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Suppose $\tilde{X}=(X_1,X_2,...)$ is a sequence of RVs on $(\Omega,\mathcal{B})$. Prove that $\sigma(\tilde{X})$ is generated by events of the form:

$\bigcap_{i=1}^m \{X_i\leq x_i\}$ for $x_1,...,x_n \in\mathbb{R}, n\geq 1$.

Then, prove $\inf_{n\geq 1}X_n, \sup_{n\geq 1}X_n$ are $\sigma(\tilde{X})$-measurable.

Finally, show that $\{\lim_{n\rightarrow\infty}X_n$ exists} $\in \sigma(\tilde{X})$ and if $\lim_{n\rightarrow\infty}X_n(\omega)$ exists $\forall\omega$, then it is $\sigma(\tilde{X})$-measurable.


As I move deeper and deeper into this self-study, some of these ideas get jumbled together.

I do know that if $X:(\Omega,\mathcal{B})\rightarrow(\Omega^',\mathcal{B}^')$ then $\sigma(X) = \{\{X\in B\}:B\in\mathcal{B}^'\}$.

I also worked in the past to prove that if I take a $\pi$-class $\mathcal{P}=\{(-\infty,a]:a\in\mathbb{R}\}$ and the class of cylinder events in $\mathbb{R}^{\infty}$ determined by $\mathcal{P}$. That is:

$\mathcal{C}=\{\omega=(x_1,x_2,...):x_i\leq b_i,i\leq n\}$ for some $b_i\in\mathbb{R},i\leq n, n\geq 1$ is a $\pi$-class and $\sigma(\mathcal{C})=\mathcal{B}(\mathbb{R}^{\infty})$.

Lastly, I tend to struggle with liminf and limsup, but I do know that if I have a measurable space $(\Omega,\mathcal{B})$ and a measurable sequence $f_n:\Omega\rightarrow\bar{\mathbb{R}}$, then I can say that:

$\inf_{n\geq 1}X_n, \sup_{n\geq 1}X_n$ are measurable.

I feel like the pieces are dumped out on the table, but I'm just getting a bit lost. Any help would be greatly appreciated. Though not necessary, answers that don't make any assumptions about my familiarity with some of these concepts are especially helpful. This is all very new to me.

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1 Answer 1

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Let $\mathscr{C}$ denote the collection of all finite intersections $\bigcap_{i=1}^{n}\{X_{i}\leq x_{i}\}$, where $x_{1},...,x_{n}\in\mathbb{R}$ and $n\geq 1$, and let $\mathscr{B}(X)$ denote the Borel $\sigma$-algebra of a topological pair $(X,\tau)$.

I will point the main ideas but some details will be left for you to verify.

(1.) The first question asks if two $\sigma$-algebras are equal, namely $\sigma(\tilde{X})$ and $\sigma(\mathscr{C})$. For this it suffices to show that the generator sets of $\sigma(\tilde{X})$ are in $\sigma(\mathscr{C})$, whence $\sigma(\tilde{X})\subset \sigma(\mathscr{C})$, and on the other hand that $\mathscr{C}\subset \sigma(\tilde{X})$, whence $\sigma(\mathscr{C})\subset \sigma(\tilde{X})$.

To start with, note that $\mathscr{B}(\mathbb{R})$ is generated by sets of the form $(-\infty,a]$, where $a\in\mathbb{R}$. So an arbitrary generator of $\sigma(\tilde{X})$ is $\{X_{n}\leq a\}$ for some $n\in\mathbb{N}$ and $a\in\mathbb{R}$. We start by showing that such set lies in $\sigma(\mathscr{C})$. Observe that \begin{equation*} \{X_{n}\leq a\}=\bigcup_{k=1}^{\infty}\left((\{X_{n}\leq a\})\cap(\bigcap_{i=1}^{n-1}\{X_{i}\leq k\})\right). \end{equation*} Since if $\omega\in\{X_{n}\leq a\}$, then choose $k\in\mathbb{N}$ so that $k>\max_{1\leq i \leq n-1}X_{i}(\omega)$, whence $\omega\in\{X_{i}\leq k\}$ for all $i\in\{1,...,n-1\}$. Using this $k$ we obtain the inclusion $\subset$ in the above equation. The inclusion $\supset$ is trivial. Hence $\{X_{n}\leq a\}$ is a countable union of members from $\sigma(\mathscr{C})$, which shows that $\{X_{n}\leq a\}\in\sigma(\mathscr{C})$. Moreover, $\sigma(\tilde{X})\subset \sigma(\mathscr{C})$.

