# $\sum_{A\in2^\Omega}P(A)=2^{|\Omega|-1}$ for probability space $(\Omega,2^\Omega,P)$ with finite $\Omega$

I'm looking for a combinatorial argument to complete a proof (below) of the following:

Claim: If $$(\Omega,2^\Omega,P)$$ is a probability space with finite $$\Omega,$$ then $$\sum_{A\in2^\Omega}P(A)=2^{|\Omega|-1}.$$

In other words, if $$\Omega$$ is finite and every subset of $$\Omega$$ is considered an event, then the sum of all the event-probabilities must equal $$2^{|\Omega|-1}.$$

Proof: Let $$\Omega=\{\omega_1,...,\omega_n\}$$ and $$p_i=P(\{\omega_i\}).$$ Then, by summing over subsets of successively larger size,
\begin{align*}&\sum_{A\in\cal 2^\Omega}P(A)\\ &=P(\emptyset)+\sum_{1\le i_1\le n}P\{\omega_{i_1}\}+\sum_{1\le i_1

The step marked $$\overset{(*)}{=}$$ would be justified by showing that in every sum $$\sum_{1\le i_1<... for $$t=1,...,n,$$ each of the $$p_i (i=1,...,n)$$ appears exactly $$\binom{n-1}{t-1}$$ times (noting of course that the sum of the $$p_i (i=1,...,n)$$ is $$1$$).

For example, if $$n=4$$ then for $$t=2$$ we have

$$\sum_{1\le i_1

as we see that each $$p_i$$ appears $$\binom{4-1}{2-1}=3$$ times.

Can anyone provide insight as to why this is generally the case? (Or perhaps give an alternative method of proof?)

EDIT: As mentioned in comments, the accepted answer provides a method that proves the following much more general result:

If $$(\Omega,\mathcal F,P)$$ is a probability space with finite $$\sigma$$-field $$\mathcal F$$, then $$\sum_{A\in\mathcal F}P(A)={1\over 2}|\mathcal F|.$$

I.e., "the sum of all the event-probabilities must equal half the number of events".

I think it's simpler to use that $$P(A) + P(\overline A) = 1$$. Note that $$2^\Omega = \{A | A \subseteq \Omega\} = \{\overline A | A \subseteq \Omega\}$$. So, we have
$$2 \cdot \sum_{A \in 2^\Omega} P(A) = \sum_{A \in 2^\Omega} P(A) + \sum_{A \in 2^\Omega} P(\overline A) = \sum_{A \in 2^\Omega} P(A) + P(\overline A) = \sum_{A \in 2^\Omega} 1 = 2^{|\Omega|}$$
• Very neat! If I'm not mistaken, we can do similarly even when the event-field $\cal F\ne 2^\Omega$; viz., $\mathcal F=\{A|A\in\mathcal F\}=\{\overline A|A\in\mathcal F\},$ so $2 \cdot \sum_{A \in \mathcal F} P(A) = \sum_{A \in \mathcal F} P(A) + \sum_{A \in \mathcal F} P(\overline A) = \sum_{A \in \mathcal F} 1 = |\mathcal F|.$ And I notice that the title claim still holds when $\Omega$ is not finite, as "half of infinity is infinity". In other words, the claim holds for all probability spaces $(\Omega,\mathcal F,P)$!? May 25 at 0:03
Here is a proof by switching the order of summation. First, write $$\sum_{A\subseteq\Omega } P(A) =\sum_{A\subseteq \Omega}\sum_{\omega\in A}P(\omega)$$ We can think of $$\sum_{A\subseteq \Omega}\sum_{\omega\in A}P(\omega)$$ as the sum of $$P(\omega)$$ over all ordered pairs $$(\omega, A)$$ such that $$\omega\in A$$. Switching the order of summation, we think of this as $$\sum_{\omega\in \Omega}\sum_{A\ni \omega}P(\omega)$$, where the inner summation ranges over all events $$A$$ which contain $$\omega$$. Since the summand $$P(\omega)$$ in the inner sum does depend on the summation index $$A$$, you can pull it out, resulting in $$\sum_{A\subseteq \Omega}P(\omega)\sum_{A\ni\omega}1$$ To compute this inner summation, you just need to count the number of events containing $$\omega$$. This number is obviously $$2^{n-1}$$, and the rest is easy.