# Total variation distance and mutually singular probability measure

Let $\mu$ and $\nu$ be two probability measures on $(\Omega,F)$. Let $||\mu-\nu||$ denote the total variation distance between $\mu$ and $\nu$. Show that if $||\mu-\nu||=1$ then the support of $\mu$ and $\nu$ are disjoint.

I'm not familiar yet with measure theory. I tried to bound  $$||\mu-\nu||=\mu(A)-\nu(A),A=\{E \in F : \mu(E) \ge \nu(E)\}$$

with something that is less than $1$, but I didn't succeed.
I think, it doesn't matter, but if does, we may assume $\Omega$ is finite.

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Did you mean that $\mu,\nu$ are probability measures? – Ilya Oct 7 '11 at 19:29
This result holds for every pair of probability measures on a topological measurable space $(\Omega,F)$. – Did Oct 7 '11 at 20:35

It does matter if $\Omega$ is finite. I am not sure that you can talk about the support of a measure if $\Omega$ is not metric. So let us assume that $\Omega$ is finite and there is a discrete metric on it, then the support of $\mu$ is simply $\{\omega\in \Omega:\mu(\omega)>0\}$. Moreover, $$\mu(\operatorname{supp}\mu) = \nu(\operatorname{supp}\nu)=1.$$ Again, you should assume that $\mu,\nu$ are probability measure, otherwise your statement is incorrect.
So, $\mu(\Omega) = \nu(\Omega) = 1$ and $\|\mu-\nu\| = \sup\limits_{A\in \mathcal F}|\mu(A)-\nu(A)| =1$ where $\mathcal F = 2^\Omega$. Suppose that there is a point $\omega'$ such that $\mu(\omega')>0$ and $\nu(\omega')>0$, then for any set $A$ $$|\mu(A)-\nu(A)|<1-\min(\mu(\omega'),\nu(\omega)')$$ which contradicts with the assumption $\|\mu-\nu\|=1$.
Be aware of the following example $\mu$ is measure on $\mathbb R$ with Gaussian density and $\nu = \delta_0$ is concentrated at $\{0\}$. Then $\operatorname{supp}\mu = \mathbb R$ and $\operatorname{supp}\nu = \{0\}$ so $\operatorname{supp}\mu\cap\operatorname{supp}\nu = \{0\}$ and even $\nu(\{0\})=1$. However, still $\|\mu-\nu\|=1$.