# measure theory inclusion-exclusion formula proof

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.

The principle of inclusion/exclusion is actually based on this equality of indicator functions: $$\mathbf{1}_{\bigcup_{i=1}^{n}M_{i}}=\sum_{k=1}^{n}\left(-1\right)^{k-1}\sum_{1\leq i_{1}<\cdots
If on both sides an integral is taken w.r.t. to some measure $$\mu$$ then we arrive at your formula.