Exercise 2

1. If $μ_{1}, \dots, μ_{n}$ are measures and $a_{1}, \dots, a_{n} \in (0, \infty)$ , then $\sum_{i = 1}^{n} a_{i} μ_{i}$ is a measure.

\begin{proof} Let $μ = \sum_{i = 1}^{n} a_{i} μ_{i}$ . Since $μ_{i}$ are measures and $a_{i} > 0$ , for any $A$ , $μ (A) = \sum_{i = 1}^{n} a_{i} μ_{i} (A) ⩾ 0$ and $μ_{i} (\emptyset) = 0$ . Then $μ (\emptyset) = \sum_{i = 1}^{n} a_{i} μ_{i} (\emptyset) = 0$ .

For any pairwise disjoint ${A_{i}}_{i = 1}^{\infty}$ , we have that $μ_{i} (\cup_{i = 1}^{\infty} A_{i}) = \sum_{i = 1}^{\infty} μ_{i} (A_{i})$ . Then

μ (\cup_{i = 1}^{\infty} A_{i}) = i = 1 \sum n a_{i} μ_{i} (\cup_{i = 1}^{\infty} A_{i}) = i = 1 \sum n a_{i} (i = 1 \sum \infty μ_{i} (A_{i})) = i = 1 \sum \infty i = 1 \sum n a_{i} μ_{i} (A_{i}) = i = 1 \sum \infty μ (A_{i}) .

Therefore, $\sum_{i = 1}^{n} a_{i} μ_{i}$ is a measure. \end{proof}

2. Let $(X, A, μ)$ be a measure space and ${E_{j}}_{j = 1}^{\infty} \subset A$ .

i) Show that $μ (lim inf E_{j}) \leq lim inf μ (E_{j})$ .
ii) Assume that there exists $A \in A$ with $μ (A) < \infty$ such that $E_{n} \subset A$ for $n \in N$ . Show that $μ (lim sup E_{j}) \geq lim sup μ (E_{j})$ .

\begin{proof} i) For any subsequence ${E_{k_{j}}}_{j = 1}^{\infty}$ satisfying $lim_{j \to \infty} μ (E_{k_{j}}) = a$ , we aim to show $μ (lim in f E_{j}) ⩽ a$ . If it is true, then any limit point of ${μ (E_{j})}$ is greater than or equal to $μ (lim in f E_{j})$ and so $μ (lim in f E_{j}) ⩽ lim in f μ (E_{j})$ . Note that $lim in f E_{j} \subseteq lim in f E_{k_{j}}$ , so we only need to prove $μ (lim in f E_{k_{j}}) ⩽ lim μ (E_{k_{j}})$ . Define $F_{j} = E_{k_{j}}$ , and it is equivalent to show $μ (lim in f F_{j}) ⩽ lim μ (F_{j})$ .

For any $N \in N_{+}$ , let $M_{N} = \cup_{n = 1}^{N} \cap_{m = n}^{\infty} F_{m} = \cap_{m = N}^{\infty} F_{m}$ . For any $n ⩾ N$ , there is $F_{n} \supseteq \cap_{m = N}^{\infty} F_{m} = M_{N}$ and so $μ (F_{n}) ⩾ μ (M_{N})$ . Take $n \to \infty$ , then $lim_{n \to \infty} μ (F_{n}) ⩾ μ (M_{N})$ for any $N \in N_{+}$ . Take $N \to \infty$ , then $lim_{n \to \infty} μ (F_{n}) ⩾ μ (\cup_{n = 1}^{\infty} \cap_{m = n}^{\infty} F_{m}) = μ (lim in f_{n \to \infty} F_{n})$ , that is, $μ (lim in f F_{j}) ⩽ lim μ (F_{j})$ .

ii) Similarly, note that $lim sup E_{j} \supseteq lim sup E_{k_{j}}$ for any subsequence ${E_{k_{j}}}_{j = 1}^{\infty}$ . To show any limit point of ${μ (E_{j})}$ is less than or equal to $μ (lim sup E_{j})$ , it suffices to show $μ (lim sup E_{k_{j}}) ⩾ lim μ (E_{k_{j}})$ . Define $F_{j} = E_{k_{j}}$ , then we only need to show $μ (lim sup F_{j}) ⩾ lim μ (F_{j})$ .

