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The definition of a probability measure $ P $ requires countable aditivity: $ P \left( \bigcup_{n = 1}^\infty A_n\right) = \sum_{n = 1}^\infty P (A_n)$ whenever $ A_1, A_2, \ldots $ is a sequence of disjoint sets in the relevant sigma algebra. I don't understand the proof of the countable finite aditivity property: $ P \left( \bigcup_{n = 1}^N A_n\right) = \sum_{n = 1}^N P (A_n)$. If someone could explain that to me that would be terrific.

Thanks

Michael Hardy
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Nicolás Leveroni
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    Just set $A_n=\emptyset$ for $n>N$. – ՃՃՃ May 28 '13 at 02:38
  • Thanks. I think you are right. But I don't understand why that proves the general case? Shouldn't it also be proven for sequences of including the ones where $ A_{n} \ne \emptyset $ for $ n > N $? Isn't it true for all finite sequences of $ A_{n} $? I'm confused because it seems to me that using $ A_{n} = \emptyset $ for $ n > N $ just proves it for a specific case. – Nicolás Leveroni May 28 '13 at 02:58
  • I'm afraid I don't get what you mean. You're given a finite number of disjoint sets, say, $A_1,\cdots ,A_N$, and you want to prove $P(\bigcup_{n=1}^NA_n)=\sum_{n=1}^N P(A_n)$. So you construct a sequence $(B_n)_{n\in\Bbb N}$ setting $B_k=A_k$ for $1\le k\le N$ and $B_k=\emptyset$ for $k>N$. Then you apply the countable aditivity property to $B_1,B_2,\cdots$. What do you get? – ՃՃՃ May 28 '13 at 03:12
  • Since countable additivity is a definition; there's no 'general case' to prove; countable additivity is an axiom. – Quinn Culver May 28 '13 at 03:13
  • Thanks again. I'm sure you are correct. I just don't understand why it's ok to set a sequence in which some $ B_{k} = \emptyset $ as proof. As I understand it, the property is supposed to be true for any sequence. This seems (to me) equivalent to proving something using a special case ($ B_{k} = \emptyset $ for $ k > N $). Again, I think you are correct but I don't get it. I hope I'm being clear on why. – Nicolás Leveroni May 28 '13 at 03:21
  • No, I don't understand where/why you're confused. See my answer and then please try again to explain what you don't understand. – Quinn Culver May 28 '13 at 03:29
  • You are not proving only a special case for the statement you’re trying to prove (finite additivity). You’re using a special case of a known result (countable additivity) to prove the general case of finite additivity. – Steve Kass Oct 20 '14 at 04:18

2 Answers2

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The definition of countable additivity says that $P ( \bigcup_{n = 1}^{\infty} A_{n}) = \sum_{n = 1}^{\infty} P ( A_{n} )$ whenever $ A_{1}, A_{2}, \dots $ is a sequence of disjoint sets in the relevant sigma algebra.

Assuming countable addivity, let's prove finite additivity, which says that $P ( \bigcup_{n = 1}^{N} A_{n}) = \sum_{n = 1}^{N} P ( A_{n} )$ whenever $ A_{1}, A_{2}, \dots A_{N} $ is a sequence of disjoint sets in the relevant sigma algebra. Given $ A_{1}, A_{2}, \dots A_{N} $, we note that $$\bigcup_{n=1}^{N}A_{n} = A_{1} \cup A_{2} \cup \cdots \cup A_{N} = A_{1} \cup A_{2} \cup \cdots \cup A_{N} \cup \emptyset \cup \emptyset \cup \cdots.$$

The point of including all those empty sets in the union above is to obtain an infinite sequence of disjoint sets (since the empty set is disjoint from everything) so we can apply countable additivity. Doing so yields $$ \begin{align*} P (A_{1} \cup A_{2} \cup \cdots \cup A_{N}) &= P(A_{1} \cup A_{2} \cup \cdots \cup A_{N} \cup \emptyset \cup \emptyset \cup \cdots) \\ &= P(A_{1}) + P(A_{2}) + \cdots + P(A_{N}) + P(\emptyset) + P(\emptyset) + \cdots \\ &= P(A_{1}) + P(A_{2}) + \cdots + P(A_{N}) + 0 + 0 + \cdots \\ &= P(A_{1}) + P(A_{2}) + \cdots + P(A_{N}). \end{align*} $$

Since the sequence $A_{1}, A_{2}, \ldots, A_{N}$ was arbitary, we're done.

Quinn Culver
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    Right!! They key sentence to me is the last one. Because $ A_{1}, ..., A_{N} $ is arbitrary! and by adding the empty sets one can get to an infinite sequence of disjoint sets and the starting point was that ANY infinite sequence of disjoint sets can be used in the infinite countable additivity property therefore the proof is good for ANY finite disjoint sequence!! Thanks so much!!! – Nicolás Leveroni May 28 '13 at 03:41
  • Sorry for being 11 years late. But from my understanding of union for set theory, the union can be intuitively understood as a relation of acting on a set of sets. An example of this can be seen in Brian M. Scott's answer in this post. –  Jun 15 '24 at 02:47
  • But now, if a union is acting on a set of sets, and if sets cannot have duplicate elements (as many seem to agree), then it is really okay for us to simply add an infinite number of empty sets into a union like you've just shown? –  Jun 15 '24 at 02:49
  • @AdamLee123 You're right. That means that if one wants to represent union as a relation of sets of sets formally (as opposed to intuitively), one would need to make sure the sets are distinct. This can be done by replacing each $A_i$ by $(A_i, i)$ and then adjusting the probability measure accordingly. See? Agree? – Quinn Culver Jun 16 '24 at 18:39
  • @QuinnCulver Oh I see. I think I get it now. So, just to be clear, we should view $A_1,A_2,A_3\cdots$ as an actual sequence (meaning a function $\alpha :\mathbb{N}\rightarrow \mathcal{P}(\Omega):= (A_i){i\in\mathbb{N}}$. And so the union $\bigcup{n = 1}^{\infty} A_{n}$ is simply acting on the range of $(A_i){i\in\mathbb{N}}$ while the summation $\sum{n = 1}^{\infty} P ( A_{n})$ is acting on the sequence $(\mathcal{P}(A_i))_{i\in\mathbb{N}}$? –  Jun 17 '24 at 03:24
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lets say we want to apply finite additivity to only $N$ events (such that $N$ is finite). In this case too we can construct an infinite union using $A_n=∅$ for $n> N$ (and $n< \infty$) and apply countable union axiom on it. It will prove finite union.

Ramiro
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Manish Jalan
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    Welcome to math.SE. This question is quite old, and it already has a good answer that basically states what you wrote in a more precise way. Please, answer old posts only if you are adding something substantial to them, or put your energies to better use by answering more recent questions. –  Jan 15 '16 at 16:43