$P(E_i) = 0, \forall i\in I \Leftrightarrow P(\bigcup_{i\in I} E_i) =0$ and $P(E_i) = 1, \forall i\in I \Leftrightarrow P(\bigcap_{i\in I} E_i) =1$ 
Prove that $P(E_i) = 0, \forall i\in I \Leftrightarrow P(\bigcup_{i\in I} E_i) =0$ and $P(E_i) = 1, \forall i\in I \Leftrightarrow P(\bigcap_{i\in I} E_i) =1$, where $I$ is a countable index set.

$P$ is a probability measure, and we have generalized Boole's inequality, to begin with:
$$0\le P\left(\bigcup_{i=1}^\infty E_i \right) \le \sum_{i=1}^\infty P(E_i)$$
Is the following proof okay?
$P(E_i) = 0$ for all $i\in I$ then $P(\cup_{i\in I}E_i)=0$ by Boole's inequality above. If $P(\cup_{i\in I}E_i)=0$ holds, then Boole's inequality is probably of no use. I tried a proof by contradiction. Let's suppose there exists $j\in I$ such that $P(E_j) \neq 0$. Then $P(\cup_{i\in I}E_i)=0$ is definitely absurd - but how do I put this in mathematical terms? $(\star \star \star)$
In the other case, we have $P(E_i^c) = 0$ for all $i\in I$, which tells us (with the help of the first part) that $$P\left(\bigcup_{i\in I}E_i^c\right)=0 \implies P\left(\bigcap_{i\in I}E_i\right)^c=0 \implies P\left(\bigcap_{i\in I}E_i\right) = 1$$ and the proof is complete.
If everything above sounds fine, I only need help with the line marked $(\star \star \star)$. Thanks!
 A: $P(A) \leq P(B)$ if $A \subseteq B$. [Because $P(B)=P(A)+P(B\setminus A)$]. Hence, $0\leq P(E_j) \leq P(\bigcup E_i)=0$ for each $j$.
Rest of your arguments are fine if $I$ is countable.
Counter-example when $I$ is uncountable: Let $P$ be Lebesgue measure on $(0,1)$. Let $E_x=\{x\}$ for all $x \in (0,1)$. Then $P(E_x)=0$ for ll $x$ but $P(\bigcup E_x)=1$.
A: For (1), note that for any $i,j \ P(A_i \cap A_j)=P(A_i|A_j)P(A_j)=0$ and in general, using product rule, $P(A_1 \cap A_2 \cap \ldots \cap A_n)=P(A_1)P(A_2|A_1) \ldots P(A_n|A_{n-1} A_{n-1} \ldots A_1)=0$, therefore
$$
P(\cup_{i=1}^{n}A_j) = \sum_j P(A_j) -\sum_{j <i}P(A_j \cap A_i) +\ldots+(-1)^{n}P(A_1 \cap A_2 \cap \ldots \cap A_n)
$$
The first sum are $0$ by the definition, the second by the first argument above, the remaining terms by the second argument above, so the limit of the expression is $0$.
For (2), note that $A$ \ $B \cup B= A\cup B$. Since they are disjoint, $P(A$ \ $B)+P(B)=P(A \cup B)$, and, obviously $P(A \cup B) \geq P(B)=1 \text{ so } P(A \cup B)=1$ and $P(A$ \ $B=0$). Also, from $P(A)=P(A \backslash B)+P(A \cap B)$ it follows that $P(A|B)=1=P(A)$, so A and B are pairwise independent. Using the same logic, $P(A_1 \cup A_2 \cup A_3)=P(A_1 \cap A_2 \cap A_3)=1=P(A_1)P(A_2)P(A_3)$. This is the first step in the inductive proof.
Assume by induction, $P(\cup_{n}A_j)=P(\cap_n A_j)=1$. Next step,
$$
P(\cup_{n+1}A_j) = P(\cup_n A_j)+P(A_{n+1}) -P(\cap_{n+1}A_j)= 2 - P(\cap_{n+1}A_j) 
$$
Hence $P(\cap_{n+1}A_j)=1$. Hence $A_j$ are independent, and P($\cap_n A_j)=1$.
