The collection of measurable sets in a measurable space forms a set algebra: it contains the empty set, is closed under complements, and closed under finite unions. This is the algebra-of-sets part of the σ-algebra structure.
If sets $E_1, \dots, E_n$ are measurable, then $\bigcup_{k=1}^n E_k$ is measurable.
Stated for an arbitrary finite indexing Finset.
Let $\mathcal{A}$ be an algebra (here, a σ-algebra), and let $\{E_n\}$ be a countable collection of measurable sets. Then there exists a pairwise disjoint countable collection $\{F_n\}$ of measurable sets such that $\bigcup_n E_n = \bigcup_n F_n$.
The Lebesgue (null-)measurable subsets of $\mathbb{R}$ form a σ-algebra: the empty set is measurable, complements of measurable sets are measurable, and countable unions of measurable sets are measurable.
Continuity of measure for increasing sequences: if $E_1 \subset E_2 \subset \cdots$ is an increasing sequence of measurable sets, then $\mu\left(\bigcup_{k=1}^{\infty} E_k\right) = \lim_{n \to \infty} \mu(E_n)$.
The pointwise limit of measurable functions is measurable: if $f_n : \alpha \to \overline{\mathbb{R}}$ are measurable and $f_n(x) \to g(x)$ for all $x$, then $g$ is measurable.
Simple functions are closed under scalar multiplication, addition, and multiplication: for any simple functions $f, g$ and scalar $c$, the functions $c \cdot f$, $f + g$, and $f \cdot g$ are again simple functions.
For any nonnegative measurable function $f : \alpha \to [0, \infty]$, there exists a sequence of simple functions $\{\varphi_n\}$ such that: (a) $0 \le \varphi_0(a) \le \varphi_1(a) \le \cdots \le f(a)$ for all $a$ (pointwise increasing and dominated by $f$); (b) $\varphi_n(a) \to f(a)$ pointwise; and (c) for every bound $B < \infty$ and every $\varepsilon > 0$, there is an $N$ such that $f(a) - \varphi_n(a) < \varepsilon$ for all $n \ge N$ and all $a$ with $f(a) \le B$ (uniform convergence on sets where $f$ is bounded).
For all $a \in \mathbb{R}$, the open interval $(a, \infty)$ is measurable.