Let $(X,M,\mu)$ be a some measure space (NOT necessarily finite) and denote $I := \{1,2,\ldots,n\}$. I'm having an incredible amount of trouble proving that for any collection of $n$ sets in $M$, the following equality holds:
$$\mu(\cup_{i=1}^n A_i) + \sum_{S \subset I,\\\\ |S| \text{ even}} \mu( \cap_{s \in S} A_s) = \sum_{S \subset I,\\|S| \text{ odd}} \mu(\cap_{s \in S} A_s)$$
I can't seem to find a way to write $\cup_{i=1}^n A_i$ as a disjoint union of sets so as to expand $\mu(\cup_{i=1}^n A_i)$ and cancel nicely with sum of measures of even intersection. I've only ever seen a proof of this identity under the assumption that $\mu$ is a finite measure. Am I going about this the wrong way? Any help would be appreciated.
For any $n>1$, everything is happening inside $X_0\,=\,\cup_{i=1}^n\,A_i.\,\,$ If $\infty \,=\,\mu X_0\,\leq\,\mu A_1 + \dots + \mu A_n$, your equality is easily proved. For in this case we must have some $\mu A_k\,=\,\infty,$ and this measure is one of the summands on the right. After that, you can assume $\infty \,>\,\mu X_0\,\geq\, \mu( \cap_{s \in S} A_s)$ for any non-empty subset $S$ of $I$. Since each sum appearing in the equality is a sum of a finite number of terms, it is an equality in the real numbers, and can be proved by induction as you see in books on probability. Such a proof usually introduces subtraction.
Maybe a different proof can be given using your idea that we should somehow (for each $n \geq 2$) write $A_1\,\cup\,A_2 \,\cup\dots \cup\,A_n$ as a finite disjoint union. This might be written:
$$\bigcup_{i\,\in\,n}\,A_i \,\,=\,\, \bigsqcup_{j\,\in\,m_n}\,C_j^n,$$
where $m_n = 2^n \,-\,1.$ You can check that $m_n\,=\,2m_{n-1}\,+\,1,$ which shows that the number of cells in the $n^{\text{th}}$ partition is "double plus one" the number in the previous partition. To see why, you should draw the usual Venn diagram for the $n=2$ case. Label the intersection $C_3^n.$ Label the cell $A_2 \setminus A_1$ as $C_1^n.$ The only remaining cell (the ${m_2}^\text{th}$) is then $C_2^n.$
As usual, you draw in a new circle for $A_3$ (going $n$ to $n+1$). Draw it. Notice what you did with your pencil: you split $C_1^2$ in two, then split $C_3^2$ in two, then split $C_2^2$ in two, and lastly introduced a new cell which was not there before. Hence $m_3\,=\,2m_{3-1}\,+\,1$.
It could be useful to label the cells in binary. If $n=3$, for example, we can display each $j\in m_n$ as $(j)$:
$$A'_1\,\cap\,A'_2\,\cap\,A_3 \,\,\,(001)$$
$$-$$
$$A'_1\,\cap\,A_2\,\cap\,A_3 \,\,\,(011)$$
$$A_1\,\cap\,A'_2\,\cap\,A_3 \,\,\,(101)$$
$$A_1\,\cap\,A_2\,\cap\,A_3 \,\,\,(111)$$
$$-$$
$$A'_1\,\cap\,A_2\,\cap\,A'_3 \,\,\,(010)$$
$$A_1\,\cap\,A'_2\,\cap\,A'_3 \,\,\,(100)$$
$$A_1\,\cap\,A_2\,\cap\,A'_3 \,\,\,(110).$$
Perhaps you could set up an induction by noticing that when $j$ is even, $C^n_j\,=\,C^{n-1}_{j\,'}\,\cap\,A_n'$, while for $j$ odd, $C^n_j\,=\,C^{n-1}_{j\,'}\,\cap\,A_n$ holds. Here for any even integer $j>1$, we define $j\,'\,\,=\,\,j/2,$ while for any odd integer $j>1$, we let $j\,'\,\,=\,\,(j-1)/2.$ (Notice that if $j>1$ is expressed in binary, then $j\,'$ is the number resulting from deletion of the unit's bit.)
I tried to pursue this for a short while, but then decided I was happy with the standard proof.