Proof. Define the function

$$\mathrm{\Phi }(y)=\{\begin{array}{ll}{\displaystyle \frac{f(y)-f(b)}{y-b}}& \text{if }y\ne b\\ \\ f^{\prime }(b)& \text{if }y=b\end{array}$$

This function is continuous at $b$ because $\underset{y\to b}{lim}\mathrm{\Phi }(y)=\mathrm{\Phi }(b)$. So for any $y\in I$, we have $f(y)-f(b)=\mathrm{\Phi }(y)(y-b)$. If we substitute $y=g(x)$ and divide the equality by $(x-a)$ for $x\ne a$, we get

$$\frac{f(g(x))-f(g(a))}{x-a}=\mathrm{\Phi }(g(x))\frac{g(x)-g(a)}{x-a}$$

Finally, if we take the limit as $x\to a$, we get

$$(f\circ g{)}^{\prime }(a)=f^{\prime }(g(a))g^{\prime }(a)$$

$◼$

Proof. Let $\mathrm{𝒪}$ be an open set. Take any $x\in \mathrm{𝒪}$ and define $I_x=\bigcup \{(a,b):x\in (a,b)\subset \mathrm{𝒪}\}$. $I_x$ is not empty because $\mathrm{𝒪}$ is open, and it is an open interval as a union of open intervals. For any $x$ and $y$ in $\mathrm{𝒪}$ if $I_x\cap I_y\ne \mathrm{\varnothing }$ then $I_x=I_y$. Therefore $\mathrm{𝒪}=\bigcup _{x\in \mathrm{𝒪}}{I}_x$ is a disjoint union of open intervals. Define a map $\{I_x{\}}_{x\in \mathrm{𝒪}}\to \mathrm{\mathbb{Q}}$ 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 $\mathrm{\mathbb{Q}}$. Therefore $\mathrm{𝒪}$ is a countable union of open intervals. $◼$

Proof. Let $E$ be a set with $\mu (E)=\mathrm{\infty }$. Since $\mu$ is $\sigma$-finite $E$ is also $\sigma$-finite and there exists a sequence of sets $\{E_j\}$ such that, $E=\bigcup _{j=1}^{\mathrm{\infty }}{E}_j$ where $\mu (E_j)<\mathrm{\infty }$ 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 $\mathrm{\infty }$. That is there exists a $k$ such that $0<\mu (E_k)<\mathrm{\infty }$ which makes $\mu$ a semifinite measure.

Let $E=\{x:f(x)>0\}$ and define $E_n=\{x:f(x)>\frac{1}{n}\}$ Then $E=\bigcup E_n$. Since $\mu$ is $\sigma$-finite there is a sequence of sets $\{X_j\}$ where $X=\bigcup X_j$ and $\mu (X_j)<\mathrm{\infty }$ for each $j$.For a fixed $j$ and $n$ $E_n\cap X_j$ has a finite measure and $$\mu (E_n\cap X_j)=\sum _{x\in E_n\cap X_j}f(x)\ge \sum _{x\in E_n\cap X_j}\frac{1}{n}$$ If the set $E_n\cap X_j$ were infinite, the above sum yields to infinity which is not possible since $\mu (E_n\cap X_j)\le \mu (X_j)<\mathrm{\infty }$. Therefore it must be finite. Moreover $E_n=\bigcup _{j=1}^{\mathrm{\infty }}{E}_n\cap X_j$ 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 $E=\{x:f(x)>0\}$ is countable. We want to show that $\mu$ is $\sigma$-finite.

