# Regulated functions and the regulated integral

Jordan Bell
April 3, 2014

## 1 Regulated functions and step functions

Let $I=[a,b]$ and let $X$ be a normed space. A function $f:I\to X$ is said to be regulated if for all $t\in[a,b)$ the limit $\lim_{s\to t^{+}}f(s)$ exists and for all $t\in(a,b]$ the limit $\lim_{s\to t^{-}}f(s)$ exists. We denote these limits respectively by $f(t^{+})$ and $f(t^{-})$. We define $R(I,X)$ to be the set of regulated functions $I\to X$. It is apparent that $R(I,X)$ is a vector space. One checks that a regulated function is bounded, and that $R(I,X)$ is a normed space with the norm $\left\|f\right\|_{\infty}=\sup_{t\in[a,b]}\left\|f(t)\right\|$.

###### Theorem 1.

If $I$ is a compact interval in $\mathbb{R}$ and $X$ is a normed algebra, then $R(I,X)$ is a normed algebra.

###### Proof.

If $f,g\in R(I,X)$, then $fg\in R(I,X)$ because the limit of a product is equal to a product of limits. For $t\in I$ we have

 $\left\|(fg)(t)\right\|=\left\|f(t)g(t)\right\|\leq\left\|f(t)\right\|\left\|g(% t)\right\|\leq\left\|f\right\|_{\infty}\left\|g\right\|_{\infty},$

so $\left\|fg\right\|_{\infty}\leq\left\|f\right\|_{\infty}\left\|g\right\|_{\infty}$. ∎

A function $f:I\to X$, where $I=[a,b]$, is said to be a step function if there are $a=s_{0} for which $f$ is constant on each open interval $(s_{i-1},s_{i})$. We denote the set of step functions $I\to X$ by $S(I,X)$. It is apparent that $S(I,X)$ is contained in $R(I,X)$ and is a vector subspace, and the following theorem states that if $X$ is a Banach space then $S(I,X)$ is dense in $R(I,X)$.11 1 Jean Dieudonné, Foundations of Modern Analysis, enlarged and corrected printing, p. 145, Theorem 7.6.1; Rodney Coleman, Calculus on Normed Vector Spaces, p. 70, Proposition 3.3; cf. Robert G. Bartle, A Modern Theory of Integration, p. 49, Theorem 3.17.

###### Theorem 2.

Let $I$ be a compact interval in $\mathbb{R}$, let $X$ be a Banach space, and let $f\in X^{I}$. $f\in R(I,X)$ if and only if for all $\epsilon>0$ there is some $g\in S(I,X)$ such that $\left\|f-g\right\|_{\infty}<\epsilon$.

We prove in the following theorem that the set of regulated functions from a compact interval to a Banach space is itself a Banach space.

###### Theorem 3.

If $I$ is a compact interval in $\mathbb{R}$ and $X$ is a Banach space, then $R(I,X)$ is a Banach space.

###### Proof.

Let $f_{n}\in R(I,X)$ be a Cauchy sequence. For each $t\in I$ we have

 $\left\|f_{n}(t)-f_{m}(t)\right\|\leq\left\|f_{n}-f_{m}\right\|_{\infty},$

hence $f_{n}(t)$ is a Cauchy sequence in $X$. As $X$ is a Banach space, this Cauchy sequence converges to some limit, and we define $f(t)$ to be this limit. Thus $f\in X^{I}$ and $\left\|f-f_{n}\right\|_{\infty}\to 0$. We have to prove that $f\in R(I,X)$. Let $\epsilon>0$. There is some $N$ for which $n\geq N$ implies that $\left\|f-f_{n}\right\|_{\infty}<\epsilon$; in particular, $\left\|f-f_{N}\right\|_{\infty}<\epsilon$. By Theorem 2, there is some $g_{N}\in S(I,X)$ with $\left\|f_{N}-g_{N}\right\|_{\infty}<\epsilon$. Then,

 $\left\|f-g_{N}\right\|_{\infty}\leq\left\|f-f_{N}\right\|_{\infty}+\left\|f_{N% }-g_{N}\right\|_{\infty}<2\epsilon,$

and by Theorem 2 this implies that $f\in R(I,X)$. ∎

The following lemma shows that the set of points of discontinuity of a regulated function taking values in a Banach space is countable.

