Some useful tricks:

\begin{align*} \mu(A\setminus B) = \mu(A)  \mu(B) \quad \text{if } \mu(B) < \infty .\end{align*}

When in doubt, write \(f = ff_n + f_n\) and use the triangle inequality.

Measurable sets are wellapproximated by opens: \begin{align*} G\in {\mathcal{M}}\implies \forall {\varepsilon}, \exists G = G({\varepsilon}) \text{ such that }\,\, m(G) \leq m(E) + {\varepsilon} .\end{align*}

Borels are \(F_\sigma\) or \(G_\delta\) up to null sets, i.e. if \(E\) is Borel, there are measure zero sets \(N\) such that \begin{align*} E = F_\sigma {\textstyle\coprod}N && E {\textstyle\coprod}N = G_\delta .\end{align*}
If \((X, {\mathcal{M}})\) is a measurable space, then a measure is a function \(\mu: {\mathcal{M}}\to [0,\infty]\) such that
 \(\mu(\emptyset) = 0\).
 Countable additivity: if \(\left\{{E_k}\right\}_{k\geq 1}\) is a countable union of disjoint sets in \(X\), then \begin{align*} \mu\qty{{\textstyle\coprod}_{k\geq 1} E_k} = \sum_{k\geq 1} \mu(E_k) .\end{align*}
If (2) only holds for finitely indexed sums, we say \(\mu\) is \(\sigma{\hbox{}}\)additive.
\begin{align*}m(A) = m(B) + m(C) {\quad \operatorname{and} \quad} m(C) < \infty \implies m(A)  m(C) = m(B).\end{align*}
Let \((X, {\mathcal{M}}, \mu)\) be a measure space. Then
 Monotonicity: \(E \subseteq F \implies \mu(E) \leq \mu(F)\).
 Countable subadditivity: If \(ts{E_k}_{k\geq 1}\) is a countable collection, \begin{align*} \mu\qty{\displaystyle\bigcup_{k\geq 1} E_k} \leq \sum_{k\geq 1} \mu(E_k) .\end{align*}
\begin{align*} \text{Continuity from below:} \quad E_{n} \nearrow E &\implies m(E_{n}) \overset{n\to\infty}\longrightarrow m(E) \\ \text{Continuity from above:} \quad m(E_{1}) < \infty \text{ and } E_{n} \searrow E &\implies m(E_{n}) \overset{n\to\infty}\longrightarrow m(E) .\end{align*}
Mnemonic: \(\lim_n \mu(E_n) = \mu(\lim E_n)\) where \(\lim_n E_n = E\coloneqq\cup_N E_n\) for \(E_n \nearrow E\) and \(\lim_n E_n = E \coloneqq\cap_n E_n\) for \(E_n\searrow E\).
Idea: break into disjoint annuli!

From below: break into disjoint annuli \(A_{2} = E_{2}\setminus E_{1}\),
 Apply countable disjoint additivity to \(E = {\textstyle\coprod}A_{i}\).

From above: funny step, use \(E_{1} = ({\textstyle\coprod}E_{j}\setminus E_{j+1}) {\textstyle\coprod}(\cap E_{j})\).
 Taking measures yields a telescoping sum, and use countable additivity, then finiteness to subtract.
For any measure \(\mu\), \begin{align*} \mu(F_1) < \infty,\, F_k \searrow F \implies \lim_{k\to\infty}\mu(F_k) = \mu(F) ,\end{align*} where \(F_k \searrow F\) means \(F_1 \supseteq F_2 \supseteq \cdots\) with \(\cap_{k=1}^\infty F_k = F\).

Note that \(\mu(F)\) makes sense: each \(F_k \in \mathcal{B}\), which is a \(\sigma{\hbox{}}\)algebra and closed under countable intersections.

Take disjoint annuli by setting \(E_k \coloneqq F_k \setminus F_{k+1}\)

Funny step: write \begin{align*} F_1 = F {\textstyle\coprod}\displaystyle\coprod_{k=1}^{\infty} E_k .\end{align*}
 This is because \(x\in F_1\) iff \(x\) is in every \(F_k\), so in \(F\), or
 \(x\not \in F_1\) but \(x\in F_2\), noting incidentally \(x\in F_3, F_4,\cdots\), or,
 \(x\not\in F_2\) but \(x\in F_3\) and thus \(F_4, F_4,\cdots\), and so on.

