# Fatou's lemma Proof

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:

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

• Use the fact that $\limsup f_n = -\liminf (-f_n)$ and linearity. – Adam Hughes Feb 15 '17 at 14:31
• @PedroGomes você fala português? Acredito que seu inglês seja a única coisa que esteja nos impedindo de entender sua pergunta... – Filburt Feb 15 '17 at 14:40
• Falo português. Eu não consigo provar a última expressão, tendo já provado as anteriores. – Pedro Gomes Feb 15 '17 at 15:35
• 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. – Pedro Gomes Feb 16 '17 at 12:17

## 2 Answers

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 $$\int f = \lim_{n\to \infty}\int f_n$$

• 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. – Pedro Gomes Feb 16 '17 at 20:58
• My apologies, so you are trying to prove the Dominated Convergence Theorem? – Wolfy Feb 17 '17 at 1:12
• Could you please write the proof of $\int fd\mu=\int\liminf_n f_n d\mu$? – Pedro Gomes Feb 17 '17 at 12:35
• Could you please prove $\int fd\mu=\int\liminf_{n\to \infty} f_nd\mu$? This is the equality I am striving to prove. – Pedro Gomes Feb 17 '17 at 12:44
• Yes, I'll provide a proof in about An hour in class – Wolfy Feb 17 '17 at 18:04

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.

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