0

Let $(X,\chi,\mu)$ a measurable space, and let $\lambda$ a charge in $\chi$. Let $\lambda^{+}$ the positive variation, that is, if $X=P\cup N$ is the Hahn's decomposition of $X$, $\lambda^{+}(E)=\lambda(E\cap P)$ for each $E\in\chi.$

I need to prove that the values for $\lambda$ are bounded, and $\lambda^{+}=\sup\{\lambda (F) : F\subset E, F\in\chi \}$.

Any hints? I really don't know what to do.

Mateus Rocha
  • 2,646

1 Answers1

1

For any $F \subset E$ we have $\lambda (F)=\lambda (F\cap P)+\lambda (F\cap N) \leq \lambda (F\cap P)=\lambda^{+}(E)$. Hence $\sup \{\lambda(F) : F \subset E\} \leq \lambda^{+}(E)$. On the other hand $E\cap P \subset E$ and $\lambda (E\cap P)=\lambda^{+}(E)$ so we get the reverse inequality. :

Kavi Rama Murthy
  • 311,013
  • The "other hand" is confuse to me. Are you saying that, since $E\cap P \subset E$, so $\lambda(E\cap P)\leq \lambda (E)$? But is this valid for charges? That is a measure's property. – Mateus Rocha Oct 18 '19 at 13:08
  • 1
    I am just using the fact that supremum of a set is greater than not equal to any element of the set. – Kavi Rama Murthy Oct 18 '19 at 13:18