Fatou Theorem what is lim(inf?

Question

We have the following theorem let fn converging to f almost everywhere on measurable A fn, n=1,2.... are integrable and nonnegative and integral of all fn less equal to K.Then f is integrable on A and integral of f on A is less equal K and integral of f over A is less equal lim(inf(integral fn on A)). The problem is that I have no idea what lim(inf means. fn converge to f so fn have one unique limit namely f

Answer 1

For a sequence of real numbers $(x_n)_{n\geq1}$, define $$u_n:=\inf_{k\geq n}x_k=\inf\{x_k\,|\,x\geq n\}.$$ Since $\{x_k\,|\,k\geq n\}\supseteq\{x_k\,|\,k\geq n+1\}$ it follows that the sequence $(u_n)_{n\geq1}$ is increasing (or if you like, non-decreasing). Hence it has a limit in $\bar{\mathbb{R}}=\mathbb{R}\cup\{-\infty,\infty\}$. Define $$\liminf_{n\to\infty}x_n:=\lim_{n\to\infty}u_n=\lim_{n\to\infty}\inf_{k\geq n}x_n.$$ For a sequence of real-valued functions $(f_n)$, we define $$(\liminf f_n)(x):=\liminf f_n(x).$$ Now suppose $f_n\ :\ A\longrightarrow[0,\infty]$ for each $n$. Fatou's Lemma states $$\int_A \liminf f_n\leq\liminf\int_A f_n$$ If each $\int_Af_n\leq K$ for each $n$ and $f_n\to f$ a.e. then this implies $$\int_Af\leq\liminf\int_Af_n\leq K<\infty$$ so $f$ is integrable. However, we do not know that the sequence $\left(\int_Af_n\right)_{n\geq1}$ has a limit, so this is the best we can do. As an example where this sequence indeed has no limit, consider $$f_n(x):=\begin{cases} 1&\text{if}\ x\in[n,n+1],\\ 2&\text{if}\ x\in[-n,-n+1]\ \text{and}\ n\ \text{is even},\\ 0&\text{otherwise.} \end{cases}$$