Show that $P$ is a probability measure on $(\Omega, F)$ [closed]

Conditions: Suppose $0 ≤ a_i ≤ 1$ such that $\sum_{i=1}^{\infty}a_i= 1$ and let $P_i$ be a collection of probability measures on $(\Omega, F)$.

Define a set function $P$ by

$$P(A) = \sum_{i=1}^{\infty}a_iP_i(A)\ ,$$
for $A\in F$. Show that $P$ is also a probability measure.

I understand that I will need to prove these 3 items:

i. $P(\Omega) = 1$.

ii. $P(A) \geq 0$ for any $A \in F$.

iii. If $\{A_k\}$ is a collection of pairwise disjoint elements of $F$, then $$P(\cup_{k=1}^{\infty}A_k) = \sum_{k=1}^{\infty} P(A_k)\ .$$

But how do I go about proving these 3?

Solution 1:

Assume that each $P_i$ is a probability measure on $(\Omega,F)$.

1. $P(\Omega)=\sum_{i=1}^{\infty}a_iP_i(\Omega)=\sum_{i=1}^{\infty}a_i = 1\ ,$ where we have used that $P_i(\Omega)=1$ for each $i$.

2. If $A\in F$, then $P_i(A)\geq 0$ for each $i$ and $a_iP_i(A)\geq 0$ since each $a_i\geq 0$. Thus

$$P(A) = \sum_{i=1}^{\infty} a_i P(A) \geq 0\ ,$$

as each term is nonnegative.

3. Suppose $\{A_k\}\subset F$ is a pairwise disjoint collection of sets. Then for each $i$, $$P_i(\cup_{k=1}^{\infty} A_k)=\sum_{k=1}^{\infty}P_i(A_k)$$ and hence

$$\begin{align}P(\cup_{k=1}^{\infty}A_k) &= \sum_{i=1}^{\infty}a_i P_i(\cup_{k=1}^{\infty} A_k)\\ &=\sum_{i=1}^{\infty}\sum_{k=1}^{\infty}a_i P_i(A_k)\\ &=\sum_{k=1}^{\infty}\sum_{i=1}^{\infty}a_i P_i(A_k)=\sum_{k=1}^{\infty}P(A_k) \end{align}$$

by Fubini's theorem.