# Monotone Functions and Discontinuities

## Definition 4.4.1

$E\subset \mathbb{R}, f:E\to\mathbb{R}, p$ is a limit point of $E\cap (p,\infty)$. $f$ has a right limit at $p$ if there exists a $L\in\mathbb{R}$ such that $\forall \epsilon >0, \exists \delta >0$ such that

\begin{align*} p<x<p+\delta, x\in E \Rightarrow \left\lvert f(x)-L\right\rvert <\epsilon \end{align*}

We write

\begin{align*} f(p+) = \lim_{x\to p+}f(x) = \lim_{x\to p, x> p}f(x). \end{align*}

Similarly, if $p$ is a limit point of $E\cap (-\infty, p)$, the left limit of $f$ at $p$, if it exists, is denoted by $f(p-)$, and we write

\begin{align*} f(p-) = \lim_{x\to p}f(x) = \lim_{x\to p, x<p} f(x). \end{align*}

## Remark

If any interval $I$ with Int$(I)\neq \emptyset$ and $f:I\to\mathbb{R}$, then $f$ has a limit point at $p\in\text{Int}(I)$,

\begin{align} \lim_{x\to p} f(x) = L \iff f(p+), f(p-) \text{ exist and } f(p+) = f(p-) = L \end{align}

Since $p\in E^\prime$ does not imply $p$ is a limit point of $E\cap (p, \infty)$ or $E\cap (-\infty, p)$, we need the set $E$ to be interval $I$ with $\text{Int}(I)\neq\emptyset$ and $p\in\text{Int}(I)$.

<Proof>

$\Rightarrow$ Suppose that $f$ has a limit at $p\in\text{Int}(I)$. Since $p\in\text{Int}(I)$, there exists $\epsilon>0$ such that $(p-\epsilon, p+\epsilon)\subset I$. Moreover, for every $\delta >0, N_\delta^\prime(p) \cap \text{I}\neq\emptyset$ since $p$ is a limit point of $I$. Then, $N^\prime_\delta (p) \cap (p, p+\epsilon)\neq\emptyset$, which implies that $N^\prime_\delta(p)\cap (p, \infty)\neq \emptyset$.

$\therefore p$ is a limit point of $I\cap (p,\infty)$.

Similarly, we can show that $p$ is also a limit point of $I\cap (-\infty, p)$.

$\Leftarrow$ It is trivial case.

$\tag*{\square}$

If $p$ is the left end point of $I$, then the right limit of $f$ at $p$ coincides with the the limit of $f$ at $p$. Similar for $p$ is the right end point of $I$.

## Definition 4.4.2

$E\subset \mathbb{R}, f:E\to\mathbb{R}, p\in E$. The function $f$ is right continuous (left continuous) at $p$

if $\forall \epsilon >0, \exists \delta>0$ such that

\begin{align} p\leq x <p+\delta \:(p-\delta < x \leq p) \Rightarrow \lvert f(x)-f(p) \rvert< \epsilon. \end{align}

## Definition 4.4.4

$f$ has a jump discontinuity at $p$ if $f(p+), f(p-)$ exist but $f(p+) \neq f(p-)$.

## Definition 4.4.6

$f:I\to\mathbb{R}$, where $I$ is an interval.

(a) $f$ is (monotone) increasing if for all $x,y\in I$ with $x<y$, then $f(x)\leq f(y)$. If $f$ is strictly increasing if or all $x,y\in I$ with $x<y$, then $f(x) < f(y)$.

(b) $f$ is (monotone) decreasing if for all $x,y\in I$ with $x<y$, then $f(x)\geq f(y)$. If $f$ is strictly decreasing if or all $x,y\in I$ with $x<y$, then $f(x) > f(y)$.

## Theorem 4.4.7

$I$ is an open interval and $f:I\to\mathbb{R}$ is monotone increasing on $I$.

$\Rightarrow \forall p\in I, f(p+), f(p-)$ exist and $\sup_{x<p}f(x) = f(p-) \leq f(p) \leq f(p+) = \inf_{x>p} f(x).$

Furthermore, if $p,q\in I$ with $p<q$, then $f(p+) \leq f(q-)$.

<proof>

Let $p\in I$ be given. Define $E:=\{f(x): x<p, x\in I\}.$

Since $f$ is monotone increasing, $E$ is bounded above by $f(p)$. Since $I$ is open, there exists a $\epsilon_0 >0$ such that $N_{\epsilon_0} (p) \subset I$, i.e., there is $x\in (p-\epsilon_0, p)$. Thus $E\neq \emptyset$. By the least upper bound property, $A=\sup E$ exists.

