There can't be $2$ different functions in the space of continuous and bounded functions

Question

How would one go about proving that in the space of continuous and bounded functions there do not exist $2$ different functions equal almost everywhere, meaning they can differ only in a set of nonzero measure?

Answer 1

$\{x:f(x) \neq g(x)\}$ is an open set. If it is not empty then there is an open interval $(a,b)$ contained in this open set. But then the measure of $\{x:f(x) \neq g(x)\} \geq b-a$ which contradicts the hypothesis. Hence $\{x:f(x) \neq g(x)\}$ is empty which means $f$ and $g$ are one and the same.

Answer 2

Hint:

If $f,g$ are different continuous functions then the function $x\mapsto|f(x)-g(x)|$ is continuous so that the set $\{x\mid |f(x)-g(x)|>0\}$ is a non-empty open set.