###### Lemma 4.

If $I$ is a compact interval in $\mathbb{R}$, $X$ is a Banach space, and $f\in R(I,X)$, then

 $\{t\in I:\textrm{f is discontinuous at t}\}$

is countable.

###### Proof.

For each $n$ let $g_{n}\in S(I,X)$ satisfy $\left\|f-g_{n}\right\|\leq\frac{1}{n}$, and let

 $D_{n}=\{t\in I:\textrm{g_{n} is discontinuous at t}\}.$

$g_{n}$ is a step function so $D_{n}$ is finite, and hence $D=\bigcup_{n=1}^{\infty}D_{n}$ is countable. It need not be true that $f$ is discontinuous at each point in $D$, but we shall prove that if $t\in I\setminus D$ then $f$ is continuous at $t$, which will prove the claim.

Suppose that $t\in I\setminus D$, let $\epsilon>0$, and take $N>\frac{1}{\epsilon}$. As $t\not\in D_{N}$, the step function $g_{N}$ is continuous at $t$, and hence there is some $\delta>0$ for which $|s-t|<\delta$ implies that $\left\|g_{N}(s)-g_{N}(t)\right\|<\epsilon$. If $|s-t|<\delta$, then

 $\displaystyle\left\|f(s)-f(t)\right\|$ $\displaystyle\leq$ $\displaystyle\left\|f(s)-g_{N}(s)\right\|+\left\|g_{N}(s)-g_{N}(t)\right\|+% \left\|g_{N}(t)-f(t)\right\|$ $\displaystyle\leq$ $\displaystyle 2\left\|f-g_{N}\right\|_{\infty}+\left\|g_{N}(s)-g_{N}(t)\right\|$ $\displaystyle<$ $\displaystyle\frac{2}{N}+\epsilon$ $\displaystyle<$ $\displaystyle 3\epsilon,$

showing that $f$ is continuous at $t$. ∎

## 2 Integrals of step functions

Let $I=[a,b]$ and let $X$ be a normed space. If $f\in S(I,X)$ then there is a subdivision $a=s_{0} of $[a,b]$ and there are $c_{i}\in X$ such that $f$ takes the value $c_{i}$ on the open interval $(s_{i-1},s_{i})$. Suppose that there is a subdivision $a=t_{0} of $[a,b]$ and $d_{i}\in X$ such that $f$ takes the value $d_{i}$ on the open interval $(t_{i-1},t_{i})$. One checks that

 $\sum_{i=1}^{k}(s_{i}-s_{i-1})c_{i}=\sum_{i=1}^{l}(t_{i}-t_{i-1})d_{i}.$

We define the integral of $f$ to be the above element of $X$, and denote this element of $X$ by $\int_{I}f=\int_{a}^{b}f$.

###### Lemma 5.

If $I$ is a compact interval in $\mathbb{R}$ and $X$ is a normed space, then $\int_{I}:S(I,X)\to X$ is linear.

###### Lemma 6.

If $I=[a,b]$ and $X$ is a normed space, then $\int_{I}:S(I,X)\to X$ is a bounded linear map with operator norm $b-a$.

