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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 $$\|\mu\|_{TV} = \mu^+(X) + \mu^-(X)$$$$\|\mu\|_{TV} = \mu^+(X) + \mu^-(X) \label{0}\tag{0}$$ 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: $$ \|\mu\|_{TV} \leq 2 \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace \leq 2 \|\mu\|_{TV}. $$$$ \|\mu\|_{TV} \leq 2 \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace \leq 2 \|\mu\|_{TV}. \label{2}\tag{2} $$ 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 $$ d_{TV}(P,Q):=\sup\limits_{A\in \mathcal B}|P(A) - Q(A)|,$$ which $$ d_{TV}(P,Q):=\sup\limits_{A\in \mathcal B}|P(A) - Q(A)|,\label{3}\tag{3}$$ which seems to me to contradict Bogachev. Is there something I am misunderstanding ? $$ \|\mu\|_{TV} = \frac{1}{2}\sup \limits_{f \text{msb},\|f\|_{\infty} \leq 1} \int f\ d\mu.\label{2}\tag{2} $$$$ \|\mu\|_{TV} = \frac{1}{2}\sup \limits_{f \text{msb},\|f\|_{\infty} \leq 1} \int f\ d\mu.\label{4}\tag{4} $$ I have the same question here, does property \eqref{24} 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 !

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 $$\|\mu\|_{TV} = \mu^+(X) + \mu^-(X)$$ 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: $$ \|\mu\|_{TV} \leq 2 \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace \leq 2 \|\mu\|_{TV}. $$ 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 $$ d_{TV}(P,Q):=\sup\limits_{A\in \mathcal B}|P(A) - Q(A)|,$$ which seems to me to contradict Bogachev. Is there something I am misunderstanding ? $$ \|\mu\|_{TV} = \frac{1}{2}\sup \limits_{f \text{msb},\|f\|_{\infty} \leq 1} \int f\ d\mu.\label{2}\tag{2} $$ I have the same question here, does property \eqref{2} 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 !

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 !

Added hyperlinks to formulas + minor Math Jaxing ($\|$ instead of $||$) + typo correction
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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}$$\|\mu\|_{TV}$ be the total variation norm on $\mathcal{M}(X)$ defined by $$||\mu||_{TV} = \mu^+(X) + \mu^-(X)$$$$\|\mu\|_{TV} = \mu^+(X) + \mu^-(X)$$ where $\mu^+$, $\mu^-$ is the Jordan-Hanh decomposition of $\mu$. Do we have the following properties (like in the case of probability measures):

(1) $||\mu||_{TV} = \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace $. $$ \|\mu\|_{TV} = \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace .\label{1}\tag{1}$$
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: $$||\mu||_{TV} \leq 2 \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace \leq 2 ||\mu||_{TV}.$$$$ \|\mu\|_{TV} \leq 2 \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace \leq 2 \|\mu\|_{TV}. $$ 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 $$ d_{TV}(P,Q):=\sup\limits_{A\in \mathcal B}|P(A) - Q(A)|,$$ which seems to me to contradict Bogachev. Is there something I am misunderstanding ?

(2)$||\mu||_{TV} = \frac{1}{2}\sup \limits_{f \text{msb},||f||_{\infty} \leq 1} \int f\ d\mu.$

I $$ \|\mu\|_{TV} = \frac{1}{2}\sup \limits_{f \text{msb},\|f\|_{\infty} \leq 1} \int f\ d\mu.\label{2}\tag{2} $$ I have the same question here, does property (\eqref{2)} holds for $\mu \in \mathcal{M}(X)$ ?

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

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 $$||\mu||_{TV} = \mu^+(X) + \mu^-(X)$$ where $\mu^+$, $\mu^-$ is the Jordan-Hanh decomposition of $\mu$. Do we have the following properties (like in the case of probability measures):

(1) $||\mu||_{TV} = \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace $.
According to Bogachev p.177 vol. 1, (1) is not true if both measures $\mu^+$ and $\mu^-$ are nonzero, and only the following is valid: $$||\mu||_{TV} \leq 2 \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace \leq 2 ||\mu||_{TV}.$$ 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 $$ d_{TV}(P,Q):=\sup\limits_{A\in \mathcal B}|P(A) - Q(A)|,$$ which seems to me to contradict Bogachev. Is there something I am misunderstanding ?

(2)$||\mu||_{TV} = \frac{1}{2}\sup \limits_{f \text{msb},||f||_{\infty} \leq 1} \int f\ d\mu.$

I have the same question here, does property (2) holds for $\mu \in \mathcal{M}(X)$ ?

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

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 $$\|\mu\|_{TV} = \mu^+(X) + \mu^-(X)$$ 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: $$ \|\mu\|_{TV} \leq 2 \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace \leq 2 \|\mu\|_{TV}. $$ 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 $$ d_{TV}(P,Q):=\sup\limits_{A\in \mathcal B}|P(A) - Q(A)|,$$ which seems to me to contradict Bogachev. Is there something I am misunderstanding ? $$ \|\mu\|_{TV} = \frac{1}{2}\sup \limits_{f \text{msb},\|f\|_{\infty} \leq 1} \int f\ d\mu.\label{2}\tag{2} $$ I have the same question here, does property \eqref{2} 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 !

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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 $$||\mu||_{TV} = \mu^+(X) + \mu^-(X)$$ where $\mu^+$, $\mu^-$ is the Jordan-Hanh decomposition of $\mu$. Do we have the following properties (like in the case of probability measures):

(1) $||\mu||_{TV} = \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace $.
According to Bogachev p.177 vol. 1, (1) is not true if both measures $\mu^+$ and $\mu^-$ are nonzero, and only the following is valid: $$||\mu||_{TV} \leq 2 \sup \limits_{A \in \mathcal{B}} \lbrace |\mu(A)| \rbrace \leq 2 ||\mu||_{TV}.$$ 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 $$ d_{TV}(P,Q):=\sup\limits_{A\in \mathcal B}|P(A) - Q(A)|,$$ which seems to me to contradict Bogachev. Is there something I am misunderstanding ?

(2)$||\mu||_{TV} = \frac{1}{2}\sup \limits_{f \text{msb},||f||_{\infty} \leq 1} \int f\ d\mu.$

I have the same question here, does property (2) holds for $\mu \in \mathcal{M}(X)$ ?

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