Lebesgue Integral

**dedust** · November 25th 2009, 06:42 AM

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 $% %TCIMACRO{\dint }% %BeginExpansion {\displaystyle\int} %EndExpansion \left\vert f_{n}-f\right\vert \longrightarrow0$

thks

**Jose27** · November 25th 2009, 02:22 PM

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.

**Moo** · November 25th 2009, 02:54 PM

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.

**Jose27** · November 25th 2009, 03:32 PM

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).

**dedust** · November 25th 2009, 05:24 PM

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
$% %TCIMACRO{\dint \limits_{E}}% %BeginExpansion {\displaystyle\int\limits_{E}} %EndExpansion \left\vert f(x)-f_{n}(x)\right\vert =% %TCIMACRO{\dint \limits_{E\backslash A}}% %BeginExpansion {\displaystyle\int\limits_{E\backslash A}} %EndExpansion \left\vert f(x)-f_{n}(x)\right\vert +% %TCIMACRO{\dint \limits_{A}}% %BeginExpansion {\displaystyle\int\limits_{A}} %EndExpansion \left\vert f(x)-f_{n}(x)\right\vert$

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

and

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

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

thks

**Moo** · November 27th 2009, 11:43 PM

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.

**dedust** · December 19th 2009, 08:43 AM

Originally Posted by Moo

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?

**Kat-M** · December 22nd 2009, 05:35 PM

Originally Posted by Moo

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?

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