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I am currently Working on exercise 2.18 of Oksendals SDE's. The purpose of this post is mainly to check whether my answers are correct, except for part c were I am having some difficulty.

The question is as follows:

a) let $\Omega = \{1,2,3,4,5\}$ and let $\mathcal{U}$ be the collection

$$ \mathcal{U} = \{\{1,2,3\},\{3,4,5\}\}$$ of subsets of $\Omega$. Find the smallest $\sigma$-algebra containing $\mathcal{U}$ (i.e. the $\sigma$-algebra $\mathcal{H}_{\mathcal{U}}$ generated by $\mathcal{U}$).

b) Define $X: \Omega \to \mathbb{R}$ by

$$ X(1) = X(2)=0,\ \ X(3)=10, \ \ X(4)=X(5)=1$$ is $X$ measuarable with respect to $\mathcal{H}_{\mathcal{U}}$?

c)Define $Y:\Omega \to \mathbb{R}$ by

$$Y(1)=0, \ \ Y(2)=Y(3)=Y(4)=Y(5)=1$$

Find the $\sigma$-algebra $\mathcal{H}_{Y}$ generated by $Y$.

My solutions

a) To obtain the smallest $\sigma$-algebra containing $\mathcal{U}$ I simply need to add the missing sets that make it a $\sigma$-algebra, So I need to add all the sets, such that the resulting set is closed under countable Unions, Intersections and closed under Complement

Hence is $\mathcal{H}_{\mathcal{U}}$ the following

$$\mathcal{H}_{\mathcal{U}}=\{ \{1,2,3,\},\{3,4,5\},\{1,2,3,4,5\},\{3\},\{4,5\},\emptyset, \{1,2\}, \{1,2,4,5\} \} \ ?$$

b) For a probability space $(\Omega, \mathcal{F}, P)$, then a function $X:\Omega \to \mathbb{R}^n$ is called $\mathcal{F}$-measurable if

$$X^{-1}(U) := \{ \omega \in \Omega ; X(\omega) \in U\} \in \mathcal{F}$$

for all open sets $U \in \mathbb{R}^n$.

Hence in the question above $X$ is $\mathcal{H}_{\mathcal{U}}$-measurable since the inverse abides by the definition given above ?.

c) This part I am rather quite stuck on. In Oksendals SDE's the $\sigma$-algebra $\mathcal{H}_{Y}$ generated by $Y$ is the smallest $\sigma$-algebra on $\Omega$ containing all the sets

$$ Y^{-1}(U); \ \ U\subset \mathbb{R}^n \ open.$$

i.e.

$$\mathcal{H}_{Y}=\{Y^{-1}(B);B \in \mathcal{B}\}, $$

Where $\mathcal{B}$ is the Borel $\sigma$-algebra on $\mathbb{R}^n$.

However, I am unsure how to use this to construct the $\sigma$-algebra? Part of my difficulty I think stems from the fact that I dont fully understand what a Borel $\sigma$-algebra is.

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  • $\begingroup$ b) take $\mathcal{U} = (0.5, 10.5)$ - it seems to be open, but its preimage is $\{3,4,5\}$ which is not in $\mathcal{H}_{\mathcal{U}}$ $\endgroup$
    – dEmigOd
    Commented Apr 4, 2018 at 19:13

1 Answer 1

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(a): You forgot about $\{ 3,4,5 \}$ itself.

(b): You have a typo ($X^{-1}(U)$ not just $X^{-1}$). You are otherwise right. The main point is that this $\sigma$-algebra does not know the difference between 1 and 2, or the difference between 4 and 5. So any measurable function wrt this $\sigma$-algebra must assign the same value to 1 and 2 and the same value to 4 and 5. Conversely any function with those two properties will be measurable wrt this $\sigma$-algebra.

(c): Generally you do not "construct" the Borel $\sigma$-algebra, you just work with its basic definition. Describing it in more detail is a major undertaking but it falls under the heading of descriptive set theory, not really measure theory. But this question has nothing to do with that. You just have a function and you ask for the smallest $\sigma$-algebra which makes that function measurable. (More precisely, it is the intersection of all $\sigma$-algebras making that function measurable). If you understood my comment in part (b), you should be able to figure this out.

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  • $\begingroup$ Thanks for the reply! With regards to your (b) if $\mathcal{H}_{\mathcal{U}}$ also contained the singleton sets {1} and {2} say, would X be no longer $\mathcal{F}$-measurable? since it would need to know the difference between 1 and 2? $\endgroup$
    – seraphimk
    Commented Apr 4, 2018 at 19:30
  • $\begingroup$ @seraphimk You have it backwards. If $X$ assigned different values to 1 and 2 then it would no longer be $\mathcal{H}_U$ measurable unless you added $\{ 1 \}$ and $\{ 2 \}$ to $\mathcal{H}_U$. $\endgroup$
    – Ian
    Commented Apr 4, 2018 at 19:33
  • $\begingroup$ Apologies It was a typo that is what I was trying to say. $\endgroup$
    – seraphimk
    Commented Apr 4, 2018 at 19:37
  • $\begingroup$ For part c would $\mathcal{H}_{Y} = \{ \{0\}, \{2,3,4,5\},\emptyset, \} $ $\endgroup$
    – seraphimk
    Commented Apr 4, 2018 at 19:54

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