Theorem (Chain rule) Let $I,J$ be intervals and $f,g$ be two functions on them respectively with $g(J) \sub I$. Suppose that $g$ is differentiable at a point $a$ and $f$ is differentiable at a point $b = g(a)$. Then the composition of the functions is differentiable at point $a$ and $$ (f\circ g)'(a) = f'(g(a))g'(a) $$
Proof Define the function
$$ \Phi(y) = \begin{cases} \dfrac{f(y) - f(b)}{y - b} & \text{if } y \neq b \\ f'(b) & \text{if } y = b \end{cases} $$
This function is continuous at $b$ because $\lim_{y \to b}\Phi(y) = \Phi(b)$. So for any $y \in I$, we have $f(y) - f(b) = \Phi(y)(y - b)$. If we substitute $y = g(x)$ and divide the equality by $(x - a)$ for $x \neq a$, we get
$$ \frac{f(g(x)) - f(g(a))}{x - a} = \Phi(g(x))\frac{g(x) - g(a)}{x - a} $$
Finally, if we take the limit as $x \to a$, we get
$$ (f \circ g)'(a) = f'(g(a))g'(a) $$
$\blacksquare$
Proposition Every open set of $\mathbb{R}$ can be written as countable disjoint union of open intervals.
Proof Let $\mathcal{O}$ be an open set. Take any $x \in \mathcal{O}$ and define $I_x = \bigcup \lbrace (a,b) : x \in (a,b) \sub \mathcal{O} \rbrace$. $I_x$ is not empty because $\mathcal{O}$ is open, and it is an open interval as a union of open intervals. For any $x$ and $y$ in $\mathcal{O}$ if $I_x \cap I_y \neq \empty$ then $I_x = I_y$. Therefore $\mathcal{O} = \bigcup_{x \in \mathcal{O}} I_x$ is a disjoint union of open intervals. Define a map $\lbrace I_x \rbrace _{x \in \mathcal{O}} \to \mathbb{Q}$ that assigns each connected interval to a rational number contained in it. This function is injective and hence the cardinality of domain is less than or equal to the cardinality of $\mathbb{Q}$. Therefore $\mathcal{O}$ is a countable union of open intervals. $\blacksquare$
Let $X$ be a set, $\mathcal{M} = \mathcal{P}(X)$ and $f: X \to [0, \infty] $ be a function. Then $$ \mu(E) = \sum_{x \in E}f(x) $$ defines a measure on $\mathcal{M}$.
Proposition $\mu$ is $\sigma$-finite iff $\mu$ is semifinite and $\lbrace x : f(x) > 0\rbrace$ is countable
Proof: Let $E$ be a set with $\mu(E) = \infty$. Since $\mu$ is $\sigma$-finite $E$ is also $\sigma$-finite and there exists a sequence of sets ${E_j}$ such that, $E = \bigcup_{j=1}^{\infty}E_j$ where $\mu(E_j) < \infty$ for all $j$. Moreover at least one of the sets $E_j$ should have a positive measure otherwise the measure of $E$ could not be $\infty$. That is there exists a $k$ such that $0 < \mu(E_k) < \infty$ which makes $\mu$ a semifinite measure.
Let $E = \lbrace x : f(x) > 0\rbrace$ and define $E_n = \lbrace x : f(x) > \frac{1}{n} \rbrace$ Then $E = \bigcup E_n$. Since $\mu$ is $\sigma$-finite there is a sequence of sets $\lbrace X_j \rbrace$ where $X = \bigcup X_j$ and $\mu(X_j) < \infty$ for each $j$.For a fixed $j$ and $n$ $E_n \cap X_j$ has a finite measure and $$ \mu(E_n \cap X_j) = \sum_{x \in E_n \cap X_j} f(x) \geq \sum_{x \in E_n \cap X_j}\frac{1}{n} $$ If the set $E_n \cap X_j$ were infinite, the above sum yields to infinity which is not possible since $\mu(E_n \cap X_j) \leq \mu(X_j) < \infty$. Therefore it must be finite. Moreover $E_n = \bigcup_{j=1}^{\infty}E_n \cap X_j$ is a countable union of finite sets, and is therefore countable. Since $E$ is a countable union of countable sets it is also countable.
Conversely, assume that $\mu$ is a semifinite measure and the set $E = \lbrace x : f(x) > 0 \rbrace$ is countable. We want to show that $\mu$ is $\sigma$-finite.
Since $E$ is countable, we can write its elements as a sequence, so $E = \bigcup_{n=1}^\infty \lbrace x_n \rbrace$. Let us check the measure of each singleton $\lbrace x_n \rbrace$. Suppose for the sake of contradiction that $\mu(\lbrace x_n \rbrace) = f(x_n) = \infty$ for some $n$. Since $\mu$ is semifinite, there must exist a subset $A \subset \lbrace x_n \rbrace$ such that $0 < \mu(A) < \infty$. However, the only subsets of a singleton are the empty set $\emptyset$ and the singleton itself. Since $\mu(\emptyset) = 0$, no such subset $A$ can exist. This is a contradiction. Therefore, we must have $\mu(\lbrace x_n \rbrace) = f(x_n) < \infty$ for all $n$.