###### Proof.

If $f\in S(I,X)$, let $a=s_{0} be a subdivision of $[a,b]$ and let $c_{i}\in X$ such that $f$ takes the value $c_{i}$ on the open interval $(s_{i-1},s_{i})$. Then,

 $\left\|\int_{I}f\right\|\leq\sum_{i=1}^{k}(s_{i}-s_{i-1})\left\|c_{i}\right\|% \leq\sum_{i=1}^{k}(s_{i}-s_{i-1})\left\|f\right\|_{\infty}=(b-a)\left\|f\right% \|_{\infty}.$

This shows that $\left\|\int_{I}\right\|\leq b-a$, and if $f$ is constant, say $f(t)=c\in X$ for all $t\in I$, then $\int_{I}f=(b-a)c$ and $\left\|\int_{I}f\right\|=(b-a)\left\|c\right\|=(b-a)\left\|f\right\|_{\infty}$, showing that $\left\|\int_{I}\right\|=b-a$. ∎

###### Lemma 7.

If $a\leq b\leq c$, if $X$ is a normed space, and if $g\in S([a,c],X)$, then

 $\int_{a}^{c}g=\int_{a}^{b}g+\int_{b}^{c}g.$

## 3 The regulated integral

Let $I$ be a compact interval in $\mathbb{R}$ and let $X$ be a Banach space. Theorem 2 shows that $S(I,X)$ is a dense subspace of $R(I,X)$, and therefore if $T_{0}\in\mathscr{B}(S(I,X),X)$ then there is one and only one $T\in\mathscr{B}(R(I,X),X)$ whose restriction to $S(I,X)$ is equal to $T_{0}$, and this operator satisfies $\left\|T\right\|=\left\|T_{0}\right\|$. Lemma 6 shows that $\int_{I}:S(I,X)\to X$ is a bounded linear operator, thus there is one and only one bounded linear operator $R(I,X)\to X$ whose restriction to $S(I,X)$ is equal to $\int_{I}$, and we denote this operator $R(I,X)\to X$ also by $\int_{I}$. With $I=[a,b]$, we have $\left\|\int_{I}\right\|=b-a$. We call $\int_{I}:R(I,X)\to X$ the regulated integral.

###### Lemma 8.

If $a\leq b\leq c$, if $X$ is a Banach space, and if $f\in R([a,c],X)$, then

 $\int_{a}^{c}f=\int_{a}^{b}f+\int_{b}^{c}f.$
###### Proof.

Let $I_{1}=[a,b]$, $I_{2}=[b,c]$, $I=[a,c]$, and let $f_{1}$ and $f_{2}$ be the restriction of $f$ to $I_{1}$ and $I_{2}$ respectively. From the definition of regulated functions, $f_{1}\in R(I_{1},X)$ and $f_{2}\in R(I_{2},X)$. By Theorem 2, for any $\epsilon>0$ there is some $g\in S(I,X)$ satisfying $\left\|f-g\right\|_{\infty}<\epsilon$. Taking $g_{1}$ and $g_{2}$ to be the restriction of $g$ to $I_{1}$ and $I_{2}$, we check that $g_{1}\in S(I_{1},X)$ and $g_{2}\in S(I_{2},X)$. Then by Lemma 7,

 $\displaystyle\left\|\int_{I}f-\int_{I_{1}}f_{1}-\int_{I_{2}}f_{2}\right\|_{\infty}$ $\displaystyle\leq$ $\displaystyle\left\|\int_{I}f-\int_{I}g\right\|_{\infty}+\left\|\int_{I}g-\int% _{I_{1}}g_{1}-\int_{I_{2}}g_{2}\right\|_{\infty}$ $\displaystyle+\left\|\int_{I_{1}}g_{1}+\int_{I_{2}}g_{2}-\int_{I_{1}}f_{1}-% \int_{I_{2}}f_{2}\right\|_{\infty}$ $\displaystyle=$ $\displaystyle\left\|\int_{I}(f-g)\right\|_{\infty}+0$ $\displaystyle+\left\|\int_{I_{1}}(g_{1}-f_{1})\right\|_{\infty}+\left\|\int_{I% _{2}}(g_{2}-f_{2})\right\|_{\infty}$ $\displaystyle\leq$ $\displaystyle(c-a)\left\|f-g\right\|_{\infty}+(b-a)\left\|g_{1}-f_{2}\right\|_% {\infty}$ $\displaystyle+(c-b)\left\|g_{2}-f_{2}\right\|_{\infty}.$

