Measure Theory#

$\sigma$-Field #

Question : What sets (subsets of $\Omega$) am I allowed to measure?

Definition

For some set $\Omega$, a $\sigma$-Field $\mathcal{F}$ is a collection of sets $A\in \Omega$ s.t.

1. $\varnothing, ~\Omega \in \mathcal{F}$
2. If $A\in \mathcal{F}$ then $A^c \in \mathcal{F}$
3. $\mathcal{F}$ is closed under union: for a countable collection of sets $\{A_i\}_{i=1}^\infty$ s.t. $A_i \in \mathcal{F}$ for all $i\in \mathbb{N}$ then   $\displaystyle\bigcup_{i=1}^{\infty} A_i \in \mathcal{F}$

   Equivalence  : $\mathcal{F}$ also contains countable intersections because $~\displaystyle\bigcap_{i=1}^{\infty} A_i = \displaystyle\bigcup_{i=1}^{\infty} A_i^c\in \mathcal{F}$

Measure#

Definition

A measure $\mu~:~\mathcal{F}\rightarrow\mathbb{R}_+$ is a mapping from a $\sigma$-field $\mathcal{F}$ to the non-negative real numbers s.t.

1. $\mu(\varnothing) = 0$
2. For a pairwise disjoint collection $\{A_i\}_{i=1}^\infty$ then $\mu\left(\displaystyle\bigcup_{i=1}^{\infty} A_i\right) = \displaystyle\sum_{i=1}^{\infty} \mu(A_i)$  (Countably Additivity)

Special Cases for a Measure Space $(\Omega, \mathcal{F}, \mu)$:

• If $\mu(\Omega) = 1$ then we say that $(\Omega, \mathcal{F}, \mu)$ is a Probability Measure
• If $\mu(A) < \infty$ for any $A$ then we say that $\mu$ is a Finite Measure
• If $\Omega=\displaystyle\bigcup_{i=1}^{\infty} A_i$ s.t. $\mu(A_i) < \infty ~~ \forall i~$ then $\mu$ is a $\sigma$-Finite Measure

  Example

  $\Omega=\mathbb{R}~$,  $\mu([a,b])=b-a$ and $\mathbb{R}= \displaystyle\bigcup_{i=1}^{\infty} [i-1,i]\displaystyle\cup[-i,-i+1]$ then $\mu$ is a $\sigma$-Finite Measure

Power Set#

Definition

Power set $\mathscr{P}(\Omega)$ or $2^\Omega$ is the set of all subsets of $\Omega$

Semi-Ring of Sets#

Definition

$\mathcal{A}$, a collection of subsets of $\Omega~$, is called a Semi-Ring when

1. $\varnothing \in \mathcal{A}$
2. If $A, B \in \mathcal{A}$ then $A\displaystyle\cap B \in \mathcal{A}$
3. If $A, B \in \mathcal{A}$ then there exists a finite number of pariwise disjoint sets $C_i \in \mathcal{A}$ for $~i=1,\cdots,n~$ s.t. $A\setminus B = \displaystyle\bigcup_{i=1}^{n} C_i$

Example

$\mathcal{A}$ : all intervals $(a,b]$ in $\mathbb{R}$ is a semi-ring

Ring of Sets#

Definition

$\mathcal{A}$, a collection of subsets of $\Omega~$, is called a Ring when

1. $\varnothing \in \mathcal{A}$
2. If $A, B \in \mathcal{A}$ then $A\setminus B \in \mathcal{A}$
3. If $A, B \in \mathcal{A}$ then $A\displaystyle\cup B\in \mathcal{A}~$ (This implies finite unions are also in $\mathcal{A}$)

From the defintion, we know that the intersection of two sets is also in $\mathcal{A}$ because $A\displaystyle\cap B = A\setminus (A\setminus B)$

Example

All finite unions of half open intervals $(a,b]$ in $\mathbb{R}$ is a ring

Field#

Definition

$\mathcal{A}$ is a Field if it is a Ring and $\Omega \in \mathcal{A}$

Equivalence

Since the whole set $\Omega$ is in $\mathcal{A}$, then

• the second condition of Ring is equivalent to $\mathcal{A}$ being closed under complimentation as $A \setminus B = A \displaystyle\cap A^c$ and $A^c = \Omega \setminus A$
• the third condition of Ring is equivalent to $\mathcal{A}$ being closed under finite unions as $A\displaystyle\cup B = (A^c \displaystyle\cap B^c)^c$

