I feel like this should be an easy problem, but I've been banging my head against a wall on this.
Let $X$ be a non-empty set and $\mu^*\colon 2^X\to \left[0,\infty\right]$ an outer measure. Assume that $A\subset B\subset X$ with $\mu^*(A)=\mu^*(B)<\infty$. Then $\mu^*(B\setminus A)=0$.
Is that true? Without the assumption $\mu^*(A)=\mu^*(B)<\infty$ this is easily shown to be false, but the counterexample relies on the fact that $\infty-\infty$ is undefined. In this case, however, that is no longer possible. I've tried to prove it without getting really anywhere. Any help is appreciated.
On my notation: $X\subset Y$ includes the case $X=Y$.
This is false even with the finiteness assumption. For example, take $X=\{a,b\}$, and set $\mu^*(\emptyset)=0, \mu^*(\{a\})=\mu^*(\{b\})=\mu^*(X)=1$: then $\mu^*$ is an outer measure on $X$, but $\mu^*(\{a\})=\mu^*(X)$ and $\mu^*(X\setminus \{a\})=1$.