Fatou's Lemma Strengthened to Equality I'm trying to use an example to show that Fatou's lemma can not be strengthened to equality.  I was given a hint, which I'm not quite sure how to use.  I was told that if I look at the one-dimensional case, and let $f_k(x)=\begin{cases}
       k, &\quad\text{if } - \frac{1}{k} \leq x \leq \frac{1}{k}\\
       0, &\quad\text{elsewhere} \ 
     \end{cases}$ , then $\int f_kdm=\frac{2}{k}(k)=2, \forall k$, and $g_k(x) \to 0, \forall x$, except for $x=0$, for which $g_k \to \infty$.  How can I use this to show that equality can not be achieved?  I thought specific examples couldn't be used to prove general behaviors?  Can someone please help?
 A: You cannot use an specific example to show that a general statement is true. For this you need a "proof".
But you can use an specific example to show that a general statement is not true.
For instance, the example $f(x)=|x|$ shows that the statement "All continuous functions are differentiable" is not true. But the fact that $105=3\cdot5\cdot7$ does not prove that any $n\in\mathbb{N}$ can be decomposed into the product of primes.
A: You already have that
$$
\lim_{k\to\infty}\operatorname{inf}\int f_k\,dm=\lim_{k\to\infty}2=2.
$$
Now you must find
$$
\int \lim_{k\to\infty}\operatorname{inf}f_k\,dm.
$$
But, the limit function is $0$ off a set of measure $0$.
A: There is a nice exercise on Rudin, Real and Complex Analysis (Chapter 1) about this. 
Consider a measure space $(X,\mathcal A, \mu)$ and pick a measurable subset $E$. Then consider the sequence $(f_n)_{n\in\mathbb N}$ of real valued functions defined as $f_n=\chi_E$ if $n$ is even, $f_n=\chi_{X\setminus E} = 1-\chi_E$ 
otherwise. Then it is easy to prove that $\liminf_{n}f_n = 0$ though 
$$
\liminf_n \int_Xf_n d\mu = \min\{\mu(E), \mu(X\setminus E)\} > 0
$$
if both $\mu(E)>0$ and $\mu(X\setminus E)>0$.Hope this helps.  
