---- > [!definition] Definition. ([[total variation measure]]) > Suppose $\nu$ is a [[complex measure]] on a [[σ-algebra|measurable space]] $(X, \Sigma)$. The **total variation measure** on $(X, \Sigma)$ is the [[measure]] $|\nu|: \Sigma \to [0, \infty]$ given by $\begin{align} E \mapsto \sup \left\{ \sum_{k=1}^{n} |\nu(E_{k})| : n \in \mathbb{N}, E_{k} \subset E \text{ and } E_{j} \cap E_{k}=\emptyset \text{ if } j \neq k \right\} \end{align}$ where the sets $E_{k}$ belong to $\Sigma$. ^definition > [!equivalence] Equivalence for real/signed measures. > For real measures, we can consider only $n=2$ in the definition of total variation measure. > Suppose $\nu$ is a [[signed measure|real measure]] on a [[σ-algebra|measurable space]] $(X, \Sigma)$ and $E \in \Sigma$. Then $|\nu|(E)=\sup \{ |\nu(A) |+ |\nu(B)| : A, B \subset E \text{ and } A \cap B= \emptyset \}.$ > > > > [!proof]- Proof. > > If $E_{1},\dots,E_{n} \subset E$ are members of $\Sigma$, consider $A= \bigcup_{\{ k : \nu(E_{k})>0 \}}E_{k}$, $B=\bigcup_{\{ k : \nu(E_{k})<0 \}}$. Clearly $A \cap B=\emptyset$ and $A \sqcup B= E_{1} \sqcup \dots \sqcup E_{n}$. Hence $|\nu(A)|+ |\nu(B)|=\sum_{k=1}^{n} |\nu(E_{k})|$. > > [!basicexample] For $h \in \mathcal{L}^{1}(\mu)$, the total variation measure of $d\nu=h \, d\mu$ is the [[measure with a density|measure with density]] $|h|$ with respect to $\mu$. That is, $|\nu|(E)=\int _{E} |h| \, d\mu =\|h\|_{L^{1}(E)}.$ for all $E \in \Sigma$. > [!basicproperties] > - $|\nu(E)| \leq |\nu|(E)$ > > > [!proof]- Proof. > > It suffices to find one disjoint collection $E_{1},\dots ,E_{n}$ of subsets of $E$ satisfying $|\nu(E)| \leq \sum_{k=1}^{n} |\nu(E_{k})|$. Letting $n=1$ and $E_{1}=E$ does the trick. > > - $|\nu|(E)=\nu(E)$ if $\nu$ is a [[finite measure|finite (positive) measure]]. > > > [!proof]- Proof. > > The inequality $|\nu|(E) \geq \nu(E)$ is shown above. The reverse follows from additivity and monotonicity of $\nu$: given disjoint $E_{1},\dots,E_{n} \subset E$, one has $\nu(E) \geq \nu(E_{1} \sqcup \dots \sqcup E_{n})=\sum_{k=1}^{n}\nu(E_{k})$. Taking the [[supremum]] preserves this inequality. > > - $|\nu|(E)=0$ if only if $\nu(A)=0$ for every $A \in \Sigma$ satisfying $A \subset E$. > > > [!proof]- Proof. > > One direction is obvious. For the reverse, suppose $|\nu|(E)=0$. Then if $A \in \Sigma$ is such that $A \subset E$, applying the first property above together with monotonicity of the (positive) total variation measure gives $|\nu(A)| \leq |\nu|(A) \leq |\nu|(E)=0$, hence $\nu(A)=0$. > > > [!justification] > - [ ] bring over the justification that $|\nu |$ is indeed a measure ^justification ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```