Note : Field + countable unions $=$ $\sigma$-Field

Set Function#

Definition

A set function $\mu: ~\mathcal{A}\rightarrow \mathbb{R}^+$ (not necessarily a measure) is a mapping from a set of functions $\mathcal{A}$ to $\mathbb{R}^+$. For $A, B \in \mathcal{A}$, we say that

• $\mu$ is monotone if $A\subset B$ implies $\mu(A)\leq \mu(B)$
• $\mu$ is additive if $\mu(A\displaystyle\cup B) = \mu(A) + \mu(B)$ for $A, B$ disjoint
• $\mu$ is countably additive if $\mu\left(\displaystyle\bigcup_{i=1}^{\infty} A_i\right) = \displaystyle\sum_{i=1}^{\infty} \mu(A_i)$ for $\{A_i\}_{i=1}^\infty$ pairwise disjoint and $\displaystyle\bigcup_{i=1}^{\infty} A_i \in \mathcal{A}$
• $\mu$ is countably sub-additive if $\mu\left(\displaystyle\bigcup_{i=1}^{\infty} A_i\right) \leq \displaystyle\sum_{i=1}^{\infty} \mu(A_i)$ for $\displaystyle\bigcup_{i=1}^{\infty} A_i \in \mathcal{A}$ (not nessarily pairwise disjoint)
• $\mu$ is a pre-measure if $\mu$ is countably additive, $\mu(\varnothing)=0$ and $\mathcal{A}$ is a ring. For a pre-measure, it is also
  • monotone because for $A\subset B$ we have $\mu(B)=\mu(A\displaystyle\cup (B\setminus A)) = \mu(A) + \mu(B\setminus A) \geq \mu(A)$
  • sub-additive because for $\displaystyle\bigcup_{i=1}^{\infty} A_i \in \mathcal{A}$ let $B_i=A_i \setminus \left(\displaystyle\bigcup_{j=1}^{i-1} A_j\right)$ then $B_i$ are pairwise disjoint and by monotonicity, $\mu\left(\displaystyle\bigcup_{i=1}^{\infty} A_i\right) = \mu\left(\displaystyle\bigcup_{i=1}^{\infty} B_i\right) = \displaystyle\sum_{i=1}^{\infty} \mu(B_i) \leq \displaystyle\sum_{i=1}^{\infty} \mu(A_i)$

Outter Measure#

Definition

For a pre-measure $\mu$ on a ring $\mathcal{A}$, the outter measure (Not necessarily a Measure) $\mu^* : \mathcal{M} \rightarrow \mathbb{R}^+$ is defined as

$$\mu^*(E) = \inf \left\{ \displaystyle\sum_{i=1}^{\infty} \mu(A_i) ~:~ E\subset \displaystyle\bigcup_{i=1}^{\infty} A_i, ~A_i\in \mathcal{A} \right\}$$

for any $E\subseteq \Omega$. This is equal to the smallest possible sum of the pre-measure over all finite or countable collections of $A_i$ in $\mathcal{A}$ that cover $E$.

Can we “Measure” (Outer Measure) any $E\subset \Omega$ ? Not necessarily

Denote $\mathcal{M}$ to be the collection of all $\mu^*$ measurable sets where we say that $B\subseteq \Omega$ is $\mu^*$ measureable when

Caratheodory Extension Theorem#

Caratheodory Extension Theorem

Let $\mathcal{A}$ be a ring on $\Omega$ and $\mu$ be a pre-measure, then $\mu$ extends to a measure on $\sigma(\mathcal{A})$.

Note

• $\sigma(\mathcal{A})$ is the $\sigma$-field comes from extend $\mathcal{A}$ by including coutable unions and $\Omega$ itself.
• The “corret” extension is the outter measure $\mu^*$.

Proof

Assume $B\subset \Omega$ and $\mu^*(B)<\infty$

1. Prove stuff about $\mu^*$

   • $\mu^*(\varnothing)=0$ as $\mu$ is a pre-measure

   • $\mu^*$ is non-negative as $\mu$ is non-negative

   • $\mu^*$ is monotone (increasing)

     Let $B_1, B_2 \subset \Omega$ and $B_1\subset B_2$ then for any ${A_i}$ s.t. $B_2 \subseteq \displaystyle\bigcup_i A_i$. Then $B_1 \subseteq \displaystyle\bigcup_i A_i$ thus $\mu^*(B_1)\leq \mu^*(B_2)$