Now take measures, and note that we get a telescoping sum: \begin{align*} \mu(F_1) &= \mu(F) + \sum_{k=1}^\infty \mu(E_k) \\ &= \mu(F) + \lim_{N\to\infty} \sum_{k=1}^N \mu(E_k) \\ &\coloneqq\mu(F) + \lim_{N\to\infty} \sum_{k=1}^N \mu(F_k \setminus F_{k+1} ) \\ &\coloneqq\mu(F) + \lim_{N\to\infty} \sum_{k=1}^N \mu(F_k)  \mu(F_{k+1} ) \hspace{5em}\text{to be justified}\\ &= \mu(F) + \lim_{N\to\infty} [ (\mu(F_1)  \mu(F_2)) + (\mu(F_2)  \mu(F_3)) + \cdots \\ & \hspace{8em} + (\mu(F_{N1})  \mu(F_N)) + (\mu(F_N)  \mu(F_{N+1})) ] \\ \\ &= \mu(F) + \lim_{N\to\infty} \mu(F_1)  \mu(F_{N+1}) \\ &= \mu(F) + \mu(F_1)  \lim_{N\to\infty} \mu(F_{N+1}) .\end{align*}

Justifying the measure subtraction: the general statement is that for any pair of sets \(A\subseteq X\), \(\mu(X\setminus A) = \mu(X)  \mu(A)\) when \(\mu(A) < \infty\): \begin{align*} X &= A {\textstyle\coprod}(X\setminus A) \\ \implies \mu(X) &= \mu(A) + \mu(X\setminus A) && \text{countable additivity} \\ \implies \mu(X) \mu(A) &= \mu(X\setminus A) && \text{if } \mu(A) < \infty .\end{align*}

Now use that \(\mu(F_1)<\infty\) to justify subtracting it from both sides: \begin{align*} \mu(F_1) &= \mu(F) + \mu(F_1)  \lim_{N\to\infty} \mu(F_{N+1}) \\ \implies 0 &= \mu(F_1)  \lim_{N\to\infty} \mu(F_{N+1}) \\ \lim_{N\to\infty} \mu(F_{N+1}) &= \mu(F_1) .\end{align*}

Now use that \(\lim_{N\to\infty}\mu(F_{N+1}) = \lim_{N\to\infty} \mu(F_N)\) to conclude.
Outer Measure
 Monotonicity: \(E\subseteq F \implies m_*(E) \leq m_*(F)\).
 Countable Subadditivity: \(m_*(\cup E_{i}) \leq \sum m_*(E_{i})\).
 Approximation: For all \(E\) there exists a \(G \supseteq E\) such that \(m_*(G) \leq m_*(E) + \varepsilon\).
 Disjoint ^{1} Additivity: \(m_*(A {\textstyle\coprod}B) = m_*(A) + m_*(B)\).
A set \(E\) is measurable iff it can be approximated by an open set in \(m_*\), so there exists \(G\supseteq E\) with \(m_*(G\setminus E) < {\varepsilon}\).
Measures on \({\mathbb{R}}^d\)
Every open subset of \({\mathbb{R}}\) (resp \({\mathbb{R}}^n\)) can be written as a unique countable union of disjoint (resp. almost disjoint) intervals (resp. cubes).
Lebesgue measure is translation and dilation invariant.

