Here is my homework 2

Problem 2.4: Let \(F_1, F_2, \ldots\) be classes of sets in a common space \(\Omega\).

Part a: Suppose that \(F_n\) are fields satisfying \(F_n \subset F_{n+1}\). Show that \(\cup_{n=1}^{\infty} F_n\) is a field.

Proof: If \(A \in \cup_{n=1}^{\infty} F_n\), then there exists some field \(F_k, k \geq 1\) such that \(A \in F_k\). Then \(F^c \in F_k\) and so \(F^c \in \cup_{n=1}^{\infty} F_n\). Since \(\emptyset \in F_n\) for all \(n>0\), we get that \(\Omega \in F_n\) for all \(n\) and thus \(\Omega \in \cup_{n=1}^{\infty} F_n\). Given a finite union \(\cup A_j \in \cup F_n\). Each \(A_j\) is an element of some \(F_k\), \(k \in \{1, \ldots, \}\). Since this union is finite, we can find the largest such \(F_k\) containing the other \(F_i\)'s. This is a field, and thus \(\cup A_j \in F_k \subset \cup F_n\). Thus, the union of fields is also a field.

Part b: Suppose that \(F_n\) are \(\sigma\)-fields satisfying \(F_n \subset F_{n+1}\). Show by example that \(\cup_{n=1}^{\infty} F_n\) need not be a \(\sigma\)-field.

Proof: Let \(F_n\) be the set of all possible subsets of \(\{1, \ldots, n\}\) where \(n \in \mathbb{N}\). Each \(F_n\) is a \(\sigma\)-field. However, if we consider \(E_n =\{2n\}\), then clearly \(E_n \in F_{2n}\). However, \(\cup_{n=1}^{\infty} E_n \in \cup F_n\) but \(\cup_{n=1}^{\infty} E_n \notin F_n\) for any \(n \in \mathbb{N}\). Thus, the union of \(\sigma\)-fields is not always a \(\sigma\)-field.

Problem 2.9: Show that if \(B \in \sigma(A)\), then there exists a countable subclass \(A_B\) of \(A\) such that \(B \in \sigma(A_b)\).

Proof: Let \(C = \sqcup \{ \sigma(F) | F \subseteq A \textrm{ and } F \textrm{ is countable} \}\). Then we claim that \(C\) is a \(\sigma\)-field. Proving the claim, notice that \(\Omega \in C\) since each \(\sigma(F)\) contains it. If \(E \in C\), then \(E\) is in one of the \(\sigma(F)\), and thus \(E^c \in \sigma(F)\) and so \(E^c \in C\). Finally, if \(E_1, \ldots, \in C\), then there is a subclass \(F\) that consists of the countable \(E_1, \ldots\). Thus, the \(\sigma\)-field generated by that subclass is also in \(C\). Thus, the union of the \(E_1, \ldots\) is also in \(C\). Thus, \(C\) is a \(\sigma\)-field.

Clearly, we can see that \(C \subseteq \sigma(A)\). If \(B \in A\), then \(B \in \sigma(B) \subseteq C\). Since \(B\) was arbitrary, we conclude that \(\sigma(A) \subseteq C\). Thus, \(C=\sigma(A)\). So then, given any \(B \in \sigma(A)\), we have that \(B \in C\). So then \(B\) lies in one of the possible \(\sigma(F)\), where \(F\) is countable. Set \(F = A_B\), then \(B \in \sigma(A_B)\).

Problem 2.12: Show that a \(\sigma\)-field cannot be countably infinite -- its cardinality must be finite or else at least that of the continuum. Show by example that a field can be countably infinite.

Proof: Assume for sake of contradiction, that there was a countably infinite \(\sigma\)-field, \(F\), over a base space \(\Omega\). If we consider the map \(f: \Omega \rightarrow F\) where \(f(x) = \cap_{x \in A} A\) where \(A \in S\). This sends an element, \(x\), to the smallest set containing it in \(S\). This map is injective. If \(f(x) \cap f(y) \neq \emptyset\) and \(x \notin f(y)\), then \(f(x) \setminus f(y) \in F\) which is a smaller set containing \(x\). Thus, \(x \in f(y)\) and \(y \in f(x)\). Since \(f\) maps our point to the smallest set in \(F\) containing it, \(x \in f(x) \subseteq f(y)\) and \(y \in f(y) \subseteq f(x)\). Thus, \(x=y\). Since each point gets mapped uniquely to a set containing it in \(\Omega\), we could write \(f(\Omega)\) as a union of sets in \(F\). Similarly, we can write any set in \(F\) as a union of sets in the image \(f(\Omega)\). So then \(|F| = |P(f(\Omega))|\). But this is uncountable, so we're done.

