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I am studying Measure Theory, using the Bartle's book "Elements of Integration and Lebesgue Measure" and I couldn't do the exercise 3.Q:

"If $\mu$ is a charge on $X$, let $\mathcal{v}$ be defined for $E \in X$ by $$ \mathcal{v}(E) = \sup \sum_{j=1}^n | \mu(A_j)|$$

Where the supremum is taken over all finite disjoint collections $\{ A_j\}$ in $X$ with $E=\cup_{j=1}^n A_j$. Show that $\mathcal{v}$ is a measure in $X$."

Ok, how can I prove that $\mathcal{v}$ is countably additive?

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  • $\begingroup$ What does charge mean? A countably additive signed measure? $\endgroup$
    – tomasz
    Commented Apr 17, 2013 at 18:09
  • $\begingroup$ Yes, exactly. Sorry, my first book in this area, I have thought that was a common designation. $\endgroup$
    – Kernel
    Commented Apr 18, 2013 at 0:30

1 Answer 1

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Hint: the collection $A_j$ need not be finite; you will get an equivalent definition if you use countable collections instead. Pick a countable collection of disjoint $E_j$ and try to show two inequalities $$\nu\left(\bigcup E_j\right)\leq/\geq\sum \nu(E_j)$$ instead of trying to show equality directly.

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  • $\begingroup$ Yes, I tried this, but using finite collections and that was a mess ... Are you sure the definition is equivalent? $\endgroup$
    – Kernel
    Commented Apr 18, 2013 at 0:36
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    $\begingroup$ @LeandroChiarini: Yes, you you can always extend the sequence to an infinite one. On the other hand, if you have an infinite sequence, then the sequence of sum of measures of finite subsequences will tend to the measure of the infinite sum, hence the result. $\endgroup$
    – tomasz
    Commented Apr 18, 2013 at 2:21

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