For any $N \in N_{+}$ , let $L_{N} = \cap_{n = 1}^{N} \cup_{m = n}^{\infty} F_{m} = \cup_{m = N}^{\infty} F_{m}$ . For any $n ⩾ N$ , there is $F_{n} \subseteq \cup_{m = N}^{\infty} F_{m} = L_{N}$ and so $μ (F_{n}) ⩽ μ (L_{N})$ . Take $n \to \infty$ , then $lim_{n \to \infty} μ (F_{n}) ⩽ μ (L_{N})$ for any $N \in N_{+}$ . Since $L_{n} \subseteq A$ with $μ (A) < \infty$ , take $N \to \infty$ , then $lim_{n \to \infty} μ (F_{n}) ⩽ μ (\cap_{n = 1}^{\infty} \cup_{m = n}^{\infty} F_{m}) = μ (lim sup_{n \to \infty} F_{n})$ , that is, $μ (lim sup F_{j}) ⩾ lim μ (F_{j})$ . \end{proof}

3. If $(X, A, μ)$ is a measure space and $E, F \in A$ , then $μ (E) + μ (F) = μ (E \cup F) + μ (E \cap F)$ .

\begin{proof} Since $μ (E) = μ (E \cap F^{c}) + μ (E \cap F)$ and $μ (F) = μ (F \cap E) + μ (F \cap E^{c})$ , we have that

μ (E) + μ (F) = μ (E \cap F) + μ (E \cap F^{c}) + μ (E \cap F) + μ (E^{c} \cup F) = μ (E \cap F) + μ ((E \cap F^{c}) \cup (E \cap F) \cup (E^{c} \cup F)) = μ (E \cap F) + μ (E \cup F) .

So we finish the proof. \end{proof}

4. Let $A$ be a $σ$ -algebra on $X$ . Let $μ : A \to [0, \infty]$ be a finite additive set function. Show that if ${A_{n}}_{n = 1}^{\infty}$ is a family of disjoint subsets in $A$ then

μ (\cup_{n = 1}^{\infty} A_{n}) \geq n = 1 \sum \infty μ (A_{n}) .

\begin{proof} Since $μ$ is a finite additive set function, then for any $N \in N_{+}$ , there is $μ (\cup_{n = 1}^{N} A_{n}) = \sum_{n = 1}^{N} μ (A_{n})$ . Let $B_{N} = \cup_{n = N + 1}^{\infty} A_{n}$ . Since $A$ is a $σ$ -algebra, then $B_{N} \in A$ . Then $μ (\cup_{n = 1}^{\infty} A_{n}) = μ (\cup_{n = 1}^{N} A_{n}) + μ (B_{N}) ⩾ \sum_{n = 1}^{N} μ (A_{n})$ for any $N \in N_{+}$ . Take $N \to \infty$ , then $μ (\cup_{n = 1}^{\infty} A_{n}) \geq \sum_{n = 1}^{\infty} μ (A_{n})$ . \end{proof}

5. Let $(X, A, μ)$ be a probability space and ${A_{j}}_{j = 1}^{\infty} \subset A$ be a sequence of sets such that $μ (A_{j}) = 1, \forall j$ . Show that $μ (\cap_{j = 1}^{\infty} A_{j}) = 1$ .

\begin{proof} As $μ (X) = μ (A_{j}) = 1$ for all $j$ , then $μ (A_{j}^{c}) = 0$ for all $j$ . It follows that

μ (\cap_{j = 1}^{\infty} A_{j}) = μ ((\cup_{j = 1}^{\infty} A_{j}^{c})^{c}) = 1 - μ (\cup_{j = 1}^{\infty} A_{j}^{c}) = 1 - 0 = 1.

