Properties of the total variation norm on space of totally finite measure (from Bogachev)

Question

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 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 !

Answer 1

(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 $f = 1_{B^+} - 1_{B^-}$.

Answer 2

Let me just give a simple counter-example to your equation (1): take the real line as space (any metric space with at least two points will do) and use the delta measures $\delta_1$ and $\delta_2$ at two distinct points. Then the real measure $\mu = \delta_1 - \delta_2$ will have total variation norm $2$ but $|\mu(A)| \le 1$ for all measurable subsets $A \subseteq \mathbb{R}$.

As a reliable reference, you can take a look at Rudin's book on Real and Complex Analysis.

For the second question: take $P$ and $Q$ be the two delta measures at different points. Then again $\mu = P - Q$ as above has total variation norm $2$. However, the supremum in your equation (3) is $1$, strictly smaller that $2$: if $A$ contains only one of the points (no matter which), then $|P(A) - Q(A)| = 1$. If $A$ contains none, we have $|P(A) - Q(A)| = 0$. If $A$ contains both points, both measures give $P(A) = 1 = Q(A)$, hence also no contribution to the sup.