Proving simple statement from conditional probability

Question

QUESTION: if $A_1,A_2,...$ are mutually exclusive then $$P(A_1 \cup A_2 \cup ... | C ) = \sum_i P(A_i | C)$$

SO I figure I must use that $P(A_1 \cup A_2 \cup ...) = P(A_1) + P(A_2) + ...$ somewhere. but I am not sure how:

$$P(A_1 \cup A_2 \cup ... | C) = \dfrac{P(\bigcup_i A_i \cap C)}{P(C)}$$

$$ = \dfrac{P(A_1 \cap C \cup A_2 \cap C ...)}{P(C)}$$ but now I am stuck

Answer 1

It is a special case. Consider $C$ being the set of all events...

Answer 2

If the $A_i$ are mutually exclusive, then so are $A_i \cap C$. So continuing from your last step, $$\frac{P((A_1 \cap C) \cup (A_2 \cap C) \cup \cdots)}{P(C)} = \frac{\sum_i P(A_i \cap C)}{P(C)} =\sum_i P(A_i \mid C).$$