Since $E$ is countable, we can write its elements as a sequence, so $E=\bigcup _{n=1}^{\mathrm{\infty }}\{x_n\}$. Let us check the measure of each singleton $\{x_n\}$. Suppose for the sake of contradiction that $\mu (\{x_n\})=f(x_n)=\mathrm{\infty }$ for some $n$. Since $\mu$ is semifinite, there must exist a subset $A\subset \{x_n\}$ such that $0<\mu (A)<\mathrm{\infty }$. However, the only subsets of a singleton are the empty set $\mathrm{\varnothing }$ and the singleton itself. Since $\mu (\mathrm{\varnothing })=0$, no such subset $A$ can exist. This is a contradiction. Therefore, we must have $\mu (\{x_n\})=f(x_n)<\mathrm{\infty }$ 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:

$$\mu (X\setminus E)=\sum _{x\in X\setminus E}f(x)=0<\mathrm{\infty }$$

We can express the entire space $X$ as the disjoint union of $X\setminus E$ and the singletons from $E$:

$$X=(X\setminus E)\cup \bigcup _{n=1}^{\mathrm{\infty }}\{x_n\}$$

Since $X$ is written as a countable union of sets ($X\setminus E$ and all $\{x_n\}$), each of which has finite measure, we conclude that $\mu$ is a $\sigma$-finite measure. $◼$

Proof. Let $M=supS$ and $\epsilon >0$. Since $M-\epsilon <M$, the number $M-\epsilon$ is not an upper bound of $S$, so some $x\in S$ satisfies $x>M-\epsilon$.

Conversely, assume the condition and let $M^{\prime }<M$. With $\epsilon =M-M^{\prime }$ there is an $x\in S$ with $x>M-\epsilon =M^{\prime }$, so $M^{\prime }$ is not an upper bound of $S$. Hence $M$ is the least upper bound. $◼$

Proof. Let $M=supS$. For each $n$ the lemma with $\epsilon =1/n$ gives $x_n\in S$ with $M-\frac{1}{n}<x_n\le M$, so $|x_n-M|<\frac{1}{n}$ and $x_n\to M$.

If $M\notin S$, every $x_n$ is strictly less than $M$. Choose $x_1$ as above and, having chosen $x_n<M$, apply the lemma with $\epsilon =min(M-x_n,\frac{1}{n+1})>0$ to get $x_{n+1}\in S$ with $x_{n+1}>M-\epsilon \ge x_n$ and $x_{n+1}>M-\frac{1}{n+1}$. $◼$

Proof. Suppose $nx\le y$ for every $n$. Then $A=\{nx:n\in \mathrm{\mathbb{N}}\}$ is nonempty and bounded above by $y$; let $\alpha =supA$. Since $x>0$, $\alpha -x<\alpha$ is not an upper bound of $A$, so $\alpha -x<mx$ for some $m$. But then $\alpha <(m+1)x\in A$, a contradiction.

For the second claim, $0$ is a lower bound of $\{1/n\}$. If some $\epsilon >0$ were a lower bound, we would have $n\epsilon \le 1$ for all $n$, contradicting the first part with $x=\epsilon$, $y=1$. $◼$

Proof.

1. For any $a\in A$, $infA\le a\le supA$. Every upper bound of $B$ is an upper bound of $A$, so $supA\le supB$; similarly $infB\le infA$.
2. $m$ is a lower bound of $A$ if and only if $-m$ is an upper bound of $-A$, so the greatest lower bound of $A$ is minus the least upper bound of $-A$.
3. For $a\in A$, $b\in B$ we have $a+b\le supA+supB$, so $supA+supB$ is an upper bound of $A+B$. Given $\epsilon >0$, the lemma gives $a\in A$, $b\in B$ with $a>supA-\frac{\epsilon }{2}$ and $b>supB-\frac{\epsilon }{2}$, so $a+b>supA+supB-\epsilon$. By the lemma, $sup(A+B)=supA+supB$.
4. For $c>0$, $M$ is an upper bound of $A$ if and only if $cM$ is an upper bound of $cA$. For $c<0$ write $cA=|c|(-A)$ and use 2.
5. Every $b\in B$ is an upper bound of $A$, so $supA\le b$ for all $b\in B$. Thus $supA$ is a lower bound of $B$, and $supA\le infB$. $◼$

