Probability disjunction: Sum probabilities greater than actual probability Is it true that $$P(A \lor B \lor C \lor ...) \leq P(A) + P(B) + P(C) + ...$$?
I saw the general form of the probability disjunction here: https://stats.stackexchange.com/questions/87533/whats-the-general-disjunction-rule-for-n-events , but the right-hand side of the formula contains positive terms beyond simply $P(A)$, $P(B)$, etc.
My attempt by induction:
$$P(A \lor B) = P(A) + P(B) - P(A \land B) \leq P(A) + P(B)$$
Assume now $P(\bigvee_{i=1}^n C_i) \leq \sum_{i=1}^nP(C_i)$
Then $$P(C_{n+1}\lor \bigvee_{i=1}^n C_i) = P(\bigvee_{i=1}^n C_i) + P(C_{n+1}) - P(C_{n+1}\land \bigvee_{i=1}^n C_i) \leq \sum_{i=1}^nP(C_i) + P(C_{n+1}) - P(C_{n+1}\land \bigvee_{i=1}^n C_i) \leq \sum_{i=1}^nP(C_i) + P(C_{n+1})$$
 A: Yes the statement is true. Let $(A_i)_{i\geq 1}$ be a collection of events. Then note that
$$
\begin{align}
P\left(\bigcup_{i=1}^\infty A_i\right)
&=P(A_1)+P(A_2\bar{A_1})+P(A_3\bar{A_2}\bar{A_1})+P(A_4\bar{A_3}\bar{A_2}\bar{A_1})+\dotsb\\
&\leq P(A_1)+P(A_2)+P(A_3)+P(A_4)+\dotsb
\end{align}
$$
where the first equality is by countable additivity and the second by the fact that $A\subseteq B\implies P(A)\leq P(B)$.
A: Your work is correct.
Perhaps an easier way to do it: choose $A=C_{n+1}$ and $B=\bigvee_{i=1}^n C_i$. The induction step then follows from the case with only $A$ and $B$.
A: Hint: Let $A_1,A_2 \ldots, A_r$ be a collection of events. Then
$$A_1 \lor A_2 \lor A_3 \lor \ldots \lor A_r = A'_1 \lor A'_2 \ldots \lor A'_r,$$,
where
$$A_1 \doteq A_1; \ A'_l \doteq A_l \setminus (A_1 \lor A_2 \ldots \lor A_{l-1}) \ \text{for each } l=1,2,\ldots,r $$
So
$$\mathbb{P}[A_1 \lor A_2 \lor A_3 \lor \ldots \lor A_r] = \mathbb{P}[A'_1 \lor A'_2 \ldots \lor A'_r].$$
And as the $A'_l$s are mutually disjoint,
$$ \mathbb{P}[A'_1 \lor A'_2 \ldots \lor A'_r] = \sum_{l=1}^r \mathbb{P}[A'_l]$$
Yielding 
$$\mathbb{P}[A_1 \lor A_2 \lor A_3 \lor \ldots \lor A_r] = \mathbb{P}[A'_1 \lor A'_2 \ldots \lor A'_r] = \sum_{l=1}^r \mathbb{P}[A'_l].$$
But note that $\mathbb{P}[A_l] \ge \mathbb{P}[A'_l]$ for each $l$, because $A'_l \subseteq A_l$. Can you finish from here.
