Measure space and inclusion-exclusion Theorem

Recently, I have been studying real analysis through the book Real Analysis: Measures, Integrals and Applications, and I have come across a generalization of the inclusion-exclusion theorem. I sensed something important there, so I skipped to find the specific results I needed. As a result, I lack a solid understanding of some of the concepts I mentioned.

Let $(X,\mathcal{A},\mu )$$(X,\mathcal A,\mu)$ is a measure space, Let $E\subseteq \mathcal{A}$$E\subseteq\mathcal A$

And $\begin{array}{c}{\chi }_E:=\{\begin{array}{l}1,x\in E\\ 0,x\notin E\end{array}\end{array}$$\chi_{E}:=\begin{cases}1,x\in E\\0,x\notin E\end{cases}$ , according to the definition of integral,

we have $\begin{array}{c}\mu (E)={\int }_X{\chi }_{E}d\mu \end{array}$$\mu(E)=\int_X\chi_E d\mu\\$

And according to De Morgen Law

${\displaystyle {\chi }_{E_1\cup E_2\cup ...E_n}=1-{\chi }_{\overline{E_1\cup E_2\cup ...E_n}}=1-{\chi }_{\overline{E_1}\cap \overline{E_2}\cap ...\overline{E_n}}=1-{\chi }_{\overline{E_1}}{\chi }_{\overline{E_2}...\chi \overline{E_n}}}$$\displaystyle \chi_{E_1\cup E_2\cup...E_n}=1-\chi_{\overline{E_1\cup E_2\cup ...E_n}}=1-\chi_{\overline{E_1}\cap\overline{E_2}\cap...\overline{E_n} }=1-\chi_{\overline{E_1}}\chi_{\overline{E_2}...\chi{\overline{E_n}}}$

$\begin{array}{c}=1-\prod _{i=1}^n(1-{\chi }_{E_i})=\sum {\chi }_{E_i}-\sum _{i<j}{\chi }_{Ei}{\chi }_{E_j}+...(-1{)}^{n+1}{\chi }_{E_1}{\chi }_{E_2}...{\chi }_{E_n}\end{array}$$=1-\prod_{i=1}^n(1-\chi_{E_i})=\sum\chi_{E_i}-\sum_{i<j} \chi_{Ei}\chi_{E_j}+...(-1)^{n+1}\chi_{E_1}\chi_{E_2}...\chi_{E_n}\\$

And integral two sides, we can get

$\mu (E_1\cup E_2\cup ...E_n)$$\mu(E_1\cup E_2\cup...E_n)$

$=\sum \mu (E_i)-\sum _{i<j}\mu (E_i\cap E_j)+...(-1{)}^{n+1}\mu (E_1\cap E_2\cap ...E_n)$$=\sum\mu(E_i)-\sum_{i<j}\mu(E_i\cap E_j)+...(-1)^{n+1}\mu(E_1\cap E_2\cap... E_n)$

And observe that the cardinality function ${\mu }_C$$\mu_C$ is a measure

${\mu }_C\ge 0$$\mu_C\ge 0$

${\mu }_C(\mathrm{\varnothing })=0$$\mu_C(\empty)=0$

$\begin{array}{c}{\mu }_C(\underset{i=1}{\overset{\mathrm{\infty }}{⨆}}{E}_i)=\sum _{i=1}^n{\mu }_C(E_i)\end{array}$$\mu_C(\bigsqcup_{i=1}^{\infty} E_i)=\sum_{i=1}^n \mu_C(E_i)\\$

We can get the count version inclusion-exclusion theorem