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

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 !

• There's no contradiction, since for probability measures $\mu^-=0$. Also, $\|P\|_{TV}=d_{TV}(P,0)$. The "problem" arises when BOTH $\mu^\pm$ are nonzero, as clearly illustrated by Stefan's counterexample. – leo monsaingeon Jun 28 at 10:41
• Thanks, if I understand: for $P$ and $Q$ probability measures, $d_{TV}(P,Q) \neq ||P-Q||_{TV}$ where $d_{TV}$ is defined in \eqref{3} and $||\cdot||_{TV}$ is defined in \eqref{0}. This is where my mistake comes from I think. – Léo D Jun 28 at 11:38
• I have difficulties to understand the problem. Let $\|\mu\| := \sup_{A \in \cal{B}} |\mu(A)$. Then as noted in (2) $\|,\|$ and $\|,\|_{TV}$ are equivalent norms. Further $\|\mu\| = \max\{\mu^+(X),\mu^-(X)\}$. Where is the problem? In (3) there is no problem, both sides define equivalent metrics. – Dieter Kadelka Jun 28 at 13:56

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

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.

• Thanks. I have the same question as for @leo monsaingeon comment. To be sure: does this mean that $d_{TV}(P,Q) \neq ||P−Q||_{TV}$ where $d_{TV}$ is defined in \eqref{3} and $||\cdot||_{TV}$ is defined in \eqref{0} ? – Léo D Jun 28 at 13:15
• This seems to be the case, I added a few lines. – Stefan Waldmann Jun 28 at 13:43