Measure Theory

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 $$f = f-f_n + f_n$$ and use the triangle inequality.

• Measurable sets are well-approximated 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_{N-1}) - \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 non-measurable 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,
• “Convolution-ish” transformations $$(x,y) \mapsto f(x-y)$$ 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

Footnotes
1.
This holds for outer measure iff $$\mathrm{dist}(A, B) > 0$$.
#todo