So far my experience has been that in algebra classes the well-definedness of a function has been stressed in most of the exercises while in analysis/measure theory there are usually so many details to talk about that well-definedness is not usually considered that much. Therefore I was wondering what would be a diabolic example of a mapping $m$ which satisfies all properties of a measure/outer-measure except the well-definedness.
That is, if you were given the task: "Prove or disprove: $m$ is a measure/an outer-measure on the set $X$", then you could show that 1.) $m:2^X\to [0, \infty]$, 2.) $m(\varnothing) = 0$, 3.) $A\subset B \subset X\implies m(A)\leq m(B)$ and 4.) $m\left(\bigcup_{i=1}^\infty A_i\right)\leq \sum_{i=1}^\infty m\left(A_i\right)$ but for some $A = B \subset X: m(A) \neq m(B)$ making it impossible for $m$ to be a measure as it is not a function.