Real Analysis, 2.18 (Fatou's Lemma) Integration of Nonnegative functions

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.

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 $(1)$, $$\int \lim_{k\rightarrow \infty}\inf_{n\geq k}f_n = \lim_{k\rightarrow \infty}\int \inf_{n\geq k}f_n \leq \lim_{k \to +\infty}\inf_{n\geq k}\int f_n$$

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 $k$".

• 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? – Wolfy Jun 23 '16 at 23:03
• 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. – fonini Jun 23 '16 at 23:05
• Ok, I will take your advice thanks – Wolfy Jun 23 '16 at 23:07
• 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. – fonini Jun 23 '16 at 23:08
• Why is $\int f_k\leq\inf_{n\geq k}\int f_n$? – fonini Jun 23 '16 at 23:09