# Math Help - Lebesgue Integral

1. ## 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 $%
%TCIMACRO{\dint }%
�ginExpansion
{\displaystyle\int}
%EndExpansion
\left\vert f_{n}-f\right\vert \longrightarrow0$

thks

2. 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.

3. 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.

4. 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).

5. 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
$%
%TCIMACRO{\dint \limits_{E}}%
�ginExpansion
{\displaystyle\int\limits_{E}}
%EndExpansion
\left\vert f(x)-f_{n}(x)\right\vert =%
%TCIMACRO{\dint \limits_{E\backslash A}}%
�ginExpansion
{\displaystyle\int\limits_{E\backslash A}}
%EndExpansion
\left\vert f(x)-f_{n}(x)\right\vert +%
%TCIMACRO{\dint \limits_{A}}%
�ginExpansion
{\displaystyle\int\limits_{A}}
%EndExpansion
\left\vert f(x)-f_{n}(x)\right\vert$

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

and

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

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

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.

7. ## need clarifications

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?

8. ## lebesgue integral

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?