Therefore, $μ (\cap_{j = 1}^{\infty} A_{j}) = 1$ . \end{proof}

8. Assume that $E_{1}, E_{2}, \dots$ are measurable subsets of the measure space $(X, A, μ)$ with the property that for $n \neq = m, μ (E_{n} \cap E_{m}) = 0$ . Show that

μ (\cup_{n = 1}^{\infty} E_{n}) = n = 1 \sum \infty μ (E_{n}) .

\begin{proof} Since $E_{i}$ are measurable subsets, then $μ (E) = μ (E \cap A) + μ (E \cap A^{c})$ for all $A \subseteq X$ . It yields that

μ (E_{1} \cup E_{2}) = μ ((E_{1} \cup E_{2}) \cap E_{1}) + μ ((E_{1} \cup E_{2}) \cap E_{2}^{c}) = μ (E_{1}) + μ (E_{2}) .

Repeat the procedure above, then we have that $μ (\cup_{n = 1}^{N} E_{n}) = \sum_{n = 1}^{N} μ (E_{n})$ . Since $\cup_{n = 1}^{\infty} E_{n} \supseteq \cup_{n = 1}^{N} E_{n}$ , then $μ (\cup_{n = 1}^{\infty} E_{n}) ⩾ μ (\cup_{n = 1}^{N} E_{n}) = \sum_{n = 1}^{N} μ (E_{n})$ . Take $N \to \infty$ , there is

μ (\cup_{n = 1}^{\infty} E_{n}) ⩾ n = 1 \sum \infty μ (E_{n}) .

On the other hand, $\sum_{n = 1}^{N} μ (E_{n}) ⩽ \sum_{n = 1}^{\infty} μ (E_{n})$ and so $μ (\cup_{n = 1}^{N} E_{n}) ⩽ \sum_{n = 1}^{\infty} μ (E_{n})$ . Take $N \to \infty$ , there is

μ (\cup_{n = 1}^{\infty} E_{n}) ⩽ n = 1 \sum \infty μ (E_{n}) .

Therefore, $μ (\cup_{n = 1}^{\infty} E_{n}) = \sum_{n = 1}^{\infty} μ (E_{n})$ . \end{proof}

9. Let $(X, A)$ be a measurable space and assume that $μ : A \to [0, \infty]$ finitely additive and $σ$ -subadditive. Show that $μ$ is $σ$ -additive.

\begin{proof} Note that for a family ${A_{i}}_{i = 1}^{\infty}$ which is pairwise disjoint, there is

μ (\cup_{i = 1}^{n} A_{i}) μ (\cup_{i = 1}^{\infty} A_{i}) = i = 1 \sum n μ (A_{i}), ⩽ i = 1 \sum \infty μ (A_{i}) .

Since

μ (\cup_{i = 1}^{\infty} A_{i}) = μ (\cup_{i = 1}^{n} A_{i}) + μ (\cup_{i = n + 1}^{\infty} A_{i}) ⩾ μ (\cup_{i = 1}^{n} A_{i}) = i = 1 \sum n μ (A_{i}),

then take $n \to \infty$ and $μ (\cup_{i = 1}^{\infty} A_{i}) ⩾ \sum_{i = 1}^{\infty} μ (A_{i})$ . Therefore, $μ (\cup_{i = 1}^{\infty} A_{i}) = \sum_{i = 1}^{\infty} μ (A_{i})$ , that is, $μ$ is $σ$ -additive. \end{proof}

10. Let $(X, A, μ)$ be a finite measure space and $S = {A \in M : μ (A) = 0 \mbox or μ (A) = μ (X)}$ . Prove that $S$ is a $σ$ -algebra of subsets of $X$ .