But $\left\|g_{1}-f_{1}\right\|_{\infty}\leq\left\|g-f\right\|_{\infty}$ and $\left\|g_{2}-f_{2}\right\|_{\infty}\leq\left\|g-f\right\|_{\infty}$, hence we obtain

 $\left\|\int_{I}f-\int_{I_{1}}f_{1}-\int_{I_{2}}f_{2}\right\|_{\infty}<(c-a)% \epsilon+(b-a)\epsilon+(c-b)\epsilon=2(c-a)\epsilon.$

Since $\epsilon>0$ was arbitrary, we get

 $\left\|\int_{I}f-\int_{I_{1}}f_{1}-\int_{I_{2}}f_{2}\right\|_{\infty}=0,$

so

 $\int_{I}f=\int_{I_{1}}f_{1}+\int_{I_{2}}f_{2},$

proving the claim. ∎

We prove that applying a bounded linear map and taking the regulated integral commute.22 2 Jean-Paul Penot, Calculus Without Derivatives, p. 124, Proposition 2.18.

###### Lemma 9.

Suppose that $I$ is a compact interval in $\mathbb{R}$ and that $X$ and $Y$ are Banach spaces. If $f\in R(I,X)$ and $T\in\mathscr{B}(X,Y)$, then $T\circ f\in R(I,Y)$ and

 $\int_{I}T\circ f=T\int_{I}f.$
###### Proof.

Because $T$ is continuous we have $T\circ f\in R(I,Y)$. For $\epsilon>0$, there is some $g\in S(I,X)$ satisfying $\left\|f-g\right\|_{\infty}<\epsilon$. Write $I=[a,b]$. Because $g$ is a step function, there is a subdivision $a=s_{0} of $I$ and there are $c_{i}\in X$ such that $g$ takes the value $c_{i}$ on the open interval $(s_{i-1},s_{i})$. Furthermore, $T\circ g$ takes the value $Tc_{i}$ on the open interval $(s_{i-1},s_{i})$ so $T\circ g\in S(I,Y)$, and then because $T$ is linear,

 $\int_{I}T\circ g=\sum_{i=1}^{k}(s_{i}-s_{i-1})Tc_{i}=T\sum_{i=1}^{k}(s_{i}-s_{% i-1})c_{i}=T\int_{I}g.$

Using this,

 $\displaystyle\left\|\int_{I}T\circ f-T\int_{I}f\right\|$ $\displaystyle\leq$ $\displaystyle\left\|\int_{I}T\circ f-\int_{I}T\circ g\right\|+\left\|\int_{I}T% \circ g-T\int_{I}g\right\|$ $\displaystyle+\left\|T\int_{I}g-T\int_{I}f\right\|$ $\displaystyle=$ $\displaystyle\left\|\int_{I}T\circ(f-g)\right\|+\left\|T\int_{I}(f-g)\right\|$ $\displaystyle\leq$ $\displaystyle(b-a)\left\|T\circ(f-g)\right\|_{\infty}+\left\|T\right\|\left\|% \int_{I}(f-g)\right\|$ $\displaystyle\leq$ $\displaystyle(b-a)\left\|T\right\|\left\|f-g\right\|_{\infty}+\left\|T\right\|% (b-a)\left\|f-g\right\|_{\infty}$ $\displaystyle<$ $\displaystyle 2(b-a)\left\|T\right\|\epsilon.$

As $\epsilon>0$ is arbitrary, this means that

 $\left\|\int_{I}T\circ f-T\int_{I}f\right\|=0,$

and so

 $\int_{I}T\circ f=T\int_{I}f.$

## 4 Left and right derivatives

Suppose that $I$ is an open interval in $\mathbb{R}$, $X$ is a normed space, $f\in X^{I}$, and $t\in I$. We say that $f$ is right-differentiable at $t$ if $\frac{f(t+h)-f(t)}{h}$ has a limit as $h\to 0^{+}$, and that $f$ is left-differentiable at $t$ if $\frac{f(t+h)-f(t)}{h}$ has a limit as $h\to 0^{-}$. We call these limits respectively the right derivative of $f$ at $t$ and the left derivative of $f$ at $t$, denoted respectively by $f_{+}^{\prime}(t)$ and $f_{-}^{\prime}(t)$. For $f$ to be differentiable at $t$ means that $f_{+}^{\prime}(t)$ and $f_{-}^{\prime}(t)$ exist and are equal.

