An exercise on compactness on $L^1$ If $g$ is a nonnegative measurable function on $[0,1]$, and let $K=\{f\in L^1([0,1]) :|f|\leq g \,\,\,\,  a.e.\}$, prove the following:
(a)$K$ is closed,
(b)If $K$ is compact then $g\in L^1$.
Part (a) is easy, any help with part (b)?
Actually the exercise has a part (c) also, which i initially thought it was trivial but it seems it has some work. Any suggestion is welcome!
(c)If $g=1$ prove that $K$ is not compact.
 A: Hint: Consider the functions $f_n = g\cdot \chi_{\{g\le n\}}.$

Added later: Since the problem has a part (c) now, let me give a hint for that: Let
$$f(x) = 0, x\in[-1,0), \,\,f(x) = 1, x\in[0,1).$$
Then extend $f$ to $\mathbb {R}$ by making it periodic with period $2.$ Consider the functions $f_n(x) = f(2^nx), x\in [0,1].$ (I see now Ian earlier had also suggested "oscillation" as the key to this one.)
A: Consider bounded cutoffs of $g$, i.e. $g_n(x)=\begin{cases} g(x) & \text{if } g(x) \in [0,n] \\ 0 &  \text{otherwise} \end{cases}$. These are all in $K$, and converge pointwise to $g$. By compactness there is an $L^1$-convergent subsequence whose limit is in $K$. Can you use this to argue that in fact $g \in K$, hence $g \in L^1$?
In part c, the basic recipes for making a sequence of $L^p$ functions with no $L^p$-convergent subsequence are:


*

*A sequence whose $L^p$ norms blow up monotonically. For example, $f_n(x)=n$. This one comes directly from the general theory of metric spaces, nothing special here.

*A sequence which converges pointwise to zero but has a fixed $L^p$ norm. To make this actually work, you must violate the hypotheses of the Vitali Convergence Theorem. There are basically only two ways to do this:


*

*You can make a sequence which blows up on small sets. For example, $n \chi_{[0,1/n]}$.

*You can make a sequence which moves a positive amount of "mass" to infinity. For example, $\chi_{[n,n+1]}$. This is only possible in an infinite measure space.


*A sequence which oscillates more and more rapidly as $n$ increases.


The first two won't work in your case. Play with the third one.
A: Suppose that $\int_0^1 g(t)\,{\rm d}t = \infty$. Now decompose $[0,1]$ into sets where $g\leqslant n$ What can you say about the the sequence of restrictions of $g$ to these sets? 
