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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?

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@Julián Aguirre, so basically if I just show that this specific example leads to strict inequality, then I've achieved my goal? –  Angelo Christophell Nov 19 '12 at 16:08

3 Answers 3

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.

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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 $0$.

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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.

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