Now we want to show that $A = f(p-)$.

Let $\epsilon >0$ be given. Then there is $x_0 \in I$ with $x_0 <p$ s.t. $A-\epsilon < f(x_0) \leq A$ since $A$ is the least upper bound of $E$. For $x_0 < x< p,$

\begin{align} A-\epsilon< f(x_0) \leq f(x) \leq A < A+\epsilon \end{align}

$\therefore \lvert f(x)-A\rvert< \epsilon$ for all $x_0<x<p$.

$\therefore f(p-) = A$

Similarly, $S:=\{f(x): x>p, p\in I \}$. Since $f$ is monotone increasing, $S$ is bounded below by $f(p)$. Since $I$ is open, there is $\epsilon_0>0$ such that $N_{\epsilon_0} (p) \subset I$, i.e. there is $x\in (p, p+\epsilon_0)$. Thus, $S\neq \emptyset$. By the least upper bound property, $B:=\inf S$ exists.

Now we want to show that $B=f(p+)$.

Let $\epsilon >0$ be given. There is $x_0 \in I$ with $x_0 >p$ such that $B\leq f(x_0) < B +\epsilon$. For $p<x<x_0$,

\begin{align} B-\epsilon < B \leq f(x) \leq f(x_0) < B+\epsilon \end{align}

$\therefore \lvert f(x) - B\rvert < \epsilon$ for all $x\in I$ with $p<x<x_0$.

$\therefore f(p+) = B$

Therefore

\begin{align*} \sup_{x<p} f(x) = f(p+) \leq f(p) \leq f(p-) = \inf_{x>p} f(x) \end{align*}

Furthermore, for $p,q\in I$ with $p<q$,

\begin{align} f(p+) = \inf_{x<p} f(x) \leq \inf_{p<x<q} f(x) \leq \sup_{p<x<q} f(x) \leq \sup_{x<q} f(x) = f(q-) \end{align}

$\therefore f(p+) \leq f(q-)$

$\tag*{\square}$

## Definition 4.4.9

The unit jump function $I:\mathbb{R}\to\mathbb{R}$ defined by

\begin{align*} I(x) = \begin{cases}0 \quad (x<0)\\ 1\quad (x\geq 0) \end{cases} \end{align*}

Then

\begin{align*} I_a (x) := I(x-a) = \begin{cases} 0 \quad (x<a)\\ 1\quad (x\geq a) \end{cases} \end{align*}

## Theorem

Let $f: (a,b)\to\mathbb{R}$ be a monotone function. Then the cardinality for the set of discontinuities is at most countable.

<Proof>

Say $f$ is discontinuous at some point $x\in (a,b)$. Then by Theorem 4.4.7, $f(x^-)<f(x^+)$. Otherwise, $f$ is continuous at $x$. By density of $\mathbb{Q}$, there exists $r(x)\in\mathbb{Q}$ such that

$\begin{equation*} r(x)\in \left( f(x^-), f(x^+)\right). \end{equation*}$

Further, if $x_1 < x_2$, then $f(x^+_1)\leq f(x^-_2)$ by Theorem 4.4.7. So, $r(x_1)\neq r(x_2)$.

$\therefore$ The number of discontinuity is less than the cardinality of $\mathbb{Q}$.

$\tag*{\square}$

## Cauchy Criterion

Suppose that $\sum_{k=1}^\infty a_k$ converges, where $a_k\in\mathbb{R}$ for all $k=1,2,\ldots$. Since $s_n= \sum_{k=1}^n a_k$ is convergent sequence in $\mathbb{R}$, it is Cauchy sequence. So for all $\epsilon >0, \exists N\in\mathbb{N}$ such that,

\begin{align} m>n\geq N \Rightarrow \left\lvert\sum_{k=1}^m a_k -\sum_{k=1}^n a_k\right\rvert =\left\lvert\sum_{k=n+1}^m a_k \right\rvert < \epsilon. \end{align}

So, for all $\epsilon >0, \exists N\in\mathbb{N}$ such that,

\begin{align} m\geq N \Rightarrow \left\lvert\sum_{k=N+1}^m a_k \right\rvert < \epsilon. \end{align}

Since $b_m=s_m-\sum_{k=1}^Na_k$,

\begin{align} \begin{split} \lim_{m\to\infty}b_m&= \lim_{m\to\infty}\left( s_m -\sum_{k=1}^Na_k\right) \\ &=\lim_{m\to\infty}s_m-\sum_{k=1}^Na_k \end{split} \end{align}

$b_m\to L$ as $m\to L$ for some $L\in\mathbb{R}$.