This is obvious for cubes:
 For translation, if \(Q_i \rightrightarrows E\) then \(Q_i + k \rightrightarrows E + k\). One can then show \(m_*(E + k) \leq \sum {\left\lvert {Q_i + k} \right\rvert} = \sum {\left\lvert {Q_i} \right\rvert}\leq m_*(E) + {\varepsilon}\) for all \({\varepsilon}\), and get the reverse inequality by writing \(E = (E+y)y\).
 For dilation, use that \(m_*(t(A{\textstyle\coprod}B)) = tm_*(A{\textstyle\coprod}B)\), which is useful because we cover with disjoint cubes. Then use that \(tQ_i \rightrightarrows tE\) to get \(tm_*(E) \leq t\sum {\left\lvert {Q_i} \right\rvert} = \sum {\left\lvert {tQ_i} \right\rvert} \leq m_*(tE) + {\varepsilon}\) and similarly reverse to get equality.
There is a nonmeasurable set \(A\subseteq {\mathbb{R}}\).
 Use AOC to choose one representative from every coset of \({\mathbb{R}}/{\mathbb{Q}}\) on \([0, 1)\), which is countable, and assemble them into a set \(N\)
 Enumerate the rationals in \([0, 1]\) as \(q_{j}\), and define \(N_{j} = N + q_{j}\). These intersect trivially.
 Define \(M \coloneqq{\textstyle\coprod}N_{j}\), then \([0, 1) \subseteq M \subseteq [1, 2)\), so the measure must be between 1 and 3.
 By translation invariance, \(m(N_{j}) = m(N)\), and disjoint additivity forces \(m(M) = 0\), a contradiction.
If \(A_{n}\) are all measurable, \(\limsup A_{n}\) and \(\liminf A_{n}\) are measurable.
Measurable sets form a sigma algebra, and these are expressed as countable unions/intersections of measurable sets.
Let \(\{E_{k}\}\) be a countable collection of measurable sets. Then \begin{align*} \sum_{k} m(E_{k}) < \infty \implies \text{ almost every } x\in {\mathbb{R}}\text{ is in at most finitely many } E_{k} \iff m(\limsup_k E_k) = 0 .\end{align*}
In words, interpreting \(E_k\) as events and \(m(E_k) = {\mathbb{P}}(E_k)\) as a probability: if the sum of probabilities of events is finite, the probability of infinitely many events occurring is zero.
 If \(E = \limsup_{j} E_{j}\) with \(\sum m(E_{j}) < \infty\) then \(m(E) = 0\).
 If \(E_{j}\) are measurable, then \(\limsup_{j} E_{j}\) is measurable.
 If \(\sum_{j} m(E_{j}) < \infty\), then \(\sum_{j=N}^\infty m(E_{j}) \overset{N\to\infty}\to 0\) as the tail of a convergent sequence.
 \(E = \limsup_{j} E_{j} = \cap_{k=1}^\infty \cup_{j=k}^\infty E_{j} \implies E \subseteq \cup_{j=k}^\infty\) for all \(k\)
 \(E \subset \cup_{j=k}^\infty \implies m(E) \leq \sum_{j=k}^\infty m(E_{j}) \overset{k\to\infty}\to 0\).
 Characteristic functions are measurable
 If \(f_{n}\) are measurable, so are \({\left\lvert {f_{n}} \right\rvert}, \limsup f_{n}, \liminf f_{n}, \lim f_{n}\),
 Sums and differences of measurable functions are measurable,
 Cylinders \(F(x,y) = f(x)\) are measurable,
 Compositions \(f\circ T\) for \(T\) a linear transformation are measurable,
 “Convolutionish” transformations \((x,y) \mapsto f(xy)\) are measurable
Take the cone on \(f\) to get \(F(x, y) = f(x)\), then compose \(F\) with the linear transformation \(T = [1, 1; 1, 0]\).
A measure space \((X, {\mathcal{M}}, \mu)\) is \(\sigma{\hbox{}}\)finite if \(X\) can be written as a union of countably many measurable sets with finite measure.
If \((X, {\mathcal{B}}, \mu)\) is a Borel measure space where \(\mu\) is finite on all balls of finite radius, then for any \(E \in {\mathcal{B}}\) and any \({\varepsilon}>0\),
 There exists an open set \(O\) with \(E \subset O\) and \(\mu(O\setminus E) < {\varepsilon}\)
 There exists a closed set \(F\) with \(F\subset E\) and \(\mu(E\setminus F) < {\varepsilon}\).
Measurability is not preserved by homeomorphisms.
By counterexample: there is a homeomorphism that takes that Cantor set (measure zero) to a fat Cantor set.
#todo Expand