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Confusing
science forum beginner
Joined: 11 Dec 2005
Posts: 2
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 Posted: Sun Dec 11, 2005 3:47 pm Post subject:
Difference Between "a.e bounded function" and "a.e finite function"
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I am studying Real analysis and probality myself.
And sometimes they say if a function is in L^inf then it is a.e.
bounded.
And sometimes they say if a functions is in L^1 so it is a.e. finite.
So they should not be the same, right?
What is the difference between these 2 ?
Thanks. |
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gowan4@hotmail.com
science forum addict
Joined: 10 May 2005
Posts: 86
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| Posted: Sun Dec 11, 2005 5:49 pm Post subject:
Re: Difference Between "a.e bounded function" and "a.e finite function"
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The function f(x) = x^(-1/2), with f(0) = 0, is in L^1([0,1]). f(x) is
finite everywhere in [0,1] but it is not bounded a.e. |
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Stephen J. Herschkorn
science forum Guru
Joined: 24 Mar 2005
Posts: 641
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| Posted: Sun Dec 11, 2005 6:14 pm Post subject:
Re: Difference Between "a.e bounded function" and "a.e finite function"
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Confusing wrote:
| Quote: | I am studying Real analysis and probality myself.
And sometimes they say if a function is in L^inf then it is a.e.
bounded.
And sometimes they say if a functions is in L^1 so it is a.e. finite.
So they should not be the same, right?
What is the difference between these 2 ?
|
In analysis, we sometimes let functions take values in the extended real
numbers. That is, there may exist x such that f(x) equals positive
or negative infinity.
A function f is a.e. bounded if and only if there is a finite positive
real number c such that {x: |f(x)| > c} is contained in a set of
measure 0. By definitition. f is in L^infty iff f is a.e. bounded.
A function f is a.e. finite if and only if {x: f(x) = +/- infty} is
contained in a set of measure 0.
By definition, f is in L^1 iff the integral of |f| is finite. f
in L^1 implies f is a.e. finite, but the converse is not true.
--
Stephen J. Herschkorn sjherschko@netscape.net
Math Tutor on the Internet and in Central New Jersey and Manhattan |
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Confusing
science forum beginner
Joined: 11 Dec 2005
Posts: 2
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| Posted: Sun Dec 11, 2005 6:32 pm Post subject:
Re: Difference Between "a.e bounded function" and "a.e finite function"
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Very good explanation, thanks. |
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David C. Ullrich
science forum Guru
Joined: 28 Apr 2005
Posts: 2250
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| Posted: Sun Dec 11, 2005 7:42 pm Post subject:
Re: Difference Between "a.e bounded function" and "a.e finite function"
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On 11 Dec 2005 07:47:25 -0800, "Confusing" <pingkai@gmail.com> wrote:
| Quote: | I am studying Real analysis and probality myself.
And sometimes they say if a function is in L^inf then it is a.e.
bounded.
And sometimes they say if a functions is in L^1 so it is a.e. finite.
So they should not be the same, right?
What is the difference between these 2 ?
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Well, "ae bounded" is extremely bad terminology.
A person might use this phrase informally; it
doesn't actually appear in the book, does it?
Anyway, saying f is "ae finite" means exactly
what it says: for almost every x, f(x) is finite.
Or: there exists a set E of measure zero such
that f(x) is finite for every x not in E.
Similarly, "ae bounded" should mean "f(x) is
bounded for almost every x". But that doesn't
make much sense, and it's also not what is
meant; when a person says that f is ae
bounded he means that there is a set E of
measure zero such that f is bounded on
the complement of E.
************************
David C. Ullrich |
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