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

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.