书上传了我们光下载不钻研是不好的.以后我们应该合力把这些经典书的习题都做一遍, 整理一套解答出来.
我先来第一道.
Exercise 1.1.1
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If <bblatex>{{\cal F}_i},i \in I</bblatex> are <bblatex>\sigma</bblatex>-fields then <bblatex>{ \cap _{i \in I}}{\cal F}_i</bblatex> is. Here I <bblatex>\ne \emptyset</bblatex> is an arbitrary index set (i.e., possibly uncountable).
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Use the result in (1) to show if we are given a set <bblatex>\Omega</bblatex> and collection <bblatex>{\cal A}</bblatex> of subsets of <bblatex>\Omega </bblatex>, then there is a smallest <bblatex>\sigma </bblatex>-field containing <bblatex>{\cal A}</bblatex>. We will call this the <bblatex>\bf \sigma </bblatex>-field generated by <bblatex>\bf{\cal A}</bblatex> and denote it by <bblatex>\sigma \left( {\cal A} \right)</bblatex>.
1. Proof.
The empty set is in <bblatex>{{\cal F}_i}</bblatex>, <bblatex>\forall i \in I</bblatex>. Therefore it is also in <bblatex>{ \cap _{i \in I}}{{\cal F}_i}</bblatex>.
<bblatex>\forall A \in { \cap _{i \in I}}{{\cal F}_i} \Rightarrow \forall i \in I, A \in {{\cal F}_i} \Rightarrow {A^c} \in {{\cal F}_i},\forall i \in I \Rightarrow {A^c} \in { \cap _{i \in I}}{{\cal F}_i}</bblatex>
On the other hand,
<bblatex>\forall k \in \Bbb Z ^ + </bblatex> if <bblatex>{A_k} \in { \cap _{i \in I}}{\cal F}</bblatex>, then for this <bblatex>k</bblatex>, we first have
<bblatex>{\forall i \in I,{A_k} \in {{\cal F}_i}}</bblatex> and then also have
<bblatex> \cup _{k = 1}^\infty {A_k} \in {{\cal F}_i}</bblatex> for any particular <bblatex>i</bblatex>, because <bblatex>{{{\cal F}_i}}</bblatex> is a <bblatex>\sigma</bblatex>-field.
Since this last belonging relation holds for all <bblatex>{i \in I}</bblatex>, we must have
<bblatex> \cup _{k = 1}^\infty {A_k} \in { \cap _{i \in I}}{{\cal F}_i}</bblatex>
This completes the check for the satisfaction of <bblatex>{ \cap _{i \in I}}{{\cal F}_i}</bblatex> against the 3 <bblatex>\sigma</bblatex>-field axioms.
2. Proof.
Existence: Obviously <bblatex>{2^\Omega}</bblatex> is a <bblatex>\sigma</bblatex>-field containing <bblatex>{\cal A}</bblatex>.
Uniqueness: Suppose <bblatex>\left\{ {{{\cal F}_i}:i \in I} \right\}</bblatex> is the collection of all <bblatex>\sigma</bblatex>-fields containing <bblatex>{\cal A}</bblatex>.
Then by (1), <bblatex> \cap \left\{ {{{\cal F}_i}:i \in I} \right\}</bblatex> is also a <bblatex>\sigma</bblatex>-field; and it contains <bblatex>{\cal A}</bblatex>.
(By convention, the intersection symbol applied to a collection gives the intersection of all sets in the collection.)
Since, by definition, this intersection cannot have a proper subset that belongs to the original collection. (Any <bblatex>\sigma</bblatex>-field in the collection must contain the intersection.), the intersection becomes the smallest <bblatex>\sigma</bblatex>-field containing <bblatex>{\cal A}</bblatex>.
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