Results 1 to 8 of 8

Math Help - Lebesgue Integral

  1. #1
    Senior Member
    Joined
    Nov 2009
    Posts
    263

    Lebesgue Integral [need clarifications]

    Hi all,..
    i try to solve this problem, but i don't know what is the connection between convergence in measure and the integral. can someone help me. . .


    Let \left\{  f_{n}\right\} be a sequence of function and f_{n}\longrightarrow f in measure.
    Let g be an integrable function such that \left\vert f_{n}\right\vert \leq g,\forall n.
    Show that %<br />
%TCIMACRO{\dint }%<br />
%BeginExpansion<br />
{\displaystyle\int}<br />
%EndExpansion<br />
\left\vert f_{n}-f\right\vert \longrightarrow0

    thks
    Last edited by dedust; December 19th 2009 at 09:25 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    By Lebesgue's dominated convergence f is integrable, since L^1 is a vector space f-f_n is integrable for all n and so \vert f-f_n\vert is also integrable and \vert f-f_n\vert \rightarrow 0 and \vert f-f_n\vert \leq \vert f\vert + \vert f_n \vert \leq \vert f\vert +\vert g \vert so applying Lebesgue again you get the result.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    But hmmm where do you use the assumption that fn converges in measure to f ?
    It's not an almost everywhere convergence, so we can't say that |fn-f| goes to 0, almost everywhere...

    Are there more precisions about the measure we're working with ? If it's a finite measure, then it's possible to get a proof by splitting the space into {|fn-f|<e} and {|fn-f|>e}, for some given e.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Apr 2009
    From
    México
    Posts
    721
    Yeah, I read what I wanted. Anyway, if the measure is \sigma finite the result holds, I think, because Lebesgue dominated convergence is still valid in this setting, and I assume this is the case since the nature of the problem clearly suggests that particular theorem (although it could be one of those tricky questions).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2009
    Posts
    263
    yes, it's a finite measure.

    this is my work :
    Let E be the whole space.
    Define A=\left\{  x\mid\left\vert f(x)-f_{n}(x)\right\vert \geq\epsilon\right\}

    by splitting the space, we get
    %<br />
%TCIMACRO{\dint \limits_{E}}%<br />
%BeginExpansion<br />
{\displaystyle\int\limits_{E}}<br />
%EndExpansion<br />
\left\vert f(x)-f_{n}(x)\right\vert =%<br />
%TCIMACRO{\dint \limits_{E\backslash A}}%<br />
%BeginExpansion<br />
{\displaystyle\int\limits_{E\backslash A}}<br />
%EndExpansion<br />
\left\vert f(x)-f_{n}(x)\right\vert +%<br />
%TCIMACRO{\dint \limits_{A}}%<br />
%BeginExpansion<br />
{\displaystyle\int\limits_{A}}<br />
%EndExpansion<br />
\left\vert f(x)-f_{n}(x)\right\vert

    notice that
    %<br />
%TCIMACRO{\dint \limits_{E\backslash A}}%<br />
%BeginExpansion<br />
{\displaystyle\int\limits_{E\backslash A}}<br />
%EndExpansion<br />
\left\vert f(x)-f_{n}(x)\right\vert \leq\epsilon m\left(  E\backslash<br />
A\right)

    and

    %<br />
%TCIMACRO{\dint \limits_{A}}%<br />
%BeginExpansion<br />
{\displaystyle\int\limits_{A}}<br />
%EndExpansion<br />
\left\vert f(x)-f_{n}(x)\right\vert \geq\epsilon m\left(  A\right)

    is it true that \epsilon m\left(  A\right) and \epsilon m\left(  E\backslash<br />
A\right) converge to 0?
    where do we use the fact that \left\vert f_{n}\right\vert \leq g,\forall n?


    thks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    Here is a solution (not from me)

    Let's fix \epsilon
    Let A_n=\left\{ x\mid\left\vert f(x)-f_{n}(x)\right\vert \geq\epsilon\right\}
    Because we need n in it...