   • $\mu^*$ is countably sub-additive

     For $\{B_i\}_{i=1}^\infty$ and a given $\epsilon > 0$, let $B_i\subseteq \displaystyle\bigcup_i A_{ij}$ for $A_{ij}\in \mathcal{A}$ s.t.

     $$\displaystyle\sum_i \mu(A_{ij}) \leq \mu^*(B_i) + \epsilon\cdot 2^{-i}$$

     This is possible because $\mu^*$ is the infimum. As $\displaystyle\bigcup_{i=1}^\infty B_i \subseteq \displaystyle\bigcup_{i,j} A_{ij}$ and as $\mu^*$ is monotone and $\mu$ is sub-additive, then

     $$\begin{aligned} \mu^*\left(\displaystyle\bigcup_{i=1}^\infty B_i\right) &\leq \mu\left(\displaystyle\bigcup_{i,j} A_{ij}\right) \\ &\leq \displaystyle\sum_{i,j} \mu(A_{ij}) \\ &\leq \displaystyle\sum_i \mu^*(B_i) + \epsilon\cdot 2^{-i} \end{aligned}$$

     Take $\epsilon\rightarrow 0$ to get that $\mu^*$ is countably sub-additive

2. Check that $\mu$ and $\mu^*$ coincide for all $A\in \mathcal{A}$
   • For any $A\in \mathcal{A}$ we have $\mu^*(A) \leq \mu(A)$ because $\{A\}$ is a cover of $A$ and $\mu^*$ is the infimum
   • For the reverse, if $A\subset \displaystyle\bigcup_i A_i$ then by countable sub-additivity and monotonicity
   $$\mu(A) \leq \displaystyle\sum_i \mu(A\displaystyle\cap A_i) \leq \displaystyle\sum_i \mu(A_i)$$ This implies that $\mu(A)\leq \mu^*(A)$ because the right hand side is the cover of A.
   • Thus $\mu(A)=\mu^*(A) ~~ \forall A\in \mathcal{A}$
3. Check that the ring $\mathcal{A}\subset \mathcal{M}$ (all $\mu^*$ measurable sets)
   • i.e. $\forall A \in \mathcal{A}$ we want to show that $A$ is $\mu^*$-measurable: $$\mu^*(E\displaystyle\cap A) + \mu^*(E\displaystyle\cap A^c) = \mu^*(E) ~~ \forall E\subseteq \Omega$$
   • Note $\mu^*(E \displaystyle\cap A) + \mu^*(E\displaystyle\cap A^c) \geq \mu^*(E)$ as $\mu^*$ is sub-additive.
   • Next, for some $\epsilon>0~$, choose $\{A_i\}$ s.t. $E\subset \displaystyle\bigcup_i A_i$ and $\displaystyle\sum_i \mu(A_i) \leq \mu^*(E) + \epsilon$
   • Furthermore, $E\displaystyle\cap A\subset \displaystyle\bigcup_i(A\displaystyle\cap A_i)$ and $E\displaystyle\cap A^c \subset \displaystyle\bigcup_i(A^c\displaystyle\cap A_i)$ thus $$\begin{aligned} \mu^*(E\displaystyle\cap A) + \mu^*(E\displaystyle\cap A^c) &\leq \displaystyle\sum_i \mu(A\displaystyle\cap A_i) + \displaystyle\sum_i \mu(A^c\displaystyle\cap A_i) \\ &= \displaystyle\sum_i \mu(A_i) \leq \mu^*(E) + \epsilon \end{aligned}$$ and take $\epsilon\rightarrow 0$.
4. Show that $\mathcal{M}$ is a $\sigma$-Field
   • $\varnothing \in \mathcal{M}$ since $\varnothing \in \mathcal{A}$
   • $\Omega \in \mathcal{M}$ since $\forall E\subset \Omega ~~ \mu^*(E\displaystyle\cap \Omega) + \mu^*(E\displaystyle\cap\varnothing)=\mu^*(E)$
   • $\mathcal{M}$ is closed under coplimentation since $\mu^*(E\displaystyle\cap B^c) + \mu^*(E\displaystyle\cap (B^c)^c) = \mu^*(E)$ which implies that $B^c \in \mathcal{M}$
   • $\mathcal{M}$ is closed under finite intersections since for $B_1, B_2\in\mathcal{M}$ and any $E \subset \Omega$ $$\begin{aligned} \mu^*(E)&=\mu^*(E\displaystyle\cap B_1) + \mu^*(E\displaystyle\cap B_1^c) \\ &= \mu^*(E\displaystyle\cap B_1\displaystyle\cap B_2) + \mu^*(E\displaystyle\cap B_1^c\displaystyle\cap B_2) + \mu^*(E\displaystyle\cap B_1\displaystyle\cap B_2^c) + \mu^*(E\displaystyle\cap B_1^c\displaystyle\cap B_2^c) \\ &\geq \mu^*(E\displaystyle\cap B_1\displaystyle\cap B_2) + \mu^*(\{E\displaystyle\cap B_1^c\displaystyle\cap B_2\}\displaystyle\cup\{E\displaystyle\cap B_1\displaystyle\cap B_2^c\}\displaystyle\cup \{E\displaystyle\cap B_1^c\displaystyle\cap B_2^c\}) \\ &=\mu^*(E\displaystyle\cap \{B_1\displaystyle\cap B_2\}) + \mu^*(E\displaystyle\cap \{B_1\displaystyle\cap B_2\}^c) \geq \mu^*(E) ~~~~~(\text{sub-additivity}) \end{aligned}$$ thus $B_1\displaystyle\cap B_2 \in \mathcal{M}$. Properties above show that $\mathcal{M}$ is a Field.
   • To get to a $\sigma$-Field, let $\{B_i\}$ in $\mathcal{M}$ be countable and parwise disjoint and $\displaystyle\bigcup_i B_i \in \mathcal{M}$. Let $B = \displaystyle\bigcup_{i=1}^\infty B_i$, then $$\begin{aligned} \mu^*(E) &= \mu^*(E\displaystyle\cap B_1) + \mu^*(E\displaystyle\cap B_1^c) \\ &= \mu^*(E\displaystyle\cap B_1) + \mu^*(E\displaystyle\cap B_2) + \mu^*(E\displaystyle\cap B_1^c\displaystyle\cap B_2^c) \\ &~~\vdots \\ &= \displaystyle\sum_{i=1}^n \mu^*(E\displaystyle\cap B_i) + \mu^*\left(E\displaystyle\cap\bigg\{\displaystyle\bigcap_{i=1}^n B_i^c\bigg\}\right) \\ \end{aligned}$$ Then by monotonicity and sub-additivity and let $n\rightarrow \infty$, we get $$\begin{aligned} \mu^*(E) &= \displaystyle\sum_{i=1}^{\infty} \mu^*(E\displaystyle\cap B_i) + \mu^*(E\displaystyle\cap B^c) \\ &\geq \mu^*(E\displaystyle\cap B) + \mu^*(E\displaystyle\cap B^c) \\ &\geq \mu^*(E) \end{aligned}$$ Thus $\mathcal{M}$ is closed under countable unions ($\mathcal{M}$ is a $\sigma$-Field！). Choose $E = B$ we have $\mu^*(E)=\displaystyle\sum_{i=1}^\infty \mu^*(E\displaystyle\cap B_i)$ which means $\mu^*$ is countably additive.
5. Conclusion: $\mu^*$ is a set function $2^\Omega\rightarrow \mathbb{R}^+$ and it’s also a measure on $\mathcal{M}$. Since $\mathcal{A}\subset \mathcal{M}$, then $\sigma(\mathcal{A})\subseteq \mathcal{M}$. Lastly, as $\mu^*$ is a measure on $\mathcal{M}$, it is also a measure on $\sigma(\mathcal{A})$. $\square$

