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mona123

Invité

integration

hi can someone tell me please is it ok to use the difference  ''Thus, |F| = |G|−|G−F|'' written in the solution because i think it is wrong if we have (+∞)+(-∞)
this is the problem :
define the inner measure of E by |E|i = sup|F|, where the supremum
is taken over all closed subsets F of E. Show that |E|i  |E|e and if |E|e < 1, then E is
measurable if and only if |E|i = |E|e.
Solution.
Let E ⊂ Rn, let G be an open set containing E, and F ⊂ E closed. Since G is open, it is
measurable, thus |G| = |G\ F|+|G\ FC| = |F|+|G−F|. Thus, |F| = |G|−|G−F|, and
since Lebesgue measure is nonnegative, we clearly have |F|| ≤ |G|. Since |F| is a lower bound
for |G|, it is clearly less than or equal to the greatest lower bound, i.e., |F| < inf |G| = |E|e.
thanks.

Fred

Administrateur

Inscription : 26-09-2005

Messages : 7 352

Re : integration

Hello,

  I think that you should stop at [tex] |G|=|F|+|F-G| [/tex]
which is sufficient to deduce that [tex] |F|\leq |G| [/tex] even if [tex] |F-G|=+\infty [/tex].

F.