    We know that \lim_n \mu(A_n)=0.


    We have |f|\leq |f_n|+\epsilon, except over A_n (this deals with inclusion of sets), which is a set which measure goes to 0 as n goes to infinity.

    By taking the limit, we can say that |f|\leq |g|+\epsilon, except over a set of measure 0.
    Thus it is true almost everywhere.

    Since this is true for any \epsilon, we can take \epsilon=1/p and consider that \mu\left(\bigcap_{p\geq 1} \{|f|\leq |g|+\tfrac 1p\}\right)=\mu(E)
    But the measure of this intersection is the measure of \{|f|\leq |g|\}

    Hence |f|\leq |g| almost everywhere.


    And finally, we can use Lebesgue's dominated convergence theorem.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Nov 2009
    Posts
    263

    Question need clarifications

    Quote Originally Posted by Moo View Post
    Hello,

    Here is a solution (not from me)

    Let's fix \epsilon
    Let A_n=\left\{ x\mid\left\vert f(x)-f_{n}(x)\right\vert \geq\epsilon\right\}
    Because we need n in it...

    We know that \lim_n \mu(A_n)=0.


    We have |f|\leq |f_n|+\epsilon, except over A_n (this deals with inclusion of sets), which is a set which measure goes to 0 as n goes to infinity.

    By taking the limit, we can say that |f|\leq |g|+\epsilon, except over a set of measure 0.
    Thus it is true almost everywhere.

    Since this is true for any \epsilon, we can take \epsilon=1/p and consider that \mu\left(\bigcap_{p\geq 1} \{|f|\leq |g|+\tfrac 1p\}\right)=\mu(E)
    But the measure of this intersection is the measure of \{|f|\leq |g|\}

    Hence |f|\leq |g| almost everywhere.


    And finally, we can use Lebesgue's dominated convergence theorem.
    so we have |f_{n}| \leq g and |f| \leq |g| = g a.e, then

    |f - f_{n}| \leq |f| + |f_{n}| \leq 2g,

    and

    \lim_{n \to \infty} |f - f_{n}| = 0.

    from Lebesgue's dominated convergence theorem,

    \lim \int |f - f_{n}| = \int |f - f_{n}| = 0

    \therefore \int |f - f_{n}| \to 0

    am I doing right?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Dec 2008
    Posts
    154

    lebesgue integral

    Quote Originally Posted by Moo View Post
    Hello,

    Here is a solution (not from me)

    Let's fix \epsilon
    Let A_n=\left\{ x\mid\left\vert f(x)-f_{n}(x)\right\vert \geq\epsilon\right\}
    Because we need n in it...

    We know that \lim_n \mu(A_n)=0.


    We have |f|\leq |f_n|+\epsilon, except over A_n (this deals with inclusion of sets), which is a set which measure goes to 0 as n goes to infinity.

    By taking the limit, we can say that |f|\leq |g|+\epsilon, except over a set of measure 0.
    Thus it is true almost everywhere.

    Since this is true for any \epsilon, we can take \epsilon=1/p and consider that \mu\left(\bigcap_{p\geq 1} \{|f|\leq |g|+\tfrac 1p\}\right)=\mu(E)
    But the measure of this intersection is the measure of \{|f|\leq |g|\}

    Hence |f|\leq |g| almost everywhere.


    And finally, we can use Lebesgue's dominated convergence theorem.
    i am not sure how to apply lebesgue dominated convergence by knowing |f|\leq |g| almost everywhere. we still cant say lim|f-f_n| \rightarrow 0 a.e right? could you explain this please?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Lebesgue Integral
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 3rd 2010, 12:52 PM
  2. Lebesgue Integral
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: March 29th 2010, 03:00 AM
  3. Lebesgue integral
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: January 9th 2010, 06:01 PM
  4. Lebesgue integral
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: December 23rd 2009, 01:12 AM
  5. Lebesgue Integral
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: November 9th 2009, 04:19 AM

Search Tags


/mathhelpforum @mathhelpforum