHenderson I don't see how it's clear that the property is preserved under monotone limits.
– David C. Ullrich
Nov 22 at 17:12
@XanderHenderson I don't see how it's clear that the property is preserved under monotone limits.
– David C. Ullrich
Nov 22 at 17:12
@DavidC.Ullrich To be honest, I didn't give this problem much thought---my first instinct would be to try that approach. Since the question doesn't indicate what has been tried (if anything), that seemed like a good place to start.
– Xander Henderson
Nov 22 at 17:23
@DavidC.Ullrich To be honest, I didn't give this problem much thought---my first instinct would be to try that approach. Since the question doesn't indicate what has been tried (if anything), that seemed like a good place to start.
– Xander Henderson
Nov 22 at 17:23
add a comment |
1 Answer
1
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oldest
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up vote
0
down vote
I don't see a "simple" elementary proof. It's easy if you've recently covered some non-trivial results on vector-valued measures; for example here it says "Finally, a non-atomic $X$-valued measure on a $sigma$-field has compact and convex range if $dim(X)<infty$. This is Lyapunov's theorem. It fails for infinite-dimensional $X$."
answered Nov 22 at 17:30
David C. Ullrich
57.5k43791
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I don't see a "simple" elementary proof. It's easy if you've recently covered some non-trivial results on vector-valued measures; for example here it says "Finally, a non-atomic $X$-valued measure on a $sigma$-field has compact and convex range if $dim(X)<infty$. This is Lyapunov's theorem. It fails for infinite-dimensional $X$."
answered Nov 22 at 17:30
David C. Ullrich
57.5k43791
add a comment |
up vote
0
down vote
I don't see a "simple" elementary proof. It's easy if you've recently covered some non-trivial results on vector-valued measures; for example here it says "Finally, a non-atomic $X$-valued measure on a $sigma$-field has compact and convex range if $dim(X)<infty$. This is Lyapunov's theorem. It fails for infinite-dimensional $X$."
answered Nov 22 at 17:30
David C. Ullrich
57.5k43791
add a comment |
up vote
0
down vote
up vote
0
down vote
I don't see a "simple" elementary proof. It's easy if you've recently covered some non-trivial results on vector-valued measures; for example here it says "Finally, a non-atomic $X$-valued measure on a $sigma$-field has compact and convex range if $dim(X)<infty$. This is Lyapunov's theorem. It fails for infinite-dimensional $X$."
answered Nov 22 at 17:30
David C. Ullrich
57.5k43791
I don't see a "simple" elementary proof. It's easy if you've recently covered some non-trivial results on vector-valued measures; for example here it says "Finally, a non-atomic $X$-valued measure on a $sigma$-field has compact and convex range if $dim(X)<infty$. This is Lyapunov's theorem. It fails for infinite-dimensional $X$."
answered Nov 22 at 17:30
David C. Ullrich
57.5k43791
answered Nov 22 at 17:30
David C. Ullrich
57.5k43791
answered Nov 22 at 17:30
David C. Ullrich
57.5k43791
answered Nov 22 at 17:30
David C. Ullrich
57.5k43791
57.5k43791
add a comment |
add a comment |
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If you have already shown that it works for simple functions, then the next step of the bootstrapping argument is to show that it works for nonnegative functions, which can be done by considering monotone limits of simple functions. Once that is done, you can generalize to measurable functions by considering the positive and negative parts as nonnegative functions.
– Xander Henderson
Nov 21 at 15:56
Thank you very much.
– A.M.
Nov 21 at 16:51
I have already tried to show it following these steps but it fails. Could you please be more specific?
– A.M.
Nov 22 at 12:00
@XanderHenderson I don't see how it's clear that the property is preserved under monotone limits.
– David C. Ullrich
Nov 22 at 17:12
@DavidC.Ullrich To be honest, I didn't give this problem much thought---my first instinct would be to try that approach. Since the question doesn't indicate what has been tried (if anything), that seemed like a good place to start.
– Xander Henderson
Nov 22 at 17:23