Proof. Let $s=\underset{n}{sup}{a}_n<\mathrm{\infty }$ and $\epsilon >0$. By the lemma there is $N$ with $a_N>s-\epsilon$, and for $n\ge N$, $s-\epsilon <a_N\le a_n\le s$. So $a_n\to s$. If $(a_n)$ is not bounded above, for every $M$ there is $N$ with $a_N>M$, and then $a_n\ge a_N>M$ for all $n\ge N$, so $a_n\to +\mathrm{\infty }$. The decreasing case follows by applying this to $(-a_n)$. Convergent sequences are bounded, which gives the last sentence. $◼$

Proof. Let $L=lim{S}_N\in \mathrm{ℝ}$ and $\epsilon >0$. There is an $N$ with $S_N<L+\epsilon$, and then $a_n\le S_N<L+\epsilon$ for all $n\ge N$: this is 1. Since $(S_N)$ decreases to $L$, $S_N\ge L>L-\epsilon$ for every $N$, so by the approximation property each tail contains some $a_k$, $k\ge N$, with $a_k>L-\epsilon$. As $N$ is arbitrary, there are infinitely many such $k$: this is 2.

Conversely, assume 1 and 2 and fix $\epsilon >0$. By 1 there is $N_0$ with $a_n<L+\epsilon$ for $n\ge N_0$, so $S_N\le L+\epsilon$ for $N\ge N_0$. By 2 every tail contains a term $>L-\epsilon$, so $S_N>L-\epsilon$ for every $N$. Hence $|S_N-L|\le \epsilon$ for $N\ge N_0$, and $S_N\to L$.

If $(a_n)$ is not bounded above, neither is any tail (only finitely many terms were removed), so $S_N=+\mathrm{\infty }$ for all $N$; conversely, a decreasing sequence with limit $+\mathrm{\infty }$ is constantly $+\mathrm{\infty }$, so $S_1=+\mathrm{\infty }$. Finally, $S_N\to -\mathrm{\infty }$ means: for every $M$ there is $N$ with $a_n\le S_N<M$ for $n\ge N$, which is $a_n\to -\mathrm{\infty }$; the converse is the same argument read backwards. The $lim inf$ statements follow by applying these to $(-a_n)$, since $lim inf{a}_n=-lim sup(-a_n)$ (see the Arithmetic proposition below). $◼$

Proof. We prove the statement for $L=lim sup{a}_n$; for $lim inf$ apply it to $(-a_n)$.

Every $x\in E$ satisfies $x\le L$. Let $a_{n_k}\to x$. For a fixed $N$ we have $n_k\ge N$ for all large $k$, so $a_{n_k}\le S_N$; letting $k\to \mathrm{\infty }$ gives $x\le S_N$. Letting $N\to \mathrm{\infty }$ gives $x\le L$.

$L\in E$. If $L=+\mathrm{\infty }$, the sequence is not bounded above, so every tail contains arbitrarily large terms; choose $n_1<n_2<\cdots$ with $a_{n_k}>k$. If $L=-\mathrm{\infty }$, then $a_n\to -\mathrm{\infty }$ by the previous theorem. If $L$ is finite, apply the previous theorem with $\epsilon =\frac{1}{k}$: infinitely many $n$ satisfy $a_n>L-\frac{1}{k}$, and all but finitely many satisfy $a_n<L+\frac{1}{k}$, so infinitely many satisfy both. Choose such an $n_k>n_{k-1}$. Then $|a_{n_k}-L|<\frac{1}{k}$, so $a_{n_k}\to L$. $◼$

Proof. If $m\le a_n\le M$ for all $n$, then $m\le I_N\le S_N\le M$, so $lim sup{a}_n\in [m,M]$ is finite. By the theorem it is the limit of a subsequence. $◼$

Proof. Let $L$ be finite. If $a_n\to L$ and $\epsilon >0$, then $L-\epsilon <a_n<L+\epsilon$ for $n\ge N_0$, so $L-\epsilon \le I_N\le S_N\le L+\epsilon$ for $N\ge N_0$. Hence both limits equal $L$.

