I'm asked to show that $$((C[0,1], \mathbb{R}), ||f||_1), \quad ||f||_1 = \int_0^1{|f(x)|dx}$$ is not a Banach space.
$||f||_1$ is a norm on this space so therefore I need to show that the space is not complete.
How do I prove completeness / incompleteness on a space of functions? How do I even construct a cauchy sequence of functions? What does it mean for a sequence of functions to converge? Is completeness defined in another way?

Given a norm $||\cdot||$, you can define a metric $d$ on the space as $d(x,y) := ||x-y||$.
Here, the norm is given by $||f||_{1} = \int_0^1 |f(x)| dx$. Thus, the metric is $d(f,g) = \int_0^1 |f(x)-g(x)| dx$.
A sequence of functions $\{f_n\}$ is said to be cauchy, if for every $\epsilon>0$, there exists $N \in \mathbb{N}$ such that $d(f_n,f_m) = \int_0^1 |f_n(x) - f_m(x)| < \epsilon$ for all $n,m \geq N$.
A sequence of functions $\{f_n\}$ is said to converge to $f$, if for every $\epsilon>0$, there exists $N \in \mathbb{N}$ such that $d(f_n,f) = \int_0^1 |f_n(x) - f(x)| < \epsilon$ for all $n \geq N$.
$(C[0,1], \mathbb{R}), || \cdot||_1)$ is not Banach by the example in the answer above.