\begin{proof} Since $μ (\emptyset) = 0$ and $μ (X) = μ (X)$ , then $\emptyset, X \in S$ . For any $A \in S$ , if $μ (A) = 0$ , then $μ (A^{c}) = μ (X) - μ (A) = μ (X)$ ; if $μ (A) = X$ , then $μ (A^{c}) = μ (X) - μ (A) = 0$ . It yields that $A^{c} \in S$ if $A \in S$ . For ${A_{i}}_{i = 1}^{\infty} \subseteq S$ , if $μ (A_{i}) = 0$ for all $i$ , then $μ (\cup_{i = 1}^{\infty} A_{i}) ⩽ \sum_{i = 1}^{\infty} μ (A_{i}) = 0$ ; otherwise, there is a $μ (A_{k}) = μ (X)$ , $μ (\cup_{i = 1}^{\infty} A_{i}) ⩾ μ (A_{k}) = μ (X)$ and $μ (\cup_{i = 1}^{\infty} A_{i}) = μ (X)$ . Thus for any ${A_{i}}_{i = 1}^{\infty} \subseteq S$ , $\cup_{i = 1}^{\infty} A_{i} \in S$ . Therefore, $S$ is a $σ$ -algebra. \end{proof}

11. Let $A$ be an algebra of subsets of $X, μ$ a nonnegative set function on $A$ such that $μ (\emptyset) = 0$ , and $C = {C \in A : μ (A) = μ (A \cap C) + μ (A - C) \mbox f or a ll A \in A}$ . Prove:

(a) $C$ is an algebra of subsets of $X$ .
(b) The restriction of $μ$ to $C$ is finitely additive,
(c) If ${C_{1}, \dots, C_{n}} \subset C$ are pairwise disjoint and $A \in A$ , then $μ (\cup_{k = 1}^{n} (C_{k} \cap A)) = \sum_{k = 1}^{n} μ (C_{k} \cap A)$ .

\begin{proof} (a) Note that $μ (A) = μ (A \cap \emptyset) + μ (A - \emptyset) = μ (A \cap X) + μ (A - X)$ for all $A \in A$ . It yields that $\emptyset, X \in A$ . For any $C \in C$ ,

μ (A) = μ (A \cap C) + μ (A - C) = μ (A - C^{c}) + μ (A \cap C^{c})

for all $A \in A$ and so $C^{c} \in C$ . Let $C_{1}, C_{2} \in C$ . Notice that

= = = = = μ (A \cap (C_{1} \cup C_{2})) + μ (A \cap (C_{1} \cup C_{2})^{c}) μ ((A \cap C_{1}) \cup (A \cap C_{2})) + μ (A \cap (C_{1}^{c} \cap C_{2}^{c})) μ (((A \cap C_{1}) \cup (A \cap C_{2})) \cap C_{1}) + μ (((A \cap C_{1}) \cup (A \cap C_{2})) \cap C_{1}^{c}) + μ (A \cap C_{1}^{c}) - μ (A \cap C_{1}^{c} \cap C_{2}) μ (A \cap C_{1}) + μ (A \cap C_{2} \cap C_{1}^{c}) + μ (A \cap C_{1}^{c}) - μ (A \cap C_{1}^{c} \cap C_{2}) μ (A \cap C_{1}) + μ (A \cap C_{1}^{c}) μ (A) .

Therefore, $C_{1} \cup C_{2} \in C$ . For ${C_{i}}_{i = 1}^{n} \subseteq C$ , we have that $D_{2} := C_{1} \cup C_{2} \in C$ , $D_{3} = D_{2} \cup C_{3} \in C, \dots, D_{n} = D_{n - 1} \cup C_{n} \in C$ . That is, $D_{n} = \cup_{i = 1}^{n} C_{i} \in C$ . Hence, $C$ is an algebra.

(b) For any pairwise disjoint family ${C_{i}}_{i = 1}^{n}$ , we have that

μ (\cup_{i = 1}^{n} C_{i}) = μ ((\cup_{i = 1}^{n} C_{i}) \cap C_{n}) + μ ((\cup_{i = 1}^{n} C_{i}) \cap C_{n}) = μ (\cup_{i = 1}^{n - 1} C_{i}) + μ (C_{n}) = \dots = μ (C_{1}) + \dots + μ (C_{n}) .

Therefore, $μ$ is finitely additive.

