measure-theory
Fatou's lemma says that
if $f_n:X \rightarrow [0,\infty]$ are measurable,then
$$\liminf_{n\rightarrow \infty}\left(\int_X f_n \,\mathrm{d} \mu\right) \geq \int_X \liminf_{n\rightarrow \infty} f_n \,\mathrm{d}\mu$$
I like to know what this Lemma really says. That is, how can I express in words (rather informally) what this lemma actually says?
Here is one proof based on the bounded convergence theorem, adapted from Durrett.
Define $g_n(x) = \inf_{m\geq n} f_m(x)$. So, $f_n \geq g_n$ and $g_n \uparrow g(x) := \liminf_n f_n(x)$ as $n \to \infty$.
By monotonicity of the integral, we know that $\newcommand{\du}{\,\mathrm d \mu} \int f_n \du \geq \int g_n \du$, whence $$\liminf_n \int f_n \du \geq \liminf_n \int g_n \du \>.$$
Suppose $X_n \uparrow X$ where $\mu(X_n) < \infty$. By the bounded convergence theorem, for fixed $m$, we have $$ \liminf_n \int g_n \du \geq \int_{X_m} g_n \wedge m \du \to \int_{X_m} g \wedge m \du \>, $$ since the integrand in the middle is bounded and converges to the integrand on the right.
But, then $$ \liminf_n \int g_n \du \geq \sup_m \int_{X_m} g \wedge m \du = \int \liminf_n f_n \du \>. $$
Since $\liminf_n \int f_n \du \geq \liminf_n \int g_n \du$, we are done.
I like to think of the following pictures. The first two are $\int f_1$ and $\int f_2$ respectively, but even the smaller of these is larger than the area in the third picture, which is $\int \inf f_n$. Of course, Fatou's lemma is more subtle since we're talking about the limit infimum rather than just the minimum, but for the purpose of intuition this helps to make sure the inequalities go the right way.
Best Answer
$$ (0,1),\quad(1,0),\quad(0,1),\quad(1,0),\quad\ldots $$ The terms of a sequence are alternately $(0,1)$ and $(1,0)$. In either case, the sum of the two components of each pair is $1$, so the lim inf of the sum of the two is $1$. But the lim inf of the sequence of pairs is $(0,0)$, and the sum of the two components of that pair is $0$.
In other words, for every value of $x$, $f_n(x)$ may be small for infinitely many $n$, but the values of $n$ for which $f_n(a)$ is small are not the same ones for which $f_n(b)$ is small.