# Lebesgue Integral

• Nov 25th 2009, 06:42 AM
dedust
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 $\displaystyle \left\{ f_{n}\right\}$ be a sequence of function and $\displaystyle f_{n}\longrightarrow f$ in measure.
Let $\displaystyle g$ be an integrable function such that $\displaystyle \left\vert f_{n}\right\vert \leq g,\forall n$.
Show that $\displaystyle % %TCIMACRO{\dint }% %BeginExpansion {\displaystyle\int} %EndExpansion \left\vert f_{n}-f\right\vert \longrightarrow0$

thks
• Nov 25th 2009, 02:22 PM
Jose27
By Lebesgue's dominated convergence $\displaystyle f$ is integrable, since $\displaystyle L^1$ is a vector space $\displaystyle f-f_n$ is integrable for all $\displaystyle n$ and so $\displaystyle \vert f-f_n\vert$ is also integrable and $\displaystyle \vert f-f_n\vert \rightarrow 0$ and $\displaystyle \vert f-f_n\vert \leq \vert f\vert + \vert f_n \vert \leq \vert f\vert +\vert g \vert$ so applying Lebesgue again you get the result.
• Nov 25th 2009, 02:54 PM
Moo
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.
• Nov 25th 2009, 03:32 PM
Jose27
Yeah, I read what I wanted. Anyway, if the measure is $\displaystyle \sigma$ 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).
• Nov 25th 2009, 05:24 PM
dedust
yes, it's a finite measure.

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

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

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

and

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

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

thks
• Nov 27th 2009, 11:43 PM
Moo
Hello,

Here is a solution (not from me)

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

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

We have $\displaystyle |f|\leq |f_n|+\epsilon$, except over $\displaystyle A_n$ (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 $\displaystyle |f|\leq |g|+\epsilon$, except over a set of measure 0.
Thus it is true almost everywhere.

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

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

And finally, we can use Lebesgue's dominated convergence theorem.
• Dec 19th 2009, 08:43 AM
dedust
need clarifications
Quote:

Originally Posted by Moo
Hello,

Here is a solution (not from me)

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

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

We have $\displaystyle |f|\leq |f_n|+\epsilon$, except over $\displaystyle A_n$ (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 $\displaystyle |f|\leq |g|+\epsilon$, except over a set of measure 0.
Thus it is true almost everywhere.

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

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

And finally, we can use Lebesgue's dominated convergence theorem.

so we have $\displaystyle |f_{n}| \leq g$ and $\displaystyle |f| \leq |g| = g$ a.e, then

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

and

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

from Lebesgue's dominated convergence theorem,

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

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

am I doing right?
• Dec 22nd 2009, 05:35 PM
Kat-M
lebesgue integral
Quote:

Originally Posted by Moo
Hello,

Here is a solution (not from me)

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

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

We have $\displaystyle |f|\leq |f_n|+\epsilon$, except over $\displaystyle A_n$ (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 $\displaystyle |f|\leq |g|+\epsilon$, except over a set of measure 0.
Thus it is true almost everywhere.

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

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

And finally, we can use Lebesgue's dominated convergence theorem.

i am not sure how to apply lebesgue dominated convergence by knowing $\displaystyle |f|\leq |g|$ almost everywhere. we still cant say $\displaystyle lim|f-f_n| \rightarrow 0$ a.e right? could you explain this please?