Problem 2.15: On the field \(B_0\) in \((0,1]\) define \(P(A)\) to be 1 or 0 according as there does or does not exist some positive \(\epsilon_A\) such that \(A\) contains the interval \((\frac{1}{2}, \frac{1}{2} + \epsilon_A]\). Show that \(P\) is finitely but not countably additive.

Proof: Let \(A_k, k \in \mathbb{N}\) be sets that for any \(\epsilon_k > 0\), we have that \(A_k\) does not contain \((\frac{1}{2}, \frac{1}{2} + \epsilon_k]\). Notice that \(P(\cup_{k=1}^{\infty} A_k) = P((\cap_{k=1}^{\infty} A_{k}^{c})^c) = 1 - P(\cap_{k=1}^{\infty} A_{k}^{c}) = 1\). The thing to note here is that as \(n \rightarrow \infty\) we get that \(P(\cap_{k=1}^{n} A_{k}^{c}) = 0\). However, we have that \(\sum_{k=1}^{\infty} P(A_k) = 0 \). Thus, \(P\) is not countably additive.

Problem 3.11:

Part a: Show that a \(\lambda\)-system satisfies the conditions:

\((\lambda_4) A,B \in L\) and \(A \cap B = \emptyset\) imply \(A \cup B \in L\)

\((\lambda_5) A_1, A_2, \ldots \in L\) and \(A_n \uparrow A\) imply \(A \in L\)

\((\lambda_6) A_1, A_2, \ldots \in L\) and \(A_n \downarrow A\) imply \(A \in L\)

Proof:

\((\lambda_4)\): If \(A,B \in L\) and \(A\cap B = \emptyset\). Then \(A \cup B \in L\) by \((\lambda_3)\).

\((\lambda_5)\): If \(A_1, \ldots \in L\) with \(A_n \subset A_{n+1}\) and \(A_n \uparrow A\). Let \(B_n = A_n \setminus A_{n-1}\). Then \(B_{n+1} \cap B_n = \emptyset\) and \(\cup B_i = \cup A_i\). Since each \(B_i\) is disjoint, then \(\cup B_i \in L\), so \(\cup A_i \in L\). Thus \(A \in L\).

\((\lambda_6)\): We know that \(\cap A_n = A\), so then \(\cup A_{n}^{c} = A^c\). Meaning that \(A_{n}^{c} \subset A_{n+1}^{c}\). Let \(B_n = A_{n+1}^{c} \setminus A_{n}^c\). Note that the \(B_n\)'s are disjoint and that \(\cup B_n \in F\). However, we also have that \(\cup B_n = A^c\). Then this means that \(A^c \in F\) and thus \(A \in F\).

Part b: Show that \(F\) is a \(\lambda\)-system if and only if it satsifies \((\lambda_1), (\lambda_2 '),\) and \((\lambda_5)\).

Proof: Since \(\Omega \in F\) and \(A \subset \Omega\), then \(\Omega \setminus A = A^c \in F\). Next, given a sequence, \(A_1, \ldots\) of sets with \(A_m \cap A_n = \emptyset\). Note that \(A_m \subset A_{n}^{c}\) so then \(A_{n}^c - A_{m} = A_{n}^{c} \cap A_{m}^{c} \in F\). Then we see that \(A_m \cup A_n = (A_{n}^{c} \cap A_{m}^{c})^c \in F\). So the finite union of sets is still within \(F\). Now, let \(B_n = \cup_{i=1}^{n} A_i \in F\). The sequence \(B_1, \ldots\) is increasing and thus converges by \((\lambda_5)\) to \(A\). We have already shown that a \(\lambda\)-system satisfies the previous conditions before. So we are done.

Problem 3.14: Prove the existence of a Lebesgue set of Lebesgue measure 0 that is not a Borel set.

Proof:

Problem 3.16: Deduce directly by countable additivity that the Cantor set has Lebesgue measure 0.

Proof: At every iteration, the Cantor set has length \(\frac{2^n}{3^n}\) for its \(n\)th iteration. So then as \(n \rightarrow \infty\), we have that \(\lambda(C_n) = 0\).