1
$\begingroup$

I have to show that if $\Omega \subseteq \mathbb{R}$ measurable, and $\{f_n\}$ is a sequence of Lebesgue integrable functions on $\Omega$ then exists an integrable function $f$ on $\Omega$ such that

$$\lim_{n\to\infty}\int_{\Omega}|f_n - f|dx = 0$$

if $$\forall \epsilon > 0, \exists N | \forall n,m > N, \int_{\Omega}|f_n - f_m|dx < \epsilon$$.

This is equivalent to a Cauchy sequence in Measure.

I was trying to use the fact that each $f_n$ is Lebesgue integrable, and claim that for each $f_n$, exists a sequence of simple function $\phi_k$ such that $\phi_k \to f_n$ as $k\to \infty$, and $|\phi_k| \leq |f_n|$ and then use the dominated convergence theorem, since each $f_n$ is integrable and dominate $\phi_k$ then

$$\lim_{k\to \infty}\int_{\Omega}\phi_k = \int_{\Omega}f_n$$

I want to show that $\phi_k$ is the limit of $f_n$ and the integral of that integrable function is the limit of the Cauchy sequence in measure. But I get stuck here. Could anyone helps me finish the proof or I'm not going in the right direction?

$\endgroup$
1
$\begingroup$

For your question, it is equivalent to showing that $L^{1}(\Omega)$ is complete with respect to $||\cdot||_{1}$-norm. Recall that for an abstract normed space $(X,||\cdot||)$, the norm is complete iff for any sequence $(x_{n})$ in $X$, $\lim_{n\rightarrow\infty}\sum_{k=1}^{n}x_{k}$ exists whenever $\sum_{k=1}^{\infty}||x_{k}||<\infty$. (The proof is elementary and is not difficult).

We assume the above proposition without proof (if you really need it, I can re-produce it for you.). Let $(f_{n})$ be a sequence in $L^{1}(\Omega)$ such that $\sum_{k=1}^{\infty}||f_{k}||_{1}<\infty$. For each $n$, define $S_{n}=\sum_{k=1}^{n}|f_{k}|$. Clearly, for each $x\in\Omega$, $0\leq S_{1}(x)\leq S_{2}(x)\leq\ldots$. Therefore $S(x)=\lim_{n\rightarrow\infty}S_{n}(x)$ exists and is measurable (but $S(x)$ may be $+\infty$). By monotone convergence theorem, $$ \int S(x)dx=\lim_{n\rightarrow\infty}\int S_{n}(x)dx=\lim_{n\rightarrow\infty}\sum_{k=1}^{n}||f_{k}||_{1}<\infty. $$ That is, $S$ is integrable. In particular, $S(x)<\infty$ for $x$-a.e. Denote $A=\{x\in\Omega\mid S(x)<\infty\}$, which is a measurable set. For each $x\in A$, the series $\sum_{k=1}^{\infty}|f_{k}(x)|$ converges, so $\sum_{k=1}^{\infty}f_{k}(x)$ also converges (which is a consequence of the completeness of $\mathbb{R}$). Define $f(x)=\sum_{k=1}^{\infty}f_{k}(x)$ if $x\in A$ and $f(x)=0$ if $x\in A^{c}$, then $f$ is measurable. Lastly, verify that $||\sum_{k=1}^{n}f_{k}-f||_{1}\rightarrow0$ as $n\rightarrow\infty$.

$\endgroup$
  • $\begingroup$ I already got a similar solution, but it was too elaborated. However, this proof is a simpler proof thanks to the monotone convergence theorem. Thank you! Great answer. $\endgroup$ – Richard Clare Mar 26 '18 at 12:49
1
$\begingroup$

HINT

I suggest to take a look at the proof of this theorem:

A normed space $X$ is complete iff every absolutely convergent series converges in $X$

You can use the technique of the proof of this theorem to construct a convergence subsequence of your Cauchy sequence.

Then use the fact that if a Cauchy sequence has a convergent subsequence to some limit $x$ than it converges to the same limit $x$

After you find the limit prove that is integrable.

$\endgroup$
  • 1
    $\begingroup$ I could prove your statement my self but i though that would be better for you to give you hints $\endgroup$ – Marios Gretsas Aug 24 '17 at 15:10
  • 1
    $\begingroup$ math.cuhk.edu.hk/course_builder/1516/math4010/solution3.pdf this a link to the proof of the theorem in my answer $\endgroup$ – Marios Gretsas Aug 24 '17 at 15:16
  • 1
    $\begingroup$ I believe (maybe i'm wrong of course) that this is an easier approach than using simple functions.And the technique of the theorem with the series is quite nice for someone to remember. $\endgroup$ – Marios Gretsas Aug 24 '17 at 15:17
  • 1
    $\begingroup$ I think that the completenss of the $L^p$ spaces is taught in a measure theory course.In the proof you can just instead of take a norm just take the integral and use the technique(which is based on the definition of a Cauchy sequence) in the proof theorem and you are done. $\endgroup$ – Marios Gretsas Aug 24 '17 at 15:25
  • 1
    $\begingroup$ After you find the existence of $f$ as the limit of the sequence then along with the definition of the limit of the sequence where ''$\forall \epsilon>0 \exists n_0 \in \mathbb{N}$ etc...''.use: $$\int|f| \leq \int|f_{n_0}-f|+\int |f_{n_0}| \leq \epsilon +M< \infty$$ because $f_{n_0}$ is Lebesgue integrable from hypothesis $\endgroup$ – Marios Gretsas Aug 24 '17 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.