# Measure theory and independent sets

1. Mar 1, 2007

### aresnick

1. The problem statement, all variables and given/known data
Let $$\mathscr{X}$$ be a set, $$\mathscr{F}$$ a $$\sigma-$$field of subsets of S, and $$\mu$$ a probability measure on $$\mathscr{F}$$. Suppose that $$A_{1},\ldots,A_{n}$$ are independent sets belonging to $$\mathscr{F}$$. Let $$\mathscr{F}_{k}$$ be the smallest subfield of $$\mathscr{F}$$ containing $$A_{1}, \ldots, A_{k}$$. Show that if $$A \in \mathscr{F}_{k}$$, then $$A, A_{k+1}, \ldots, A_{n}$$ are indepdendent.

2. Relevant equations
Two sets are independent iff $$\mu(A \cap B) = \mu(A)\mu(B)$$.

3. The attempt at a solution
Really, my question here is what the smallest field is. It seems that, given a set $$\mathscr{X}$$, the smallest field $$\mathscr{F}_{s}$$ containing it is simply $$\left\{\emptyset, \{ \mathscr{X}\}\right\}$$. Am I just crazy?

Last edited: Mar 1, 2007
2. Mar 1, 2007

### AKG

It asks you to find the smallest subfield containing A1, ..., Ak. If someone asks you for the smallest natural greater than 10, the answer is 11, even if 0 is the smallest natural.

3. Mar 1, 2007

### aresnick

But I'm suggesting that the smallest set that contains $$A_{1} \ldots A_{k}$$ is simply a set containing $$\left\{A_{1} \ldots A_{k}\right\}$$ and the empty set (since every set must contain the empty set). It just seemed like a very simple object, in that case.

Does that make sense?

4. Mar 1, 2007

### AKG

No.

First of all, do you recognize the difference between

$$A_1 \in \{ A_1, \dots , A_k\}$$

and

$$A_1 \subset \{ A_1 ,\dots ,A_k \}$$

The first line is always true, the second is usually not. {A1, ..., Ak} is a set, but it's elements aren't just any elements, they are sets too! It is a set of sets. Normally, to avoid confusion, we like to say "collection of sets" instead of "set of sets," but they mean the same thing. The things Ai are also sets, but their elements will be elements of X. So Ai is a subset of X, {A1, ..., Ak} is a collection/set of subsets of X, i.e. {A1, ..., Ak} is an element of the power set of X.

Another point of confusion might be the word "contain." Does "x contains y" mean $y \in x$ or $y \subset x$. It could mean either, depending on the context. So note the following:

$$A_1 \in \{ A_1 ,\dots , A_k\}$$
An element of a collection of sets is a set.

$$A_1 \not\subset \{ A_1 ,\dots , A_k\}$$
(except in some weird situations that don't concern us here)

$$\{ A_1, A_2 \} \subset \{ A_1 ,\dots , A_k\}$$
$$\{ A_1\} \subset \{ A_1 ,\dots , A_k\}$$
A subset of a collection of sets is itself a collection of sets.

$$\{ A_1 \} \notin \{ A_1 ,\dots , A_k\}$$
(except in weird situations)

So the question gives you a set X. To make things very clear, we will use lower case roman letters when denoting elements of X, like x. We will use upper case roman letters when denoting subsets of X, like Y. We will use capital script letters when denoting collections of subsets of X, like $\mathcal{F}$. So we have:

$$(X, \mathcal{F} , \mu )$$

our probability space. It then says that for i = 1, ..., n:

$$A_i \in \mathcal{F}$$

with the Ai being independent. You want to consider the subfield (not just any old subcollection) $\mathcal{F}_k \subset \mathcal{F}$ such that for all i = 1, ..., k:

$$A_i \in \mathcal{F}_k$$

with $\mathcal{F}_k$ as small as possible. Note that $\mathcal{F} \subset \mathcal{F}$ and for all i = 1, ..., k:

$$A_i \in \mathcal{F}$$

but $\mathcal{F}$ is generally not the smallest.

So what is the smallest subfield of $\mathcal{F}$ the contains A1 through Ak? It's probably not going to be {A1, ..., Ak} because although that is a collection of sets, it is probably not an $\sigma$-field. An $\sigma$-field is a special kind of collection.

$$\{ \emptyset , A_1 ,\dots , A_k \}$$

is probably not a subfield either. In order to figure out what $\mathcal{F}_k$ should be, you need to first make sure you understand the meaning of $\sigma$-field and subfield. Come back when you know those definitions and you've absorbed the above.