# Demonstrating Strict Inequality in Fatou's Lemma with Sequences of Functions

a) Use the sequence of functions $(f_{n})$ $f_{n}=\left\{\begin{matrix} \chi_{[0,1]}, \,n\,\,odd\\ \chi_{(1,2)},\,n\,even \end{matrix}\right.$
to establish that the inequality in Fatou's lemma can be strict.

b) Use the sequence of functions $(f_{n})$
$f_{n}=\frac{-1\chi_{[0,n]}}{n}$
to establish that the motto of Fatou is false in general for functions that are not nonnegative.

Recall that Fatou's lemma states that
$\int \liminf_{n\rightarrow \infty} f_n dx \leq \liminf_{n\rightarrow \infty} \int f_n dx.$

(a) Note that for the given sequence of functions
$\liminf_{n\rightarrow \infty} f_n =0.$
This is because for $x\in [0,1]$ we have $\lim_{k\rightarrow \infty} f_{2k}(x)=0$, and for $x\in (1,2)$ $\lim_{k\rightarrow \infty} f_{2k+1}=0$. So $\liminf_{n\rightarrow \infty} f_n =0$ on $[0,2)$. Note that for $x\in \mathbb{R}\backslash [0,2)$,  $f_n(x)$ is zero and hence the liminf is also zero.

On the other hande
$$\int f_n =\int_0^{1}1 dx=1, \text{if n is odd}$$
and
$$\int f_n =\int_1^{2}1 dx=1, \text{if n is even}.$$
So $\int f_n=1$ for all $n$. Thus
$\liminf_{n\rightarrow \infty}\int f_n dx =0.$
So we have
$0= \int \liminf_{n\rightarrow \infty} f_n dx < \liminf_{n\rightarrow \infty} \int f_n dx=1,$
so Fatou's lemma can be strict.

(b) Let $x\in \mathbb{R}$. Then
$\lim _{n \rightarrow \infty}|f_n(x)|\leq \lim _{n \rightarrow \infty} \frac{1}{n}=0.$
Hence
$\liminf_{n\rightarrow \infty} f_n (x)=\lim _{n \rightarrow \infty}f_n(x)=0,$
for all $x$.
On the other hand,
$\int f_n dx=\int_0^{n}-\frac{1}{n}dx=-1.$
So we have
$0= \int \liminf_{n\rightarrow \infty} f_n dx > \liminf_{n\rightarrow \infty} \int f_n dx=-1.$
This shows that Fatou is false in general for functions that are not nonnegative.

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