Question

b and c please explian thx

Let 2 be a non-empty set. Let Fo be the collection of all subsets such that either A or AC is finite. (a) Show that Fo is a field. Define for E e Fo the set function P by ¡f E is finite, 0, if E is finite 1, if Ec is finite. P(h-10, (b) If is countably infinite, show P is finitely additive but not-additive. (c) If Ω is uncountable, show P is σ-additive on F.

i post the question from the book

Answer 1

(b) Suppose $\Omega$ is countably infinite. Suppose $\{E_i\}_{i=1}^n\subset \mathscript F_0$ , be a finite collection. Then to show $P(\bigcup_{i=1}^nE_i\)=\sum_{i=1}^nP(E_i)$ .

Case -1 Assume
PUE) =0 i=1
, then
UE
is finite then each $E_i$ is finite then for all i,
P(E,) = 0
for all i=1,...n. Hence
に!
, then we have $P(\bigcup_{i=1}^nE_i\)=\sum_{i=1}^nP(E_i)$ .

Case -2 Assume
PUE)-
, then
UE.) i-1
is finite, where $E_i's$ are pairwise disjoint. Then there exists (say) $E_1$, such that $(E_1)^c$ is finite. Now note that since $E_i's$ are pairwise disjoint, this gives us $E_i\subset (E_1)^c$ , for all
i > 2
, as if $E_i\not\subset E_1$, for some i>1, then $E_i\cap E_1\not=\phi$ , this is a contradiction hence each $E_i$ is finite for all $i\ge 2.$ Hence $\sum_{i=1}^nP(E_i)=P(E_1)=1$ , hence $P(\bigcup_{i=1}^nE_i\)=\sum_{i=1}^nP(E_i)$ .

Not $\sigma-$additive part. Since $\Omega$ is countable we can take $\Omega=\{e_1,e_2,...\}$ . Consider $E_i=\{e_i\}$, then for all i, $P(E_i)=0$, now note that $\bigcup_{i=1}^{\infty} E_i=\Omega=> P(\bigcup_{i=1}^{\infty} E_i)=1$ , which does not match with $\sum_{i=1}^{\infty}P(E_i)$ . Hence not $\sigma-$additive.

(c) Assume $\Omega$ is uncountable.

Case -1 Assume
PUE)-0 i-1
, then each $\bigcup_{i=1}^{\infty}E_i$ is finite then each $E_i$ 's are finite. Then we have $P(E_i)=0$, for all i, hence $\sum_{i=1}^{\infty}P(E_i)=0$ , hence $P(\bigcup_{i=1}^{\infty}E_i\)=\sum_{i=1}^{\infty}P(E_i)$ .

Case-2 Assume $P(\bigcup_{i=1}^{\infty}E_i\)=1$ , as there are only countably many $E_i's$, this can not happen each $E_i$ is finite but the union is countable infinite. Hence there exists say $E_1$ such that $E_1$ is uncountable and since $E_i \in \mathscript{F_0}$ , $(E_1)^c$ is finite. Then by the same argument note that $E_i\subset (E_1)^c$ , for all
i > 2
. , then note that there will exists only finitely many $E_i's$ as $(E_1)^c$ is finite. Then we have for all $i\ge 2.$ Hence $\sum_{i=1}^{\infty}P(E_i)=P(E_1)=1$ . Hence $P(\bigcup_{i=1}^{\infty}E_i\)=\sum_{i=1}^{\infty}P(E_i)$ .

Feel free to comment if you have any doubts. Cheers!