Let ν be a signed measure and | ν | be its total variation. Then does ν finite ⟹ | ν | finite? If so, why?I can almost see this by appealing to the Jordan decomposition of ν and the definition of | ν | , ν = ν + − ν − | ν | = ν + + ν − but I am not completely sure that there can't be a measurable set A such that ν ( A ) = 0 but ν + ( A ) = ν − ( A ) = ∞.For instance, suppose our measurable space is ( R , B ( R ) ), where B ( R ) denotes the Borel subsets of R . Let ν + , ν − map a set to the number of integers it contains. Then ν + , ν − seem to be measures, and ν would seem to be a signed measure. And if so, the implication of interest fails.

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Let   ν be a signed measure and       |    ν      |   be its total variation. Then does   ν   finite     ⟹        |    ν      |     finite? If so, why?I can almost see this by appealing to the Jordan decomposition of   ν and the definition of       |    ν      |  ,                    ν        =                  ν          +                −                  ν          −                                              |                ν                  |                =                  ν          +                +                  ν          −                    but I am not completely sure that there can&#039;t be a measurable set   A such that   ν  (  A  )  =  0 but       ν    +    (  A  )  =      ν    −    (  A  )  =  ∞.For instance, suppose our measurable space is   (      R    ,      B    (      R    )  ), where       B    (      R    ) denotes the Borel subsets of       R  . Let       ν    +    ,      ν    −   map a set to the number of integers it contains. Then       ν    +    ,      ν    −   seem to be measures, and   ν would seem to be a signed measure. And if so, the implication of interest fails.

Jamal Hamilton · Accepted Answer

First of all: An expression of the form &quot;  ∞  −  ∞&quot; is not well defined. So for a decomposition to be well defined such an expressions may not occur (that&#039;s why your example won&#039;t work because your so constructed   ν is not well-defined)Then: Yes, your implication holds, moreover for a signed measure   ν it&#039;s true that  ν   finite    ⟺        |    ν      |     finiteTo see that consider Hahn&#039;s decomposition theorem which states the existence of a positive set   P and a disjoint negative set   N s.t.   Ω  =  P  ∪  N.From this it follows that      ν    +    (  A  )  =  ν  (  A  ∩  P  )  ,      ν          −        (  A  )  =  ν  (  A  ∩  N  )holds for all measurable sets   A and it&#039;s easy to see that this implies the equivalence above.

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