Theorem 9.13 An introduction to mathematical analysis

Question

Someone please explain me that "since $c_k$ is arbitary ,we can conclude that.."

I mean how we got $\sum_{k=1}^{n} \nu(A_k)\leq \nu(\cup_{k=1}^{n}A_k)$?

Thanks in advance!

Answer 1

I tried this way: $c_k < \nu(A_k)$ is arbitary.

I choose $c_k$ such that $\nu(A_k)=c_k + \epsilon$ $...(1)$,where $\epsilon >0$, $\epsilon \to 0$

It is shown in the proof,that $\sum_{k=1}^{n} c_k < \nu(\cup_{k=1}^{n}A_k)$

$\implies\sum_{k=1}^{n}(\nu(A_k)-\epsilon)<\nu(\cup_{k=1}^{n}A_k)$ $..from(1)$

$\implies \sum_{k=1}^{n}\nu(A_k) < \nu(\cup_{k=1}^{n}A_k) + \sum_{k=1}^{n}\epsilon$

$\implies \sum_{k=1}^{n}\nu(A_k) \leq \nu(\cup_{k=1}^{n}A_k)$ $(\because \epsilon \to 0,\implies \sum_{k=1}^{n}\epsilon \to 0) $ Please reply,am i correct?

Answer 2

You have to just take limit as $c_k$ increases to $\nu (A_k)$.