$\pi$-System and $\lambda$-System #

$\pi$-system

A collection of sets $\mathcal{A}$ is a $\pi$-system if

• $\varnothing \in \mathcal{A}$
• $\forall A, B \in \mathcal{A}$, $A\displaystyle\cap B \in \mathcal{A}$.

$\lambda$-system

A collection of sets $\mathcal{L}$ is a $\lambda$-system if

• $\Omega \in \mathcal{L}$
• If $A, B \in \mathcal{L}$ and $A\subset B$ then $B\setminus A \in \mathcal{L}$.
• If $\{A_i\}_{i=1}^\infty$ is a sequence of pairwise disjoint sets in $\mathcal{L}$ then $\displaystyle\bigcup_{i=1}^\infty A_i \in \mathcal{L}$.

Note : a Field is a $\pi$-system

Example

Let $\Omega=\{1,2,3,4\}$ and $\mathcal{L}$ contains all subsets with an even number of elements. Then $\mathcal{L}$ is a $\lambda$-system but not a $\sigma$-Field since $\{1,2\}, \{2,3\}\in \mathcal{L}$ but $\{1,2\}\displaystyle\cup\{2,3\}=\{1,2,3\}\notin \mathcal{L}$

Dynkin $\pi$-$\lambda$ Theorem #

Dynkin $\pi$-$\lambda$ Theorem

Let $\mathcal{A}$ be a $\pi$-system, $\mathcal{L}$ be a $\lambda$-system and $\mathcal{A}\subseteq \mathcal{L}$. Then $\sigma(\mathcal{A})\subseteq \mathcal{L}$.