μ (\cup_{k = 1}^{n} (C_{k} \cap A)) = μ ((\cup_{k = 1}^{n} (C_{k} \cap A)) \cap C_{n}) + μ ((\cup_{k = 1}^{n} (C_{k} \cap A)) \cap C_{n}^{c}) = μ (\cup_{k = 1}^{n - 1} (C_{k} \cap A)) + μ (C_{n} \cap A) = \dots = k = 1 \sum n μ (C_{k} \cap A) .

Now we finish the proof. \end{proof}

12. Let $(X, A, μ)$ be a finite measure space and $ν$ a positive, finitely additive set function on $A$ . Prove that if for any sequence ${A_{n}} \subset A$ with $μ (A_{n}) \to 0$ we have $ν (A_{n}) \to 0$ , then $ν$ is a measure on $A$ .

\begin{proof} Let $A_{n} = \emptyset$ for all $n \in N$ . Then ${A_{n}} \subseteq A$ and $μ (A_{n}) \to 0$ as $μ (A_{n}) = 0$ . Thus $ν (A_{n}) = c$ for a constant $c$ and $ν (A_{n}) \to 0$ yield that $c = 0$ . Hence $ν (\emptyset) = 0$ . For a pairwise disjoint ${A_{n}}_{n = 1}^{\infty}$ , define $B_{n} = \cup_{i = 1}^{n} A_{i}$ and $B = \cup_{i = 1}^{\infty} A_{i}$ . Since $ν$ is finitely additive, we have that $ν (B_{n}) = \sum_{i = 1}^{n} ν (A_{i})$ . Since $(X, A, μ)$ be a finite measure space, then $μ (B) < \infty$ and $μ (B - B_{n}) \to 0$ by $B_{n} ↑ B$ . It follows that $ν (B - B_{n}) \to 0$ . Note that $ν (B) = ν (B - B_{n}) + ν (B_{n})$ , so we have that $ν (B_{n}) \to ν (B)$ , that is, $\sum_{i = 1}^{n} ν (A_{i}) \to ν (\cup_{i = 1}^{\infty} ν (A_{i}))$ . Therefore, we prove that $\sum_{i = 1}^{\infty} ν (A_{i}) = ν (\cup_{i = 1}^{\infty} ν (A_{i}))$ for any pairwise disjoint ${A_{n}}_{n = 1}^{\infty}$ , and so $ν$ is a measure on $A$ . \end{proof}

13. Let $(X, A, μ)$ be a measure space and $T : X \to Y$ a mapping from $X$ onto $Y$ . Prove that $N = {B \subset Y : T^{- 1} (B) \in A}$ is a $σ$ -algebra of subsets of $Y$ and that the set function $ν$ on $N$ given by $ν (B) = μ (T^{- 1} (B))$ is a measure, called the measure induced by $T$ .

\begin{proof} i) Since $T^{- 1} (\emptyset) = \emptyset \in A$ and $T^{- 1} (Y) = X \in A$ , then $\emptyset, Y \in N$ . For any $B \in N$ , $T^{- 1} (B^{c}) = (T^{- 1} (B))^{c} \in A$ and so $B^{c} \in N$ . For ${B_{i}}_{i = 1}^{\infty} \subseteq N$ , we have that $T^{- 1} (\cup_{i = 1}^{\infty} B_{i}) = \cup_{i = 1}^{\infty} T^{- 1} (B_{i}) \in A$ and so $\cup_{i = 1}^{\infty} B_{i} \in N$ . Therefore, $N$ is a $σ$ -algebra.

ii) Note that $ν (\emptyset) = μ (T^{- 1} (\emptyset)) = μ (\emptyset) = 0$ . For any pairwise disjoint ${B_{i}}_{i = 1}^{\infty}$ , ${T^{- 1} (B_{i})}_{i = 1}^{\infty}$ are also pairwise disjoint and $$ \nu(\cup_{i=1}^\infty B_i)=\mu(T^{-1}(\cup_{i=1}^\infty B_i))=\mu(\cup_{i=1}^\infty T^{-1}(B_i))=\sum_{i=1}^\infty\mu(T^{-1}(B_i))=\sum_{i=1}^\infty\nu(B_i).

Therefore, $\nu$ is a measure. `\end{proof}`

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