Prove Lemma Concerning Probability of Union of Events I have been working through exercises in a book on probability theory. One of the exercises asks that the reader prove a lemma concerning the probability of a union of events. One of the lemmas in the book states :
\begin{equation}
P(A \bigcup B) = P(A) + P(B) - P(A \bigcap B) \tag{1} 
\end{equation}
Where A and B are arbitrary events. Another lemma generalizes this and states :
\begin{align} 
P\left( \bigcup_{i=1}^{n} A_{i} \right) 
      & = \sum_{i} P(A_{i}) - \sum_{i < j} P(A_{i} \bigcap A_{j}) 
          + \sum_{i<j<k} P(A_{i} \bigcap A_{j} \bigcap A_{k}) - \dots +  \\\\
      & \phantom{=} (-1)^{n+1} P(A_{1} \bigcap A_{2} \bigcap \dots \bigcap A_{n} ) \tag{2}
\end{align}
The exercise asks the reader to prove $(2)$ using induction.
I think if someone can help show how to prove this it will be easier for me to solve these kinds of problems in the book.
 A: A simple solution may be obtained using indicator functions of sets and the linear properties of the integral:
$$ \Omega\setminus \bigcup^n_{j=1}A_j =\bigcap^n_{j=1}\Omega\setminus A_j$$
we get that
$$
\mathbb{1}-\mathbb{1}_{\bigcup^n_{j=1}A_j}= \prod^n_{j=1}(\mathbb{1}-\mathbb{1}_{A_j})
$$
That is,
$$
\begin{align}
\mathbb{1}_{\bigcup^n_{j=1}A_j}= \mathbb{1}-\prod^n_{j=1}(\mathbb{1}-\mathbb{1}_{A_j})&=\sum^n_{j=1}\mathbb{1}_{A_j} -\sum_{1\leq i_1<i_2\leq n}\mathbb{1}_{A_{i_1}}\mathbb{1}_{A_{i_2}} \\
&\quad\quad +\sum^n_{1\leq i_1<i_2<i_3\leq n}\mathbb{1}_{A_{i_1}}\mathbb{1}_{A_{i_2}}\mathbb{1}_{A_{i_3}}+\ldots+(-1)^n\mathbb{1}_{A_1}\cdot\ldots\cdot\mathbb{1}_{A_n}\\
&= \sum^n_{j=1}\mathbb{1}_{A_j} -\sum_{1\leq i_1<i_2\leq n}\mathbb{1}_{A_{i_1}\cap A_{i_2}} \\
&\quad\quad +\sum^n_{1\leq i_1<i_2<i_3\leq n}\mathbb{1}_{A_{i_1}\cap A_{i_2}\cap A_{i_3}}+\ldots+(-1)^n\mathbb{1}_{\cap^n_{j=1}A_j}
\end{align}
$$
The conclusion follows by linearity of the mean operator, that is integration with respect $\mathbb{P}$.
