Subset of non-decreasing functions in $L^1[0,1]$ is compact? I'm interested in the subset of functions
\begin{equation}
A := \{f:[0,1]\rightarrow [0,1] | \text{$f$ is non-decreasing}\} \subset L^1[0,1].
\end{equation}
Do we know if $A$ is compact? Since $L^1[0,1]$ is a Banach space, I tried to use the result in one of answers in Examples of compact sets that are infinite dimensional and not bounded, namely:
$A$ is compact iff,
(1) $A$ is closed and bounded.
(2) For each $\varepsilon > 0$ there exists a finite dimensional subset $F \subset L^1[0,1]$ such that $d(f, F) < \varepsilon$ for all $f \in A$.
I think (2) should hold since any non-decreasing functions are Riemann integrable and so I can approximate any $f \in A$ by points in a space of rectangular functions (essentially Riemann sum approximation of the Riemann integral) to any $\varepsilon$ accuracy.
For (1) I think it is clear that $A$ is bounded (in a unit ball). But do we know if $A$ is closed? In other words, if we have a sequence $\{f_i\} \in A$ such that $f_i\rightarrow_{L^1} f$, do we know if $f \in A$?
I think I can see this if $L^1$ convergence on $A$ implies point-wise convergence, but I'm also not sure if this is true.
 A: It is true that $A$ is closed, if you include "almost everywhere" in its definition (otherwise, it doesn't even make sense as a subset of $L^1$). Because in that case if $f$ is in the closure of $A$, then $f$ is a limit of non-decreasing functions. As $L^1$-convergence implies almost everywhere pointwise convergence of a subsequence, $f$ is a pointwise almost everywhere limit of non-decreasing functions, and hence almost everywhere non-decreasing.
Also, as defined $A$ cannot be compact because it is not bounded. But even if you consider a "bounded version" like
$$
A=\{f\in L^1[0,1]:\ \|f\|_1\leq1\ \text{ and } f\ \text{ non-decreasin a.e.}\},
$$
it is still not compact.
the set of non-decreasing functions, it cannot be compact.
Let
$$
f_n=2^n\,1_{\big[1-2^{-n},1]}.
$$
Then, for each $n$,  $f_n$ is non-decreasing and $\|f_n\|_1=1$. And if we take $m>n$, then
\begin{align}
\|f_m-f_n\|_1
&=\int_0^1|2^m\,1_{[1-2^{-m},1]}-2^n\,1_{[1-2^{-n},1]}|\\[0.3cm]
&=\int_0^1|2^n\,1_{[1-2^{-n},1-2^{-m}]}+(2^m-2^n)\,1_{[1-2^{-m},1]}|\\[0.3cm]
&=2^n(2^{-n}-2^{-m})+(2^m-2^n)\,2^{-m}\\[0.3cm]
&=2-2^{n-m+1}\geq1
\end{align}
This shows that $\{f_n\}$ does not admit a convergent subsequence, and hence $A$ is not compact.
