4
$\begingroup$

2.18 Fatou's Lemma - If $\{f_n\}$ is any sequence in $L^+$, then $$\int \left(\lim_{n\rightarrow \infty}\inf f_n\right) \leq \lim_{n\rightarrow \infty}\inf\int f_n$$

Attempted proof - We know that $$\int \left(\lim_{n\rightarrow \infty}\inf f_n\right) = \int \sup_{k\geq 1}\left(\inf_{n\geq k}f_n\right) = \int \lim_{k\rightarrow \infty}\inf_{n\geq k}f_n$$ Then by the Monotone Convergence theorem $$\int \lim_{k\rightarrow \infty}\inf_{n\geq k}f_n = \lim_{k\rightarrow \infty}\int \inf_{n\geq k}f_n$$ Note that the Monotone Convergence theorem can be applied because $$\inf_{n\geq k} f_n \leq \inf_{n\geq k+1} f_n$$

We see that $\inf_{n\geq k}f_n \leq f_k$ for all $n\geq k$. So, \begin{align*} \int \inf_{n\geq k}f_n &\leq \int f_k \ \forall n\geq k\\ &\leq\inf_{n\geq k}\int f_n\\ &\leq \lim_{k\rightarrow \infty}\inf_{n\geq k}\int f_n \end{align*}

I may have some indexing mistakes but I think this is a sufficient proof. Any suggestions is greatly appreciated.

$\endgroup$
2
$\begingroup$

Your proof is essentially correct. Just some minor adjustments are required.

2.18 Fatou's Lemma - If $\{f_n\}$ is any sequence in $L^+$, then $$\int \left(\lim_{n\rightarrow \infty}\inf f_n\right) \leq \lim_{n\rightarrow \infty}\inf\int f_n$$

Proof - We know that $$\int \left(\lim_{n\rightarrow \infty}\inf f_n\right) = \int \sup_{k\geq 1}\left(\inf_{n\geq k}f_n\right) = \int \lim_{k\rightarrow \infty}\inf_{n\geq k}f_n$$ Then by the Monotone Convergence theorem $$\int \lim_{k\rightarrow \infty}\inf_{n\geq k}f_n = \lim_{k\rightarrow \infty}\int \inf_{n\geq k}f_n \tag{1}$$ Note that the Monotone Convergence theorem can be applied because $$\inf_{n\geq k} f_n \leq \inf_{n\geq k+1} f_n$$ in other words, $\{\inf_{n\geq k} f_n\}_k$ is a non-decreasing sequence of non-negative functions.

We see that $\inf_{n\geq k}f_n \leq f_n$ for all $n \geq k$. So, \begin{align*} \int \inf_{n\geq k}f_n &\leq \int f_n \ \forall n\geq k\\ &\leq\inf_{n\geq k}\int f_n \end{align*}

Since $\inf_{n\geq k}\int f_n$ is a non-decreasing sequence of (extended) real numbers, there is $\lim_{k \to +\infty}\inf_{n\geq k}\int f_n$ and we get, from $(1)$, $$\int \lim_{k\rightarrow \infty}\inf_{n\geq k}f_n = \lim_{k\rightarrow \infty}\int \inf_{n\geq k}f_n \leq \lim_{k \to +\infty}\inf_{n\geq k}\int f_n$$

$\endgroup$
1
$\begingroup$

Looks correct to me. You could add that the monotone convergence theorem can be applied because $$\inf_{n\geq k} f_n \leq \inf_{n\geq k+1} f_n$$

Also, your "indexing mistake" is that it should have been $\inf_{n\geq k} f_n \leq f_k$. And it's not "for all $n\geq k$", since $n$ is a variable internal to the $\inf$. It's "for all $k$".

$\endgroup$
  • $\begingroup$ I think I fixed my indexing mistake. I see what you mean by the MCT comment but isn't that obvious or should I put it in so that the casual observer would understand? $\endgroup$ – Wolfy Jun 23 '16 at 23:03
  • 2
    $\begingroup$ It might be obvious, but I have the habit of always mentioning the hypotheses of the theorems I use, however obvious they may be. At least for "Theorems" that are called so. But it's just a question of style, I think. $\endgroup$ – fonini Jun 23 '16 at 23:05
  • $\begingroup$ Ok, I will take your advice thanks $\endgroup$ – Wolfy Jun 23 '16 at 23:07
  • $\begingroup$ In the last two lines, it should have been $f_n$ instead of $f_k$, since you're inside the $\inf$, shouldn't it? Also, I don't see why these inequalities hold, now that I've looked at them again. $\endgroup$ – fonini Jun 23 '16 at 23:08
  • $\begingroup$ Why is $\int f_k\leq\inf_{n\geq k}\int f_n$? $\endgroup$ – fonini Jun 23 '16 at 23:09

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.