Help with Theorem 2.9.7 from Federer's Geometric Measure Theory

Question

Suppose $\phi$ and $\psi$ are Borel regular measures (outer measures) on a metric space $X$ such that $\phi(A),\psi(A)<\infty$ for every bounded subset $A\subseteq X$. One defines a Borel regular measure $$\psi_{\phi}(A)=\inf \{ \psi(B) \, ; \, B \, \text{is a Borel set and} \, \phi(A-B)=0 \}$$ whenever $A\subseteq X$. The first step in the proof of Theorem 2.9.7 is to show that if $A\subseteq X$ is $\phi$-measurable then it is also $\psi_{\phi}$-measurable. It's clear to me that $A$ is contained in a Borel set $B$ such that $\phi(B-A)=0$, but why does this imply $\psi_{\phi}(B-A)=0$ ? Any help would be appreciated.

Answer 1

If $\phi(B−A)=0$ then from the definition of $\psi_{\phi}$, you have: $$\psi_{\phi}(B−A)=\inf\{ψ(C)\big|ϕ((B−A)−C)=0\}$$ where $C$ is Borel.
But the empty set is a Borel set, and so if you take $C=\emptyset$ you have: $$\phi((B−A)−C)=\phi(B-A)=0$$ by assumption, and: $$\psi(C)=\psi(\emptyset)=0$$ Hence the infimum has to be less than or equal to $0$ and since it can't be negative you get: $\psi_{\phi}(B−A)=0$