In a set of lecture notes that I'm reading, uniform convergence is defined as follows. A sequence $(f_n)_{n=1}^\infty$ converges uniformly to a function $f$, if $\sup_{x} |f_n(x)-f(x)| \to 0$, as $n \to \infty$. Further, it says that if the functions are continuous, then $\sup_{x} |f_n(x)-f(x)| \to 0$ as $n \to \infty$ can be written as $||f_n-f||_\infty \to 0$ as $n \to \infty$.
My question is why the assumption of continuity is added. Is the statement $\sup_{x} |f_n(x)-f(x)| \to 0$ not simply equivalent to $||f_n-f||_\infty \to 0$ ?
I think the answer is much simpler than the above comments seem to indicate and, in particular, the rather sophisticated definition of the norm on $L^\infty$ plays no role here.
The background, I suppose, is that the lecture notes the OP refers to might have already introduced the normed space $C(X)$, where $X$ is a compact topological space, or perhaps just $[0,1]$, with the norm $$\|f\|=\sup_{x\in X}|f(x)|.$$
However, it so happens that uniform convergence can be defined for functions defined on any set, for which one does not usually define a norm, in part because such functions might not be bounded.
Once the definition of uniform convergence is given in its most general form, one realizes that, for the special case of continuous functions defined on a compact set, it is possible to interpret the supremum found there in terms of the norm on $C(X)$.
That is all there is to it!