Alternatively, we can show its convergence using the sequence

\begin{align*} \{s_n\}_{n=1}^\infty \end{align*}

is Cauchy. Since $\mathbb{R}$ is complete, it suffices to show that the sequence \begin{align*} \{b_m\}_{m=N+2}^\infty \end{align*}

is Cauchy. Let $\epsilon >0$ be given. Since

\begin{align*} \{s_n\}_{n=1}^\infty \end{align*}

is Cauchy sequence, there is $N_0\in\mathbb{N}$ such that

\begin{align*} n_1> n_2> N_0 \Rightarrow \left\lvert \sum_{k=n_2+1}^{n_1} a_k \right\rvert < \epsilon. \end{align*}

Now take

\begin{align*} M:=\max \{N_0, N\}. \end{align*}

Then for all $m_1 > m_2 >M$,

\begin{align*} \lvert b_{m_1}-b_{m_2}\rvert &= \left\lvert \sum_{k=N+1}^{m_1}a_k - \sum_{k=N+1}^{m_2}a_k\right\rvert \\ &=\left\lvert \sum_{k=m_2+1}^{m_1}a_k\right\rvert \\ &<\epsilon \end{align*}

Then the sequence is bounded by $\epsilon$ and thus its limit

\begin{align} \left\lvert\sum_{k=N+1}^\infty a_k \right\rvert < \epsilon \end{align}

is also bounded.

## Theorem 4.4.10

Let $a,b\in \mathbb{R}$ and let $\{x_n\mid n\in\mathbb{N}\}$ be a countable subset of $(a,b)$ and let $\{c_n\}_{n=1}^\infty$ be any sequence of positive numbers such that $\sum_{n=1}^\infty c_n < +\infty$.

Then there is a monotone increasing function on $[a,b]$ such that

(a) $f(a)=0, f(b) = \sum_{n=1}^\infty c_n$

(b) $f$ is continuous on $[a,b] \setminus \{x_n:n\in\mathbb{N}\}$

(c) $f(x_n+) = f(x_n)$, i.e. $f$ is right continuous at each $x_n$.

(d) $f$ is discontinuous at $x_n$ for all $n\in\mathbb{N}$ and $f(x_n+) -f(x_n-)=c_n$

<Proof>

Define $f(x):=\sum_{k=1}^\infty c_k I(x-x_k)$.

Since $0\leq c_k I(x-x_k)\leq c_k$, $s_n (x) :=\sum_{k=1}^n c_k I (x-x_k) \leq \sum_{k=1}^n c_k$.

Since $\sum_{k=1}^\infty c_k < +\infty$, $\{s_n (x)\}$ is bounded above. Since $s_n (x)$ is monotone increasing sequence, $s_n (x)$ converges.

For $x<y, I(x-x_k) \leq I(y-x_k)$ for all $k=1,2,\ldots$.

So, $f(x) = \sum_{k=1}^\infty c_k I (x-x_k) \leq \sum_{k=1}^\infty I(y-x_k) = f(y).$

$\therefore f(x)$ is a monotone increasing function.

(a) Since $x_k > a$ for all $k\in\mathbb{N}$, $I(a-x_k)=0$.

$\therefore f(a) = \sum_{k=1}^\infty c_k f(a-x_k)=0$.

Since $x_k <b$ for all $k\in\mathbb{N}$, $I(b-a_k)=1$.

$\therefore f(b) = \sum_{k=1}^\infty c_k f(b-x_k) = \sum_{k=1}^\infty c_k$.

(b) $E:=\{x_n:n\in\mathbb{N}\}$. Let $p\in [a,b]\setminus E$.

1) $p\not\in E^\prime$. There is a $\delta>0$ such that $N_\delta (p) \cap E =\emptyset$. For $x\in N_\delta (p), I(x-x_k)=I(p-x_k)$ for all $k\in\mathbb{N}$. So,

\begin{align} \begin{split} f(x) &= \sum_{k=1}^\infty c_k I(x-x_k)\\ &=\sum_{k=1}^\infty c_k I(p-x_k) \end{split} \end{align}

2) $p\in E^\prime$ Let $\epsilon >0$ be given. Since $\sum_{k=1}^\infty c_k$ converges, there exists a $N\in\mathbb{N}$ such that

\begin{align} \sum_{k=N+1}^\infty c_k <\epsilon \end{align}

by Cauchy criterion. Choose a $\delta \in\mathbb{R}$ such that

\begin{align} 0<\delta < \min\{\lvert p-x_k\rvert: k=1,2,\ldots, N\} \end{align}

Then $x_k\not\in (p-\delta, p+\delta)$ for $k=1,\ldots, N$. If $x_k\in N_\delta (p) \cap E$, then $k>N$.

