Fatou's Lemma and applications

Let $g: X \to [0,+\infty]$ be any simple function such that $g \leq \liminf_{n \to+\infty} f_n$, $\mu$.a.e. We write

where $\{E_i\}_i$ are disjoint and their union is $X$.

Step 1. The condition $g \leq \liminf f_n$ gives no direct pointwise comparison between $g$ and individual $f_n$. To obtain this, let $0 < t < 1$ be arbitrary, then one has

Since the sequence $\{\inf_{k \geq n} f_k\}_n$ is increasing to $\liminf_{n \to +\infty} f_n$, there exists $N_x > 0$ such that

Step 2. Define the set

Notice $B_{i,n} \subseteq B_{i,n+1}$ for all $i,n$. We will show that $\bigcup_{n=1}^\infty B_{i,n} = E_i$. Let $x \in E_i$, then $tg(x) = ta_i$, we choose $N_x$ as above and obtain

Thus $x \in B_{i,n}$ for all $n \geq N_x$. Therefore $E_i \subseteq \bigcup_{n=1}^\infty B_{i,n}$ and since $B_{i,n}$ is monotone by $n$, we obtain $\lim_{n\to+\infty} \mu(B_{i,n}) = \mu(E_i)$. The reverse side follows automatically since each $B_{i,n}$ is a subset of $E_i$.

Step 3. We already know that to prove $a \leq b$, when it's hard to find the relation of $a$ and $b$, we can instead prove $ca \leq b$ for all $0 < c < 1$ and let $c \to 1^-$. Following this trick, we start estimating from the right side

Since $B_{k,n}$ is increasing by $n$, we have

Calculating the right side of $(1)$

Taking $\liminf_{n \to +\infty}$ both sides from $(1)$

As $g$ is arbitrary simple function less than $\liminf_{n\to +\infty} f_n$ and $0 < t < 1$ is any number, taking $t \to 1^-$, and taking supremum over all such simple $g$, we obtain the Fatou's inequality

Since $\{f_n\}$ is increasing, we have

Taking integral both sides

Since the left side holds for all $k$ and since $\{\int f_n\}$ is also increasing by the monotone property of integral, we obtain

Applying Fatou's lemma and note that both $\{f_n\}$ and $\{\int f_n\}$ are increasing

Hence, the equality holds.

By Fatou's lemma, we have

Hence,

Let $(f_n) \subseteq L^p$ be Cauchy, then we can choose a subsequence $(f_{n_k})$ satisfying

Step 1. Let $g = \sum_{i=1}^\infty |f_{n_{i+1}} - f_{n_i}|$ be the dominating function. By Minkowski's inequality

We obtain $g \in L^p$, follows that $g < \infty$, $\mu$.a.e, the series $\sum_{i=1}^\infty (f_{n_{i+1}} - f_{n_i})$ absolutely converges, it thus converges, $\mu$.a.e pointwise. Thus we can write $f = \lim_{n\to+\infty} f_{n_k}$, where

is a convergent series, $\mu$.a.e.

Step 2. The function $f = \lim_{n\to+\infty} f_{n_k}$ is measurable. Applying the triangle inequality

Since the pointwise limit of measurable functions is measurable and finite, $\mu$.a.e, it follows that $\lim_{n\to+\infty} f_{n_k} = f < \infty$ pointwise, $\mu$.a.e, follows that $f$ is measurable.

Step 3. $f \in L^p$. We have the estimate

Note that $f_{n_k} \to f$ implies $|f_{n_k}|^p \to |f|^p$ and

Apply Fatou lemma, one has

Therefore $f \in L^p$.

Step 4. $f_n$ converges to $f$ in $L^p$. Since $f_{n_k} \to f$ $\mu$-a.e, we have

From $(*{})$,

Taking $k \to +\infty$ yields

So

Since $g, f_{n_1} \in L^p$, then $2^p(|f_{n_1}| + g)^p \in L^p$, implies that $2^p(|f_{n_1}| + g)^p \in L^1$. Apply the DCT theorem with the sequence $|f_{n_k} - f|^p \to 0$ bounded by the function $2^p(|f_{n_1}| + g)^p \in L^1$, we obtain

Since $(f_n)$ is Cauchy in $L^p$ and $f_{n_k} \to f$ in $L^p$, we conclude that $f_n \to f$ in $L^p$. Indeed, for any $\varepsilon > 0$, choose $N$ large enough such that

Choose $k$ large enough so that $n_k \geq N$ and $\|f_{n_k} - f\|_{L^p} < \frac{\varepsilon}{2}$. Then for all $n \geq N$,

Hence $\|f_n - f\|_{L^p} \to 0$.