0
$\begingroup$

I am stuck trying to prove this inequality in which $u$ is an arbitrary measure and $f_n$ converges monotonically increasing to $f$ I have proved: $$\int fdu\geqslant\limsup_n\int f_ndu$$ using Fatou´s Lemma $$\liminf_n\int fdu\leqslant\liminf_n\int f_ndu$$ . I may need to use it here, however I do not know how. I do not understand why I do not get an answer if I am being clear. My problem is with this last proof. I need the following result to conclude the proof of the dominated convergence theorem. I cannot prove this last expression:

$$\int fdu\leqslant\liminf_n\int f_ndu$$

$\endgroup$
  • $\begingroup$ Use the fact that $\limsup f_n = -\liminf (-f_n)$ and linearity. $\endgroup$ – Adam Hughes Feb 15 '17 at 14:31
  • 1
    $\begingroup$ @PedroGomes você fala português? Acredito que seu inglês seja a única coisa que esteja nos impedindo de entender sua pergunta... $\endgroup$ – Filburt Feb 15 '17 at 14:40
  • $\begingroup$ Falo português. Eu não consigo provar a última expressão, tendo já provado as anteriores. $\endgroup$ – Pedro Gomes Feb 15 '17 at 15:35
  • $\begingroup$ I want to prove the $\lim_n\int f_nd\mu=\int f d\mu$. I have already proven $\int fd\mu\geqslant\limsup_nf_nd\mu$, so I need to prove $\lim\int fd\mu\leqslant\liminf_nfd\mu$ to conclude the dominated convergence theorem proof. $\endgroup$ – Pedro Gomes Feb 16 '17 at 12:17
0
$\begingroup$

From your last comment it seems that you are trying to deduce the Monotone Convergence Theorem from Fatou's lemma. Here is a proof of that:

If $\{f_n\}$ us any sequence in $L^{+}$, then $$\int f = \int (\liminf_{n\to \infty} f_n) \leq \liminf_{n\to \infty}\int f_n$$ Let $\{f_n\}_{n\in\mathbb{N}}\subset L^{+}$, then by Fatou's lemma $$\int f = \int \lim_{n\to \infty} f_n \leq \lim_{k\to \infty} \inf_{n\geq k}\int f_n$$ We know that $f = \lim_{n\to \infty}f_n(=\sup_{n} f_n)$, and that $f_n\leq f$ hence $\int f_n\leq \int f$ for all $n\in \mathbb{N}$, it is clear that $\lim_{k\to \infty}\sup_{n\geq k}\int f_n\leq \int f$. Therefore, \begin{align*} \limsup_{n\to \infty}\int f_n = \lim_{k\to \infty}\sup_{n\geq k} &\leq \int f\\ &= \int \liminf_{n\to \infty} f_n\\ &\leq \lim_{k\to \infty}\inf_{n\geq k}\int f_n\\ &= \liminf_{n\to \infty} \int f_n \end{align*} Since we also know that $$\lim_{n\to \infty}\inf \int f_n \leq \lim_{n\to \infty}\sup \int f_n$$ From the latter above we then have $$\int f = \liminf_{n\to \infty}\int f_n = \limsup_{n\to \infty} f_n$$ Which means $$\int f = \lim_{n\to \infty}\int f_n$$

$\endgroup$
  • $\begingroup$ I guess it is called the dominated convergence theorem. I do not get when you write $\int f d\mu=\int\liminf_n\ f_n d\mu$ That is precisely what I want to prove. $\endgroup$ – Pedro Gomes Feb 16 '17 at 20:58
  • $\begingroup$ My apologies, so you are trying to prove the Dominated Convergence Theorem? $\endgroup$ – Wolfy Feb 17 '17 at 1:12
  • $\begingroup$ Could you please write the proof of $\int fd\mu=\int\liminf_n f_n d\mu$? $\endgroup$ – Pedro Gomes Feb 17 '17 at 12:35
  • $\begingroup$ Could you please prove $\int fd\mu=\int\liminf_{n\to \infty} f_nd\mu$? This is the equality I am striving to prove. $\endgroup$ – Pedro Gomes Feb 17 '17 at 12:44
  • $\begingroup$ Yes, I'll provide a proof in about An hour in class $\endgroup$ – Wolfy Feb 17 '17 at 18:04
0
$\begingroup$

I think what is missed in my question and on the last answer is that we must define $f$.

If we define $f$ as the convergence of increasing from below sequence of functions $f_n\uparrow f$. Then the $\int \liminf_{n\to \infty}\ f_n$=$\int f d\mu$, because it increases from below as the definition states. Correct me please if I am wrong.

$\endgroup$
  • $\begingroup$ Yes, I believe that is correct $\endgroup$ – Wolfy Feb 26 '17 at 22:42

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.