Suppose that $p<x<p+\delta$. Then $I(x-x_k) = I(p-x_k)$ for $k=1,\ldots, N$. Furthermore, for any $x>p, 0\leq I(x-x_k) - I(p-x_k)\leq 1$ for all $k\in\mathbb{N}$. Therefore, if $p<x<p+\delta$

\begin{align} \begin{split} f(x) - f(p) &= \sum_{k=N+1}^\infty c_k(I(x-x_k)-I(p-x_k))\\ &\leq \sum_{k=N+1}^\infty c_k \\ &< \epsilon \end{split} \end{align}

$\therefore \left\lvert f(x)-f(p)\right\rvert<\epsilon$ for all $p<x<p+\delta$.
$\therefore f$ is right continuous at $p$.

Similarly, suppose that $p-\delta < x < p$. Then $I(x-x_k) = I(p-x_k)$ for $k=1,\ldots, N$. Furthermore, for any $x<p, 0\leq I(p-x_k)-I(x-x_k) \leq 1$

\begin{align} \begin{split} f(p) - f(x) &= \sum_{k=N+1}^\infty c_k(I(p-x_k)-I(x-x_k))\\ &\leq \sum_{k=N+1}^\infty c_k \\ &< \epsilon \end{split} \end{align}

$\therefore f$ is left continuous at $p$.

$\therefore f$ is continuous at $p$.

(c) We want to show that $f(x_n+)=f(x_n)$

Fix an $x_n\in E$.

1) $x_n\not\in E^\prime$. There is a $\delta >0$ such that $N^\prime_\delta (x_n) \cap E = \emptyset$, i.e. $(x_n, x_n+\delta) \cap E = \emptyset$.

For $y\in (x_n, x_n+\delta), I(y-x_k) = I(x_n-x_k)$ for all $k\in\mathbb{N}$.

$\therefore f(y) = f(x_n)$ for all $y\in (x_n, x_n+\delta)$

$\therefore f$ is right continuous at $x_n$.

2) $x_n \in E^\prime$.
Let $\epsilon >0$ be given. Choose a $N\in\mathbb{N}$ such that

\begin{align} \sum_{k=N+1}^\infty c_k <\epsilon \end{align} by Cauchy criterion. Choose a $\delta >0$ such that

\begin{align} 0<\delta <\min\{\lvert x_n-x_k\rvert :k=1,\ldots,N, k\neq n\} \end{align}

Then, $x_k \not\in (x_n, x_n+\delta)$ for $k=1,\ldots,N$. Thus, for $y\in (x_n, x_n+\delta), I(y-x_k)=I(x_n-x_k)$ for all $k=1,\ldots,N$.

For $y>x_n, 0\leq I(y-x_k) - I(x_n-x_k)\leq 1 \text{ for all } k\in\mathbb{N}$ Thus, For all $y\in(x_n, x_n+\delta)$

\begin{align} \begin{split} f(y) - f(x_n) &= \sum_{k=1}^\infty c_k(I(y-x_k) - I(x_n-x_k))\\ &= \sum_{k=N+1}^\infty c_k(I(y-x_k) - I(x_n-x_k)) \\ &\leq\sum_{k=N+1}^\infty c_k \\ &<\epsilon \end{split} \end{align}

$\therefore f$ is right continuous at $x_n$ for all $n\in \mathbb{N}$.

(d) Fix a $x_n\in E$, Suppose $y<x_n$.

If $x_n \not\in E^\prime,$ there is a $\delta >0$ such that $(x_n-\delta, x_n) \cap E = \emptyset$. Thus, $I(y-x_k) = I(x_n-x_k)$ for all $k\in \mathbb{N}$ with $k\neq n$. For $k=n$,

\begin{align} 0=I(y-x_n) \leq I(x_n-x_n) = I(0) = 1 \end{align}

$\therefore f(x_n)-f(y) = c_n$ for all $y\in(x_n-\delta, x_n)$

Now, suppose that $x_n \in E^\prime$. Let $\epsilon >0$ be given. Choose a $N \in \mathbb{N}$ such that $\sum_{k=N+1}^\infty c_k <\epsilon$. Choose a $\delta >0$ such that

\begin{align} 0<\delta <\min\{\lvert x_k-x_n\rvert :k=1,\ldots,N, k\neq n\} \end{align}

