1
$\begingroup$

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 !

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – leo monsaingeon Jun 28 at 10:41
  • $\begingroup$ 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. $\endgroup$ – Léo D Jun 28 at 11:38
  • $\begingroup$ 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. $\endgroup$ – Dieter Kadelka Jun 28 at 13:56
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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} ? $\endgroup$ – Léo D Jun 28 at 13:15
  • $\begingroup$ This seems to be the case, I added a few lines. $\endgroup$ – Stefan Waldmann Jun 28 at 13:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.