Relationship between radon-nikodym derivatives and total variation distance

Question

Let $(\mathcal{X},\mathcal{A})$ be a measure space on which we have defined two probability measures $P$ and $Q$. Let $f$ and $g$ denote Radon-Nikodym derivatives of $P$ and $Q$ with respect to a $\sigma$-finite dominating measure $\mu$. Define the total variation distance between $P$ and $Q$ as $$\text{TV}(P,Q)=\sup_{A\in\mathcal{A}}|P(A)-Q(A)|.$$I asked another question on the relationship between these derivatives $f$ and $g$ and the distance $\text{TV}(P,Q)$ here. I am reading some notes which claim that$$\text{TV}^2(P,Q)=\frac{1}{2}\int_\mathcal{X}|f(x)-g(x)|d\mu(x)$$ seemingly without any further justification. I would greatly appreciate someone telling me why this is the case, as even having asked the seperate question above, I can't see this as a consequence.

Answer 1

Are you sure there should be a square on the $\text{TV}(P,Q)$ term? Or maybe $\text{TV}^2(P,Q)$ means something else?

From your previous question you already have that $$ \text{TV}(P,Q)= P(A)-Q(A)= \int_A |f(x)-g(x)|\, d\mu(x), $$ where $A= \{ x: f(x)-g(x)>0\}$.

By the same arguments you can get $$ \text{TV}(P,Q)= Q(A^c)-P(A^c)= \int_{A^c} |g(x)-f(x)|\, d\mu(x). $$ Summing these up we get $$ 2\text{TV}(P,Q) = \int_{\mathcal{X}} |f(x)-g(x)|\, d\mu(x). $$