$\sum_{A\in2^\Omega}P(A)=2^{|\Omega|-1}$ 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,
$$\begin{align*}&\sum_{A\in\cal 2^\Omega}P(A)\\
&=P(\emptyset)+\sum_{1\le i_1\le n}P\{\omega_{i_1}\}+\sum_{1\le i_1<i_2\le n}P(\{\omega_{i_1},\omega_{i_2}\})+...+\sum_{1\le i_1<...<i_t\le n}P(\{\omega_{i_1},...,\omega_{i_t}\})+...+P(\Omega)\\[3ex]
&=\sum_{1\le i_1\le n}p_{i_1}+\sum_{1\le i_1<i_2\le n}(p_{i_1}+p_{i_2})+...+\sum_{1\le i_1<...<i_t\le n}(p_{i_1}+...+p_{i_t})+...+P(\Omega)\\[3ex]
&\overset{(*)}{=}\binom{n-1}{1-1}+\binom{n-1}{2-1}+...+\binom{n-1}{t-1}+...+\binom{n-1}{n-1}\\[3ex]
&=2^{n-1}
\end{align*}$$
The step marked $\overset{(*)}{=}$ would be justified by showing that in every sum $\sum_{1\le i_1<...<i_t\le n}(p_{i_1}+...+p_{i_t}),$ 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
$\sum_{1\le i_1<i_2\le 4}(p_{i_1}+p_{i_2})=(p_1+p_2)+(p_1+p_3)+(p_1+p_4)+(p_2+p_3)+(p_2+p_4)+(p_3+p_4)=3,$
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".
 A: 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.
A: 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
$$2 \cdot \sum_{A \in 2^\Omega} P(A) = \sum_{A \in 2^\Omega} P(A) + \sum_{A \in 2^\Omega} P(\overline A) = \sum_{A \in 2^\Omega} P(A) + P(\overline A) = \sum_{A \in 2^\Omega} 1 = 2^{|\Omega|}$$
