In $\mathbb{R}$, define $\lambda(U):=\sum (b_k-a_k)$ where $U\subset\mathbb{R}$ is an open set and $U=\cup I_k$, and each $I_k=(a_k, b_k)$ is an open interval.
Now define a set function $\mu(A):=\inf\{\lambda(U):A\subset U, U\subset\mathbb{R}:open\}$.
I can feel this definition measures the size of a set $A$ in terms of the "size" of an open set $U$ closest to the set $A$, but I'm uncertain whether the infimum exists or not.
Since the sequence of numbers $\lambda(U)$ are decreasing as $U$ approaches to $A$, if the lower bound exists, then we can tell the limit exists.
Since $\lambda:O(\mathbb{R})\to [0,+\infty]$, where $O(\mathbb{R})$ is the set of open sets of $\mathbb{R}$, we have the lower bound $0$. but I don't feel this fact is sufficient.
Can you explain me this rigorously?
It comes from the fact that every non-empty subset of $[0,\infty]$ has an infimum.
If every $U$ containing $A$ has infinite measure, then this subset will contain only infinity, and by definition the infimum is infinity.
If there is some $U$ with finite measure that contains $A$, then you can disregard those with infinite measure, so we are talking about the infimum of a non-empty subset of $[0,\infty)$. This infimum exists and is non-negative by the axioms that define $\mathbb R$ in the first place.