A measure space \((X,\mathcal{M},\mu)\) is a triple consisting of a set \(X\), a [[sigma-algebra|\(\sigma\)-algebra]] \(\mathcal{M}\) and a measure \(\mu\).
A measure space is finite if \(\mu(X)<\infty\). A measure space is \(\sigma\)-finite if we can write \(X=\cup_{n\in\mathbb{N}}E_n\) where \(\mu(E_n)<\infty\) for all \(n\).
A measure space \((X,\mathcal{M}, \mu)\) has the following useful properties: 1. If \(E,F\in\mathcal{M}, E\subseteq F\implies \mu(E)\leq \mu(F)\) (monotonicity) 2. If \(\{E_n\}_{n\in\mathbb{N}}\subseteq\mathcal{M}\implies\mu(\cup_{n\in\mathbb{N}}E_n)\leq\sum_{n\in\mathbb{N}}\mu(E_n)\) (countable subadditivity) 3. If \(\{E_n\}_{n\in\mathbb{N}}\subseteq\mathcal{M}\) and \(E_1\subseteq E_2\subseteq ...\implies \mu(\cup_{n\in\mathbb{N}}E_n)=\lim_{n\rightarrow\infty}\mu(E_n)\) (continuity from below) 4. If \(\{E_n\}_{n\in\mathbb{N}}\subseteq\mathcal{M}\) and \(E_1\supseteq E_2\supseteq ...\implies \mu(\cap_{n\in\mathbb{N}}E_n)=\lim_{n\rightarrow\infty}\mu(E_n)\) (continuity from above)
\begin{proof} 1. Since \(E\subseteq F\) we have that \(\mu(F)=\mu(E\cup E\setminus F)=\mu(E) +\mu(E\setminus F)\geq \mu(E)\) using both the fact that \(\mu(E\setminus F)\geq 0\) and that \(E\) and \(E\setminus F\) are disjoint. 2. We can define \(F_1=E_{1}\) and \(F_n=E_n\setminus \cup_{k=1}^{n-1}E_k\) such that \(F_n\) are all disjoint and \(\cup_{k=1}^nE_k=\cup_{k=1}^nF_k\) for all \(n\in\mathbb{N}\). Then \(\mu(\cup_{n\in\mathbb{N}}E_n)=\mu(\cup_{n\in\mathbb{N}}F_n)=\sum_{n\in\mathbb{N}}\mu(F_n)\leq\sum_{n\in\mathbb{N}}\mu(E_n)\) since \(\mu(F_n)\leq\mu(E_n)\) for all \(n\in\mathbb{N}\).
3. We can define \(F_1=E_1\) and \(F_n=E_n\setminus E_{n-1}\) such that \(F_n\) are all disjoint and \(\cup_{k=1}^{n}E_k=\cup_{k=1}^{n}F_k\) for all \(n\in\mathbb{N}\). Then \(\mu(\cup_{n\in\mathbb{N}} E_n)=\mu(\cup_{n\in\mathbb{N}}F_n)=\sum_{n\in\mathbb{N}}\mu(F_n)=\lim_{n\rightarrow\infty}\sum_{k=1}^n\mu(F_k)=\lim_{n\rightarrow\infty}\mu(E_n)\).
\end{proof}