Proof

Let $\mathcal{L}_0$ be the smallest $\lambda$-system such that $\mathcal{A}\subset\mathcal{L}_0$, then $\mathcal{L}_0\subseteq\mathcal{L}$. Our goal is to show that $\mathcal{L}_0$ is also a $\pi$-system and a collection of sets that is both a $\pi$-system and a $\lambda$-system is a $\sigma$-Field. Then we necessarily have that $\sigma(\mathcal{A})\subseteq\mathcal{L}_0\subseteq\mathcal{L}$.

We only need to show that $\mathcal{L}_0$ is closed under intersections.

Let $\mathcal{L}'=\{B\in\mathcal{L}_0 : B\displaystyle\cap A\in\mathcal{L}_0, \forall A\in\mathcal{A}\}$ then $\mathcal{A}\subset\mathcal{L}'$ as $\mathcal{A}$ is a $\pi$-system and let’s show that $\mathcal{L}'$ is also a $\lambda$-system:

• $\Omega\in\mathcal{L}'$ as $\mathcal{A}\subset\mathcal{L}_0$
• If $B_1, B_2\in\mathcal{L}'$ s.t. $B_1\subset B_2$ then for any $A\in\mathcal{A}$ we have that $B_1\displaystyle\cap A, B_2\displaystyle\cap A\in\mathcal{L}_0$. Thus $(B_2\displaystyle\cap A)\setminus (B_1\displaystyle\cap A) = (B_2\setminus B_1)\displaystyle\cap A\in\mathcal{L}_0$, which implies that $B_2\setminus B_1\in\mathcal{L}'$
• If $\{B_i\}_{i=1}^\infty\in\mathcal{L}'$ are pariwise disjoint, then for any $A\in\mathcal{A}$, $A\displaystyle\cap B_i\in\mathcal{L}_0$ thus $\displaystyle\bigcup_{i=1}^\infty (A\displaystyle\cap B_i)\in\mathcal{L}_0$. This implies that $A\displaystyle\cap(\displaystyle\bigcup_{i=1}^\infty B_i)=\displaystyle\bigcup_{i=1}^\infty (A\displaystyle\cap B_i)\in\mathcal{L}_0$. Hence, $\displaystyle\bigcup_{i=1}^\infty B_i\in\mathcal{L}'$.

By definition of $\mathcal{L}'$, $\mathcal{L}'\subseteq \mathcal{L}_0$ but $\mathcal{L}_0$ is the smallest. Thus $\mathcal{L}'=\mathcal{L}_0$ and $\mathcal{L}_0$ is a $\lambda$-system. Therefore, $\mathcal{L}_0$ contains all intersections with elements of $\mathcal{A}$.

Lastly, let $\mathcal{L}''=\{B\in\mathcal{L}_0 : B\displaystyle\cap C\in\mathcal{L}_0 ~\forall~ C\in\mathcal{L}_0\}$ since $\mathcal{L}_0=\mathcal{L}'$, $\mathcal{A}\subset \mathcal{L}''$. Then do the same thing we did above to $\mathcal{L}''$ to show that $\mathcal{L}''$ is a $\lambda$-system and thus $\mathcal{L}''=\mathcal{L}_0$.

Therefore, $\mathcal{L}_0$ is closed under intersections and thus a $\sigma$-Field. This implies that $\sigma(\mathcal{A})\subseteq\mathcal{L}_0\subseteq\mathcal{L}$. $\square$

Uniquessness of Extension Thereom#

Uniqueness of Extension

Let $\mu_1, \mu_2$ be $\sigma$-Finite measures on $\sigma(\mathcal{A})$ where $\mathcal{A}$ is a $\pi$-system. Then, if $\forall A\in\mathcal{A} ~\mu_1(A)=\mu_2(A)$ then $\mu_1$ and $\mu_2$ are equal on $\sigma(\mathcal{A})$.

