Let $(X,d)$ be a metric space, $\mathcal{B}$ the Borel $\sigma$-algebra on $X$, and $\mathcal{M}(X)$ the space of totally finite measures on $\mathcal{B}$. Let $\|\mu\|\_{TV}$ be the total variation norm on $\mathcal{M}(X)$ defined by $$\|\mu\|\_{TV} = \mu^+(X) + \mu^-(X) \label{0}\tag{0}$$ where $\mu^+$, $\mu^-$ is the Jordan-Hanh decomposition of $\mu$. Do we have the following properties (like in the case of probability measures):
$$
\|\mu\|\_{TV} = \sup \limits\_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace .\label{1}\tag{1}$$  

According to Bogachev p.177 vol. 1, \eqref{1} is not true if both measures $\mu^+$ and $\mu^-$ are nonzero, and only the following is valid:
$$
\|\mu\|\_{TV} \leq 2 \sup \limits\_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace \leq 2 \|\mu\|\_{TV}. \label{2}\tag{2}
$$
But I have seen ([here](https://math.stackexchange.com/questions/246412/prokhorov-metric-vs-total-variation-norm) for example) the total variation norm used as a metric (written $d\_{TV}$) on the space of probability measure, with the following definition
$$ d\_{TV}(P,Q):=\sup\limits\_{A\in \mathcal B}|P(A) - Q(A)|,\label{3}\tag{3}$$
which seems to me to contradict Bogachev. Is there something I am misunderstanding ?
$$
\|\mu\|\_{TV} = \frac{1}{2}\sup \limits\_{f \text{msb},\|f\|\_{\infty} \leq 1} \int f\ d\mu.\label{4}\tag{4}
$$
I have the same question here, does property \eqref{4} holds for $\mu \in \mathcal{M}(X)$ ?


In addition, do you know of any reference treating these questions for totally finite measure (apart from Bogachev) ?
Thanks !



Properties of the total variation norm on space of totally finite measure (from Bogachev)
(1) is certainly not true for general signed measures $\mu$. However, if we restrict to signed measures with $\mu(X)=0$, then it is true with a factor of $2$, i.e.
$$\|\mu\|\_{TV} = 2 \sup\_{A \in \mathcal{B}} |\mu(A)| \tag{\*}.$$
That is, in this special case, the leftmost inequality in (2) is attained.


For one inequality, let $X = B^+ \cup B^-$ be the Hahn decomposition for $\mu$. Note that $\|\mu\|\_{TV} = \mu(B^+) - \mu(B^-)$, while $\mu(X) = \mu(B^+) + \mu(B^-) = 0$ so that $\mu(B^+) = -\mu(B^-) = \frac{1}{2} \|\mu\|\_{TV}$. Hence taking $A = B^+$ shows the $\le$ inequality in (\*).


Conversely, for any $A \in \mathcal{B}$, the defining property of the Hahn decomposition implies $\mu(A \cap B^-) \le 0$ and $\mu(A^c \cap B^+) \ge 0$, and therefore we have $$\mu(A) = \mu(A \cap B^+) + \mu(A \cap B^-) \le \mu(A \cap B^+) \le \mu(B^+) = \frac{1}{2} \|\mu\|\_{TV}.$$
A similar argument shows $\mu(A) \ge -\frac{1}{2} \|\mu\|\_{TV}$, so that $|\mu(A)| \le \frac{1}{2} \|\mu\|\_{TV}$. This shows the $\ge$ inequality.


In particular, taking $\mu = P-Q$ where $P,Q$ are both probability measures, we see that $d\_{TV}(P,Q)$ as defined by (3) is exactly half of $\|P-Q\|\_{TV}$. So the definitions are the same, up to a constant factor of 2.




---


Your equation (4) is also off by a factor of 1/2. The identity
$$\|\mu\|\_{TV} = \sup\_{\|f\|\_\infty \le 1} \int f\,d\mu$$
is true for every signed measure. To see one direction, write
$$\int f\,d\mu = \int f\,d\mu^+ - \int f\,d\mu^- \le \mu^+(X) + \mu^-(X) = \|\mu\|\_{TV}.$$
For the opposite inequality, take $f = 1\_{B^+} - 1\_{B^-}$.