Now, consider the complement of $E$, which is $X \setminus E$. For any $x \in X \setminus E$, we have $f(x) = 0$. Consequently, the measure of this complement is:
$$ \mu(X \setminus E) = \sum_{x \in X \setminus E} f(x) = 0 < \infty $$
We can express the entire space $X$ as the disjoint union of $X \setminus E$ and the singletons from $E$:
$$ X = (X \setminus E) \cup \bigcup_{n=1}^\infty \lbrace x_n \rbrace $$
Since $X$ is written as a countable union of sets ($X \setminus E$ and all $\lbrace x_n \rbrace$), each of which has finite measure, we conclude that $\mu$ is a $\sigma$-finite measure. $\blacksquare$
Theorem Let $(X, \mathcal{M}, \mu)$ be a measure space.
- (Monotonicity) If $E,F \in \mathcal{M}$ and $E \sub F$ then $\mu(E) \leq \mu(F)$
- (Subadditivity) If $\lbrace E_j \rbrace_{1}^{\infty} \sub \mathcal{M}$ then $\mu \big( \bigcup_{1}^{\infty} E_j \big) \leq \sum_{1}^{\infty} \mu(E_j)$
- (Continuity from below) If $\lbrace E_j \rbrace_{1}^{\infty} \sub \mathcal{M}$ and $E_1 \sub E_2 \sub \cdots $, then $\mu\big( \bigcup E_j \big) = \lim \mu(E_j)$
- (Continuity from above) If $\lbrace E_j \rbrace_{1}^{\infty} \sub \mathcal{M}$ and $E_1 \supset E_2 \supset \cdots $,and $\mu(E_j) < \infty$ then $\mu\big( \bigcap E_j \big) = \lim \mu(E_j)$
Gerald B. Folland Real Analysis Solutions
Section 1.2 $\sigma$-Algebras
1. A family of sets $\mathcal{R} \subset \mathcal{P}(X)$ is called a ring if it is closed under finite unions and differences (i.e., if $E_1,\ldots,E_n \in \mathcal{R}$, then $\bigcup_{j=1}^{n} E_j \in \mathcal{R}$, and if $E,F \in \mathcal{R}$, then $E \setminus F \in \mathcal{R}$). A ring that is closed under countable unions is called a $\sigma$-ring.
- (a) Rings (resp. $\sigma$-rings) are closed under finite (resp. countable) intersections.
- (b) If $\mathcal{R}$ is a ring (resp. $\sigma$-ring), then $\mathcal{R}$ is an algebra (resp. $\sigma$-algebra) iff $X \in \mathcal{R}$.
- (c) If $\mathcal{R}$ is a $\sigma$-ring, then $\lbrace E \subset X : E \in \mathcal{R} \text{ or } E^c \in \mathcal{R}\rbrace$ is a $\sigma$-algebra.
- (d) If $\mathcal{R}$ is a $\sigma$-ring, then $\lbrace E \subset X : E \cap F \in \mathcal{R} \text{ for all } F \in \mathcal{R}\rbrace$ is a $\sigma$-algebra.
Solution For part (a) and (b) I will prove only $\sigma$-ring case, the finite case is very similar
(a) Let $\mathcal{R}$ be a $\sigma$-ring and $\lbrace E_j \rbrace$ be a sequence of sets in $\mathcal{R}$. Since $\bigcap\limits_{1}^{\infty} E_j = E_1 \setminus \bigcup\limits_{2}^{\infty} \big( E_1 \setminus E_j \big)$ and $\sigma$-ring is closed under countable union and differences the intersection is in the $\sigma$-ring. $\blacksquare$
(b) Assume that $\mathcal{R}$ is a $\sigma$-ring and contains $X$. We just need to show that $\mathcal{R}$ is closed under complement. For any $E \in \mathcal{R}$ we have $E^c = X \setminus E$ and since a $\sigma$-ring is closed under set difference $\mathcal{R}$ is closed under complement, and we are done. For the other direction assume that $\mathcal{R}$ is a $\sigma$-algebra and $E,F \in \mathcal{R}$ two sets. Then we have $E \setminus F = E \cap F^c$ and since $\sigma$-algebra is closed under intersection and complement it is also closed in set difference. Moreover, for any $E \in \mathcal{R}$ we have $E \cup E^c = X$ and we are done. $\blacksquare$.
(c) We need to show that the given set is closed under complement and countable union. The firs one is trivial since it contains all the complements of its sets by definition. For the second one assume that $\lbrace E_i \rbrace$ is a family of subsets of the given set $\mathcal{A} = \lbrace E \subset X : E \in \mathcal{R} \text{ or } E^c \in \mathcal{R}\rbrace$. Decompose $\lbrace E_i \rbrace$ into two part indexed with $J$ $K$ there. $E_j$ and $E_k^c$ are in $\mathcal{R}$ for any $j,k$. Than we have, $$ \bigcup_{i\in I} E_i = \left(\bigcup_{j \in J}E_j \right) \cup \left(\bigcup_{k \in K} E_k \right) \ $$ if we get the complement of the lefthand side we get
$$ \left( \bigcup_{i\in I} E_i \right)^c = \left(\bigcup_{j \in J}E_j \right)^c \cap \left(\bigcup_{k \in K} E_k \right)^c = \left(\bigcup_{j \in J}E_j \right)^c \cap \left(\bigcap_{k \in K} E_k^c \right) $$
And the last intersection can be written as set differences like follows. $$ \left(\bigcap_{k \in K} E_k^c \right) \setminus \left(\bigcup_{j \in j} E_j \right) $$ Now we have intersections of sets of $\mathcal{R}$ set difference unions of sets of $\mathcal{R}$ and by part (a) and the difference is in $\mathcal{R}$ and we are done. $\blacksquare$