Question on relation between radon-nikodym derivatives and the total variation distance

Question

Let $(\mathcal{X},\mathcal{A})$ be a measure space on which we have defined two probability measures $P$ and $Q$. I am reading some notes online which makes the following jump without explanation in this setting: Let $f$ and $g$ denote Radon-Nikodym derivatives of $P$ and $Q$ with respect to a $\sigma$-finite dominating measure $\mu$ and then let $A:=\{x\in\mathcal{X}:f(x)>g(x)\}$. Then $P(A)-Q(A)=\text{TV}(P,Q)$, where $\text{TV}(P,Q)$ is the total variation distance $$\text{TV}(P,Q):=\sup_{B\in\mathcal{A}}|P(B)-Q(B)|.$$ So it appears that the event $A$ defined above maximises this distance $|P(B)-Q(B)|$, but I am not sure why this is - an explanation would be greatly appreciated.

Answer 1

Well $P(B)-Q(B)=\int_{B}\left(f(x)-g(x)\right)\mu(dx)$ is maximised by putting $B=A$ while $Q(B)-P(B)=\int_{B}\left(g(x)-f(x)\right)\mu(dx)$ is maximised by putting $B=A^{c}$. It comes out the same whichever you choose because $P(A)-Q(A)=Q(A^{c})-P(A^{c})$.