1
$\begingroup$

The dominated convergence theorem states that if we have have $(u_j)_{j\in \mathbb{N}}$ which are all integrable with respect to the measure $\mu$ and $|u_j|<w$ for an positive integrable function, then $\lim_{j \rightarrow \infty} u_j = u(x)$ is integrable as well.

Now suppose $f_1,f_2........$ converge uniformly to some $f$ in the space $(\Omega,F)$ and $\mu(\Omega)<\infty$. I want to prove that $f$ is integrable. If i can prove that there exist an positive integrable function $w$ with $|u_j| \leq w$ we are finished.

The main problem here is that i don't know how to construct this boundary. I know that $f_i + f_j$ also integrable, so i've tried to construct w as $\sum |f_i|$, but this function is not necesarrily bounded is it? an infinite sum of finite members do not need to be finite again, so i'm pretty sure this fails. But how would one need to construct this bound?

Kees

$\endgroup$
  • 1
    $\begingroup$ I suppose $f_i$ are integrable. What domain have $f_i$? You have that $\sum \int |f_i| = \int \sum |f_i|$ so if $\int |f_i|$ is a convergent sequence it converge, otherwise diverge $\endgroup$ – Davide F. Feb 8 '15 at 10:22
  • $\begingroup$ yes of course sorry, should have noted that as well. The domain is just a space $(\Omega,F)$ with $\mu(\Omega) < \infty$ $\endgroup$ – Kees Til Feb 8 '15 at 10:25
  • $\begingroup$ can you be more explicit with the last term, if $\int|f_i|$ is a convergent sequence then $\sum \int|f_i|$ is convergent. why is that? $\endgroup$ – Kees Til Feb 8 '15 at 10:30
  • 1
    $\begingroup$ Sorry, I mean that if you have that $\int |f_i|$, that is a real sequence, converge than $\sum |f_i|$ is integrable. I don't know is you can reach a solution form there, just thinking a way to bound your $w$ funtion $\endgroup$ – Davide F. Feb 8 '15 at 10:33
1
$\begingroup$

There is no need for the dominated convergence theorem:

Since $f_n$ converges uniformly to $f$, we can choose $n \in \mathbb{N}$ such that

$$\sup_{\omega \in \Omega} |f_n(\omega)-f(\omega)| \leq 1.$$

As $\mu$ is a finite measure, this implies

$$\int_{\Omega} |f_n-f| \, d\mu \leq \int_{\Omega} 1 \, d\mu = \mu(\Omega)<\infty,$$

i.e. $f_n-f$ is integrable. Finally, as $f_n$ is integrable, it follows from the triangle inequality that

$$\int |f| \, d\mu \leq \int |f_n-f| \, d\mu + \int |f_n| \, d\mu < \infty.$$

Hence, $f \in L^1(\mu)$.

$\endgroup$
  • $\begingroup$ nice! I thought i needed to use the dominated convergence theorem because the next question i need to prove $\lim_{n \rightarrow \infty}\int f_n = \int f$. $\endgroup$ – Kees Til Feb 8 '15 at 10:48
  • $\begingroup$ @KeesTil To show this, you don't need the dominated convergence theorem. Note that $$\left| \int f_n - \int f \right| \leq \int |f_n-f| \, d\mu \leq \dots $$ $\endgroup$ – saz Feb 8 '15 at 10:52
  • $\begingroup$ hmm... i don't exactly see how you get there like that, because you can set on the dots: $\int \sup(f_n-f) \leq \int 1 < \infty$? $\endgroup$ – Kees Til Feb 8 '15 at 10:58
  • 1
    $\begingroup$ @KeesTil Note that $$\int |f_n-f| \, d\mu \leq \|f_n-f\|_{\infty} \mu(\Omega).$$ So, since $\|f_n-f\|_{\infty}$ converges to $0$ .... $\endgroup$ – saz Feb 8 '15 at 10:59
1
$\begingroup$

$$\int |f_i-f_j|<\mu(\Omega)\sup_\Omega|f_i-f_j|\to0$$ so $f_i$ is a Cauchy sequenze in $L_1$ and we have to have $f_i\to f$ in the meaning of $L_1$ but $L_1$ is complete and therfore $f \in L_1$

$\endgroup$
  • 1
    $\begingroup$ Ouch: saw saz solution, my is a bit overcomplicated, if you just remove the $_j$ index is better $\endgroup$ – Davide F. Feb 8 '15 at 10:48

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.