For the converse inclusion, since sets of type $\{X_{n}\leq a\}$ for $n\in\mathbb{N}$ and $a\in\mathbb{R}$ generate $\sigma(\tilde{X})$, then every member of $\mathscr{C}$ is a finite intersection of members from $\sigma(\tilde{X})$, and thus $\sigma(\mathscr{C})\subset \sigma(\tilde{X})$. This concludes that $\sigma(\mathscr{C})= \sigma(\tilde{X})$

(2.) The measurability of $\inf X_{n}$ and $\sup X_{n}$ follows by observing that for all $a\in\mathbb{R}$: \begin{align*}\{\inf X_{n}\leq a\}=\bigcup_{n=1}^{\infty}\{X_{n}\leq a\}\,\,\,\mathrm{and}\,\,\,\{\sup X_{n}\leq a\}=\bigcap_{n=1}^{\infty}\{X_{n}\leq a\}. \end{align*} Hence $\{\inf X_{n}\leq a\},\{\sup X_{n}\leq a\}\in\sigma(\tilde{X})$ for all $a\in\mathbb{R}$.

(3.) From $(2.)$ it follows that $\limsup X_{n}$ and $\liminf X_{n}$ are also measurable, since $\limsup X_{n}=\inf_{n\geq 1}(\sup_{k\geq n} X_{n})$ and $\liminf X_{n}=\sup_{n\geq 1}(\inf_{k\geq n} X_{n})$. Hence if $\lim X_{n}$ exists, then in particular $\lim X_{n}=\liminf X_{n}=\limsup X_{n}$ and it is thus measurable.

For the first part of $(3.)$, let $X(\omega):=\lim X_{n}(\omega)$ (which is either $\limsup X_{n}(\omega)$ or $\liminf X_{n}(\omega)$, as you prefer, and thus measurable) for those $\omega$ for which the limit exists. Since the difference of two measurable functions is measurable, and taking absolute values remains measurable, then the following set \begin{align*} \{\omega\in\Omega:\lim X_{n}(\omega) &=X(\omega)\}=\{\omega\in\Omega:\forall k\in\mathbb{N}\exists m\in\mathbb{N}\,\,\mathrm{s.t.}\,\,\forall n\geq m\,\,|X_{n}(\omega)-X(\omega)|\leq\frac{1}{k}\} \\ &=\bigcap_{k\in\mathbb{N}}\bigcup_{m\in\mathbb{N}}\bigcap_{n\geq m}\{\omega\in\Omega:|X_{n}(\omega)-X(\omega)|\leq\frac{1}{k}\} \end{align*} is as well measurable. I.e. $\{ \lim X_{n}\,\,\mathrm{exists}\}$ is measurable.

I will try to answer all your questions below in the comments section. If there was something that seemed to be false, please make a remark below so I may correct it.

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thank you for the answer and comments. I have never taken topology, so I am somewhat unclear on the necessity of the topological pair $(X,\tau)$ at the outset. Is $\tau$ just the standard topology here? – Justin Jun 11 '12 at 18:30
I also read the first question as only needing to show the first inclusion, ie $\sigma(\tilde{X})\subset\sigma(\mathcal{C})$ – Justin Jun 11 '12 at 18:39
@Justin: Topology is just the family of open sets. If it confuses you, you can ignore that part and simply think of $\mathscr{B}(X)$ being the Borel $\sigma$-algebra of $X$ the way you have it defined. – T. Eskin Jun 11 '12 at 19:44
I have been working on this over the last few days and I completely understand 2 and 3 above. On 1. I am still a bit unsure. Saying $\sigma(\tilde{X})=\{X_n\le a\},n\in\mathbb{N},a\in\mathbb{R}$ it appears that there is a sigma field on one side and a single event on the other. Could you expand on part 1 at all? – Justin Jun 15 '12 at 17:41
@Justin: What I meant was, that $\sigma(\tilde{X})=\sigma(\{\{X_{n}\leq a\}:a\in\mathbb{R},\,\,n\in\mathbb{N}\})$, i.e. that $\{X_{n}\leq a\}$ form of sets are generators of $\sigma(\tilde{X})$. This is straight from the definition, as $\sigma(\tilde{X})$ is the smallest $\sigma$-algebra for which respect all the functions $X_{n}$ are measurable, and $X_{n}$ to be measurable is equivalent of saying that $\{X_{n}\leq a\}$ is a measurable set for all $a\in\mathbb{R}$ since $\{]-\infty,a]:a\in\mathbb{R}\}$ generates the Borel algebra of $\mathbb{R}$. – T. Eskin Jun 16 '12 at 23:18

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