In measure theory i heard recently a statement in my class, which says that the set of all (positive) measures does not make a Banach space ( whereas the set of signed measures makes up a Banach space ).
I know that in order to start the proof you need first to define a norm. Is it arbitrary ? I must then prove it is a norm. Then i would need to prove that every Cauchy sequence of positive measures has a limit in the space, which means that is still a positive measure. Must one construct such a sequence ?
Can somebody give me some hints ?
Thanks for the comment.
For the set if signed measures (on a sigma-algebra $\mathcal F$ on a set $\Omega$) we can use the "total variation" norm.
$$ \|\sigma\|_{TV} = \sigma(P)-\sigma(N) $$ where $P$ is the positive set, $N$ is the negative set: $$ N \cup P = \Omega,\qquad N \cap P = \varnothing \\ \sigma(A) \ge 0\qquad\text{for all } A \subseteq P \\ \sigma(A) \le 0\qquad\text{for all } A \subseteq N $$ I leave it to you to show the space of signed measures is complete in this norm.