Proof (Finite Measures)

Assuming $\mu_1(\Omega)=\mu_2(\Omega)<\infty$

Let $\mathcal{L}=\{B\subset\Omega : \mu_1(B)=\mu_2(B)\}$ and we only need to show that $\mathcal{L}$ is a $\lambda$-system because since $\mathcal{A}\subset\mathcal{L}$, by Dynkin $\pi$-$\lambda$ Theorem, $\sigma(\mathcal{A})\subset\mathcal{L}$ which means $\mu_1=\mu_2$ coincide on $\sigma(\mathcal{A})$.

• $\Omega\in\mathcal{L}$
• If $A, B\in\mathcal{L}$ with $A\subset B$ then $$\begin{aligned} \mu_1(B\setminus A) + \mu_1(A)&=\mu_1(B)\\ &=\mu_2(B)\\ &=\mu_2(B\setminus A) + \mu_2(A) < \infty \end{aligned}$$ hence $B\setminus A\in\mathcal{L}$
• $\{A_i\}_{i=1}^\infty$ are pairwise disjoint and $A_i\in\mathcal{L}$ $$\begin{aligned} \mu_1(\displaystyle\bigcup_{i=1}^\infty A_i)=\displaystyle\sum_{i=1}^\infty \mu_1(A_i)&=\displaystyle\sum_{i=1}^\infty\mu_2(A_i)\\ &=\mu_2(\displaystyle\bigcup_{i=1}^\infty A_i) < \infty \end{aligned}$$ hence $\displaystyle\bigcup_{i=1}^\infty A_i\in\mathcal{L}$.

$\mathcal{L}$ is a $\lambda$-system！ $\square$

Proof ($\sigma$-Finite Measures)

• For any $A\in\mathcal{A}$ s.t. $\mu_1(A)=\mu_2(A)<\infty$, we define $\mathcal{L}_A$ to be all $B\subseteq\Omega$ s.t. $\mu_1(A\displaystyle\cap B)=\mu_2(A\displaystyle\cap B)$.

  Proceeding as in the proof above, we can show that $\mathcal{L}_A$ is a $\lambda$-system and thus $\sigma(A)\subset\mathcal{L}_A ~ \forall A\in\mathcal{A}$.

• By $\sigma$-Finiteness we decompose $\Omega=\displaystyle\bigcup_{i=1}^\infty A_i$, $A_i\in\mathcal{A}$ and $\mu_1(A_i)=\mu_2(A_i) < \infty$. For any $B\in\sigma(\mathcal{A})$ and any $n\in\mathbb{N}$

  $$\mu_1\left(\displaystyle\bigcup_{i=1}^n(B\displaystyle\cap A_i)\right)=\displaystyle\sum_{i=1}^n \mu_1(B\displaystyle\cap A_i)-\displaystyle\sum_{i < j}\mu_1(B\displaystyle\cap A_i \displaystyle\cap A_j) + \cdots$$

  here we use inclusion-exclusion formula. This also works for $\mu_2$.

  Since $\mathcal{A}$ is a $\pi$-system, $A_i\displaystyle\cap A_j\in\mathcal{A}$ as well as futher intersections, thus

  $$\mu_1\left(\displaystyle\bigcup_{i=1}^n(B\displaystyle\cap A_i)\right)=\mu_2\left(\displaystyle\bigcup_{i=1}^n(B\displaystyle\cap A_i)\right) ~\forall n\in\mathbb{N}$$

  Let $n\rightarrow\infty$ we get $\mu_1(B)=\mu_2(B)~ \forall B\in\sigma(\mathcal{A})$ $\square$

Borel $\sigma$-Field#

Definition

$\mathcal{B}(\mathbb{R})=\sigma(\text{open sets in }\mathbb{R})$

Product Measure#

Definition

For two Measure Spaces $(\mathbb{X}, \mathcal{X}, \mu)$ and $(\mathbb{Y}, \mathcal{Y}, \nu)$, define $(\mathbb{X}\times\mathbb{Y}, \mathcal{X}\times\mathcal{Y}, \pi)$ where $\pi(A\times B)=\mu(A)\nu(B)$ for $A\in\mathcal{X}$ and $B\in\mathcal{Y}$.

Question : How are these related? $\mathcal{B}(\mathbb{X})\times\mathcal{B}(\mathbb{Y})$ and $\mathcal{B}(\mathbb{X}\times\mathbb{Y})$

From Dudley 4.1.7 $\mathcal{B}(\mathbb{X})\times\mathcal{B}(\mathbb{Y})\subset\mathcal{B}(\mathbb{X}\times\mathbb{Y})$, but these two are “usually” equal to each other. e.g. $\mathbb{X}=\mathbb{Y}=\mathbb{R}$