# Compactness of a subset of $L_1([0,1])$

Suppose that $$X = L_1([0,1]):= \{f:[0,1]\mapsto [0,\infty): \int_0^1 |f(x)|~dx <\infty\}$$. Equip $$X$$ with the standard metric $$d(f,g) = \int_0^1 |f(x)-g(x)|~dx$$. Now, define: $$C:= \{f\in X: f(x) \in [0,1]~\forall x \in [0,1] \}~.$$ In other words, $$C$$ is the set of all non-negative, integrable functions on $$[0,1]$$, bounded above by $$1$$. My question is, is $$C$$ a compact subset of $$X$$?

I have seen examples showing that the unit $$L_1$$ ball is non-compact (for example Unit sphere in $L^p([0,1])$ is not compact.), but all these examples are constructing a sequence of unbounded functions on $$[0,1]$$. In other words, what I want, is a sequence of non-negative, bounded functions in $$X$$ that does not have a convergent subsequence. Any help will be greatly appreciated!

I recall seeing this in a Real Analysis course: Define the functions: $$f_n:[0,1]\to{}\{0,1\}$$ as $$f_n(x)=$$ the $$n$$th bit of the binary expression for $$x$$. You then have that $$\int_0^1|f_n-f_m|d\mu=\frac{1}{2}$$ (since $$f_n-f_m=1$$ precisely when the $$m$$th bit is $$0$$ and the $$n$$th is $$1$$ or vice versa) and thus no subsequence converges.
I believe you can also look at the functions generated by the mapping of Alspach (A fixed point free nonexpansive map,'' Dale Alspach, Proc. of the AMS, Vol. 82, No. 3 (Jul., 1981), pp. 423-424), which is an adaptation of the Baker transform (https://en.wikipedia.org/wiki/Baker%27s_map).