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Can someone prove the following?

Let $(X,\mathcal{F})$ be a measurable space and let $\mu$ and $\lambda$ be probability measures with domain $\mathcal{F}$. Define $\text{TV}$ via $\text{TV}(\mu,\lambda) = \sup_{A\in\mathcal{F}} |\mu(A)-\lambda(A)|$.

Let $\eta$ be an arbitrary finite positive measure on $(X,\mathcal{F})$ such that $\mu$ and $\lambda$ are absolutely continuous with respect to it. (For example, $\eta=\mu+\lambda$.) Define the Radon-Nikodym derivatives $f_\mu$ and $f_\lambda$ via $f_\mu = \frac{d \mu}{d\eta}$ and $f_\lambda = \frac{d \lambda}{d\eta}$. Then, regardless of the specific choice of $\eta$, $\text{TV}(\mu,\lambda) = \frac{1}{2}\int_X|f_\mu-f_\lambda|d\eta$.

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Answer: you can. Hint: TV$(\mu,\lambda)=\int_A(f_\mu-f_\lambda)\mathrm d\eta=\mu(A)-\lambda(A)$ with $A=[f_\mu\geqslant f_\lambda]$. – Did Mar 6 '12 at 18:43

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