So, $x_k \not\in (x_n-\delta, x_n)$ for $k=1,\ldots, N$. Then $I(y-x_k) = I(x_n-x_k)$ for all $k=1,\ldots, N$ with $k\neq n$. For all $y\in (x_n-\delta, x_n)$,

\begin{align} \begin{split} c_n \leq f(x_n) - f(y) &= \sum_{k=1}^\infty c_k (I(x_n-x_k) - I(y-x_k)) \\ &=c_n + \sum_{k=N+1}^\infty c_k (I(x_n-x_k) - I(y-x_k)) \\ &\leq c_n + \sum_{k=N+1}^\infty c_k \\ &< c_n + \epsilon \end{split} \end{align}

So, \begin{align} \begin{split} c_n - \epsilon < f(x_n) - f(y) < c_n +\epsilon &\iff -c_n-\epsilon < f(y) - f(x_n) < -c_n + \epsilon \\ &\iff -\epsilon < f(y) - f(x_n) + c_n <\epsilon \\ &\iff \lvert f(y) - (f(x_n)-c_n)\rvert < \epsilon \end{split} \end{align}

$\therefore f(x_n-) = f(x_n) + c_n$

$\therefore f(x_n) = f(x_n-) - c_n$

$\tag*{\square}$

## Theorem 4.4.12

Let $I\subset \mathbb{R}$ be an interval and let $f:I\to\mathbb{R}$ be a strictly monotone continuous function on $I$. $\Rightarrow f^{-1}$ is strictly monotone continuous on $J=f(I)$.

<proof> Without loss of generality, assume that $f$ is strictly increasing function. Let $x,y\in I$ with $x<y$ be given. Since $f$ is continuous, by corollary 4.2.12, $f(I) = J$ is an interval.

Furthermore, since $f$ is strictly increasing, $f(x) < f(y)$. Thus, $f$ is one-to-one function from $I$ onto $J:=f(I)$. Thus, $f^{-1}$ is also one-to-one function from $J$ onto $I$.

Let $y_1,y_2 \in J$ with $y_1 < y_2$ be given. Then there are unique $x_1, x_2 \in I$ such that $y_1= f(x_1), y_2 = f(x_2)$. Since $f$ is strictly increasing, $f^{-1}(y_1) = x_1 < x_2 = f^{-1}(y_2)$

$\therefore f^{-1}$ is strictly increasing function.

Now, we want to show that $f^{-1}$ is continuous on $J$. First, we prove that $f^{-1}$ is left continuous.

Let $y_0\in J$ such that $(-\infty, y_0)\cap J \neq \emptyset$ be given, i.e. $y_0$ is not the left end point of $J$.

Let $x_o \in I$ such that $y_0 = f(x_0)$. Since $f^{-1}$ is strictly increasing and $(-\infty, y_0) \cap J \neq\emptyset$, $(-\infty, x_0) \cap I \neq \emptyset.$

Let $\epsilon >0$ be given.

1. $x_0-\epsilon \not\in I$ For all $y\in (-\infty, y_0] \cap J$,$x_0-\epsilon < f^{-1}(y) \leq f^{-1}(y_0)$. So, $\left\lvert f^{-1}(y) -f^{-1}(y_0)\right\rvert < \left\lvert f^{-1}(y_0)-x_0+\epsilon \right\rvert = \epsilon$
2. $x_0-\epsilon \in I$ Choose a $\delta = y_0 - f(x_0-\epsilon)=f(x_0) = f(x_0-\epsilon)>0$.

For all $y\in(y_0-\delta, y_0]$,

\begin{align} \begin{split} x_0-\epsilon = f^{-1}(y_0-\delta) &<f^{-1}(y) \\ &\leq f^{-1}(y) \\ &\leq f^{-1}(y_0) =x_0\\ &<x_0 +\epsilon \end{split} \end{align}

$\therefore \left\lvert f^{-1}(y) -x_0 \right\rvert <\epsilon$ for all $y\in (y_0-\delta, y_0]$.

$\therefore f$ is left continuous at $y_0$.

Similar argument proves that $f$ is right continuous at $y_0\in J$ such that $(y_0, +\infty) \cap J\neq\emptyset$.

$\therefore f$ is continuous at $y_0 \in J$.

$\therefore f$ is continuous at $y_0\in J$.

$\tag*{\square}$

## Reference

• Manfred Stoll, Introduction to Real Analysis, Pearson

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