Conversely, let $lim inf{a}_n=lim sup{a}_n=L$ and $\epsilon >0$. Since $I_N\to L$ and $S_N\to L$, there are $N_1,N_2$ with $I_N>L-\epsilon$ for $N\ge N_1$ and $S_N<L+\epsilon$ for $N\ge N_2$. For $n\ge N_0=max(N_1,N_2)$, $$L-\epsilon <I_{N_0}\le a_n\le S_{N_0}<L+\epsilon ,$$ so $|a_n-L|<\epsilon$ and $a_n\to L$.

Let $L=+\mathrm{\infty }$. If $a_n\to +\mathrm{\infty }$, then for every $M$ we have $a_n>M$ for $n\ge N$, so $S_N\ge I_N\ge M$; both limits are $+\mathrm{\infty }$. If $lim inf{a}_n=+\mathrm{\infty }$, then $a_n\ge I_n\to +\mathrm{\infty }$. The case $L=-\mathrm{\infty }$ is symmetric. $◼$

Proof. Write $S_N(a)=\underset{k\ge N}{sup}{a}_k$, and so on.

1. $\underset{k\ge N}{sup}(-a_k)=-\underset{k\ge N}{inf}{a}_k$ by Rule 2; let $N\to \mathrm{\infty }$.
2. If $a_k\le b_k$ for $k\ge N_0$, then $S_N(a)\le S_N(b)$ and $I_N(a)\le I_N(b)$ for $N\ge N_0$; let $N\to \mathrm{\infty }$.
3. For $k\ge N$, $a_k+b_k\le S_N(a)+S_N(b)$, so $S_N(a+b)\le S_N(a)+S_N(b)$, and letting $N\to \mathrm{\infty }$ gives the right inequality. The left one is the right one applied to $(-a_n)$, $(-b_n)$, using 1. The middle one is $lim inf\le lim sup$.
4. By 3, $lim sup(a_n+b_n)\le lim sup{a}_n+b$. Applying 3 to $a_n=(a_n+b_n)+(-b_n)$ gives $lim sup{a}_n\le lim sup(a_n+b_n)-b$.
5. $S_N(ca)=c{S}_N(a)$ by Rule 4.
6. For $k\ge N$, $0\le a_k{b}_k\le S_N(a)S_N(b)$, so $S_N(ab)\le S_N(a)S_N(b)$; let $N\to \mathrm{\infty }$. $◼$

Proof. We prove the right inequality; the left one is an exercise below. Let $\alpha =lim sup\frac{c_{n+1}}{c_n}$. If $\alpha =+\mathrm{\infty }$ there is nothing to prove. Take any $\beta >\alpha$. By the $\epsilon$-characterization there is $N$ with $\frac{c_{n+1}}{c_n}<\beta$ for $n\ge N$, hence $c_{N+p}<{\beta }^p{c}_N$ for $p\ge 1$. With $n=N+p$, $$c_n<\left(c_N{\beta }^{-N}\right){\beta }^n,\sqrt[n]{c_n}<\sqrt[n]{c_N{\beta }^{-N}}\beta (n>N).$$ Since $\sqrt[n]{K}\to 1$ for every constant $K>0$, the right side tends to $\beta$, and Rule 2 of the Arithmetic proposition gives $lim sup\sqrt[n]{c_n}\le \beta$. As $\beta >\alpha$ was arbitrary, $lim sup\sqrt[n]{c_n}\le \alpha$. $◼$

Proof.

1. $x\in \bigcap _N\bigcup _{n\ge N}{A}_n$ means: for every $N$ there is an $n\ge N$ with $x\in A_n$. This says exactly that the set $\{n:x\in A_n\}$ is unbounded, i.e. infinite.
2. $x\in \bigcup _N\bigcap _{n\ge N}{A}_n$ means: there is an $N$ with $x\in A_n$ for all $n\ge N$.

If $x\in A_n$ for all $n\ge N$, then $x$ is in infinitely many $A_n$, which gives the inclusion. $◼$

Proof.

