I am trying to understand the following equality involving probability measures in Wikipedia: ‖ μ − ν ‖ = | μ − ν | ( X ) = 2 sup { | μ ( A ) − ν ( A ) | : A ∈ Σ }where the total variation norm ‖ ⋅ ‖ and the total variation of a measure | ⋅ | are defined in the some articles. The article has been flagged for not citing sources or offering a proof.I have found a proof of a similar result for discrete probability distributions (1). But I have not been able to adapt it to continuous probability distributions, which is the result above.

Question

I am trying to understand the following equality involving probability measures in Wikipedia:  ‖  μ  −  ν  ‖  =      |    μ  −  ν      |    (  X  )  =  2  sup  {      |    μ  (  A  )  −  ν  (  A  )      |    :  A  ∈  Σ  }where the total variation norm   ‖  ⋅  ‖ and the total variation of a measure       |    ⋅      |   are defined in the some articles. The article has been flagged for not citing sources or offering a proof.I have found a proof of a similar result for discrete probability distributions (1). But I have not been able to adapt it to continuous probability distributions, which is the result above.

Valeria Wolfe · Accepted Answer

First we need to show   μ  −  ν  (  ⋅  ) is a (finite) signed measure. I am not sure if that is trivial or requires some clever trick. Then, we let   B be the measurable set   E  ∩      D    +   using the Hahn decomposition (      D    +    ⊔      D    −  ) of   X under   μ  −  ν s.t.   μ  −  ν is positive on       D    +  . It is then easy to prove that for all   A  ∈  Σ,   μ  (  A  )  −  ν  (  A  )  ≤  μ  (  B  )  −  ν  (  B  ) and   ≤  ν  (      B    c    )  −  μ  (      B    c    )  =  μ  (  B  )  −  ν  (  B  ). Then you go on to say that      sup          A      ∈      Σ            |    μ  (  A  )  −  ν  (  A  )      |    ≤  μ  (  B  )  −  ν  (  B  )and, since   B  ∈  Σ,      sup          A      ∈      Σ            |    μ  (  A  )  −  ν  (  A  )      |    ≥  μ  (  B  )  −  ν  (  B  )  .From this, it is possible to conclude                              sup                      A            ∈            Σ                                    |                μ        (        A        )        −        ν        (        A        )                  |                                    =        μ        (        B        )        −        ν        (        B        )                                          =                  1          2                (        μ        (        B        )        −        ν        (        B        )        +        ν        (                  B          c                )        −        μ        (                  B          c                )        )                                          =                  1          2                (        μ        (        A        ∪                  D          +                )        −        ν        (        A        ∪                  D          +                )        +        ν        (        A        ∪                  D          −                )        −        μ        (        A        ∪                  D          −                )        )                                          =                  1          2                (        (        μ        −        ν        )        (        A        ∪                  D          +                )        +        (        μ        −        ν        )        (        A        ∪                  D          −                )        )                                          =                  |                μ        −        ν                  |                (        A        )        .