measure theory inclusion-exclusion formula proof

Question

If we have a finite measure space and $\{M_{j}\}_{j=1}^{\infty} \in \mathbb{M}$ where $\mathbb{M}$ is a sigma algebra. Then the following holds

$$\mu\Big(\bigcup_{j=1}^{n}M_{j}\Big) = \sum_{\emptyset \neq K \subset \{1,\dots,n\}}(-1)^{|K|-1}\mu\Big(\bigcap_{j \in K}M_{j}\Big)$$

Proposed approach:

It seems that this formula is similar to an inclusion-exclusion formula? One approach I was thinking was an induction approach. Obviously if we take $|K|=1$ the formula holds. The induction step could be to assume it holds for $|K-1|-1$ and then simply prove the final result.

Does this seem a viable approach, any other suggested approaches are welcome. Thanks.

Answer 1

The principle of inclusion/exclusion is actually based on this equality of indicator functions: $$\mathbf{1}_{\bigcup_{i=1}^{n}M_{i}}=\sum_{k=1}^{n}\left(-1\right)^{k-1}\sum_{1\leq i_{1}<\cdots<i_{k}\leq n}\mathbf{1}_{M_{i_{1}}\cap\cdots\cap M_{i_{k}}}\tag1$$

For a proof of that see this answer.

If on both sides an integral is taken w.r.t. to some measure $\mu$ then we arrive at your formula.