1. De Morgan's laws: ${\left(\bigcap _N\bigcup _{n\ge N}{A}_n\right)}^c=\bigcup _N\bigcap _{n\ge N}{A}_n^c$, and similarly for the other one.
2. If the sets increase, $\bigcap _{n\ge N}{A}_n=A_N$, so $lim inf{A}_n=\bigcup _N{A}_N$; and $\bigcup _{n\ge N}{A}_n=\bigcup _n{A}_n$ for every $N$, so $lim sup{A}_n=\bigcup _n{A}_n$ too. If they decrease, $\bigcup _{n\ge N}{A}_n=A_N$ and $\bigcap _{n\ge N}{A}_n=\bigcap _n{A}_n$, which gives $lim sup{A}_n=lim inf{A}_n=\bigcap _n{A}_n$.
3. Fix $x$. The sequence ${\chi }_{A_n}(x)$ takes only the values $0$ and $1$, so its $lim sup$ is $1$ exactly when it equals $1$ infinitely often, i.e. when $x\in lim sup{A}_n$, and $0$ otherwise. Likewise its $lim inf$ is $1$ exactly when it is eventually $1$, i.e. when $x\in lim inf{A}_n$. $◼$

Proof. For every $N$, $lim sup{E}_n\subset \bigcup _{n\ge N}{E}_n$, so by monotonicity and subadditivity $$\mu (lim sup{E}_n)\le \sum _{n=N}^{\mathrm{\infty }}\mu (E_n).$$ The right side is the tail of a convergent series, so it tends to $0$ as $N\to \mathrm{\infty }$. $◼$

Proof. The sets $V_N=\bigcap _{n\ge N}{E}_n$ increase to $lim inf{E}_n$, so $\mu (lim inf{E}_n)=\underset{N}{lim}\mu (V_N)$ by continuity from below. Since $V_N\subset E_n$ for every $n\ge N$, $\mu (V_N)\le \underset{n\ge N}{inf}\mu (E_n)$, and letting $N\to \mathrm{\infty }$ gives the first inequality.

The sets $U_N=\bigcup _{n\ge N}{E}_n$ decrease to $lim sup{E}_n$ and $\mu (U_1)<\mathrm{\infty }$, so $\mu (lim sup{E}_n)=\underset{N}{lim}\mu (U_N)$ by continuity from above. Since $E_n\subset U_N$ for $n\ge N$, $\mu (U_N)\ge \underset{n\ge N}{sup}\mu (E_n)$, and letting $N\to \mathrm{\infty }$ gives the second inequality. $◼$

Proof (Step 1). The members of $\mathrm{ℝ}$ will be certain subsets of $\mathrm{\mathbb{Q}}$, called cuts. A cut is defined as any set $\alpha \subset \mathrm{\mathbb{Q}}$ such that:

• (I) $\alpha \ne \mathrm{\varnothing }$ and $\alpha \ne \mathrm{\mathbb{Q}}$.
• (II) If $p\in \alpha$, $q\in \mathrm{\mathbb{Q}}$, and $q<p$, then $q\in \alpha$.
• (III) If $p\in \alpha$, then $p<r$ for some $r\in \alpha$.

Proof (Step 2). Define $\alpha <\beta$ as $\alpha ⊊\beta$. With this relation $\mathrm{ℝ}$ is an ordered set.

• (i) For $\alpha ,\beta \in \mathrm{ℝ}$, only one is true: $\alpha <\beta$, $\alpha =\beta$, or $\beta <\alpha$.
• (ii) For $\alpha <\beta$ and $\beta <\gamma$, $\alpha <\gamma$.

Hence $\mathrm{ℝ}$ has an order relation.

Proof (Step 3). The ordered set $\mathrm{ℝ}$ has the least upper-bound property. Let $A$ be a subset of $\mathrm{ℝ}$ and $\beta$ be an upper bound of $A$. That is, for any $\alpha \in A$, $\alpha \subset \beta$. Define $\gamma =\bigcup _{\alpha \in A}\alpha$. We will prove that $supA=\gamma$.