The following is the mean value theorem for functions taking values in a Banach space.33 3 Henri Cartan, Differential Calculus, p. 39, Theorem 3.1.3.

###### Theorem 10 (Mean value theorem).

Suppose that $I=[a,b]$, that $X$ is a Banach space, and that $f:I\to X$ and $g:I\to\mathbb{R}$ are continuous functions. If there is a countable set $D\subset I$ such that $t\in I\setminus D$ implies that $f_{+}^{\prime}(t)$ and $g_{+}^{\prime}(t)$ exist and satisfy $\left\|f_{+}^{\prime}(t)\right\|\leq g_{+}^{\prime}(t)$, then

 $\left\|f(b)-f(a)\right\|\leq g(b)-g(a).$
###### Corollary 11.

Suppose that $I=[a,b]$, that $X$ is a Banach space, and that $f:I\to X$ is continuous. If there is a countable set $D\subset I$ such that $t\in I\setminus D$ implies that $f_{+}^{\prime}(t)=0$, then $f$ is constant on $I$.

## 5 Primitives

Let $I=[a,b]$, let $X$ be a normed space, and let $f,g\in X^{I}$. We say that $g$ is a primitive of $f$ if $g$ is continuous and if there is a countable set $D\subset I$ such that $t\in I\setminus D$ implies that $g$ is differentiable at $t$ and $g^{\prime}(t)=f(t)$.

###### Lemma 12.

Suppose that $I$ is a compact interval in $\mathbb{R}$, that $X$ is a Banach space, and that $f:I\to X$ is a function. If $g_{1},g_{2}:I\to X$ are primitives of $f$, then $g_{1}-g_{2}$ is constant on $I$.

###### Proof.

For $i=1,2$, as $g_{i}$ is a primitive of $f$ there is a countable set $D_{i}\subset I$ such that $t\in I\setminus D_{i}$ implies that $g_{i}$ is differentiable at $t$ and $g_{i}^{\prime}(t)=f(t)$. Let $D=D_{1}\cup D_{2}$, which is a countable set. Both $g_{1}$ and $g_{2}$ are continuous so $g=g_{1}-g_{2}:I\to X$ is continuous, and if $t\in I\setminus D$ then $g$ is differentiable at $t$ and $g^{\prime}(t)=g_{1}^{\prime}(t)-g_{2}^{\prime}(t)=f(t)-f(t)=0$. Then Corollary 11 shows that $g$ is constant on $I$, i.e., that $g_{1}-g_{2}$ is constant on $I$.∎

We now give a construction of primitives of regulated functions.44 4 Jean-Paul Penot, Calculus Without Derivatives, p. 124, Theorem 2.19.

###### Theorem 13.

If $I=[a,b]$, $X$ is a Banach space, and $f\in R(I,X)$, then the map $g:I\to X$ defined by $g(t)=\int_{a}^{t}f$ is a primitive of $f$ on $I$.

###### Proof.

For $t\in[a,b)$ and $\epsilon>0$, because $f$ is regulated there is some $0<\delta such that $0 implies that $\left\|f(t+r)-f(t^{+})\right\|\leq\epsilon$. For $0 and for any $0<\eta, using Lemma 8 we have

 $\displaystyle\left\|\int_{a}^{t+r}f-\int_{a}^{t}f-\int_{t}^{t+r}f(t^{+})\right\|$ $\displaystyle=$ $\displaystyle\left\|\int_{t}^{t+r}f-\int_{t}^{t+r}f(t^{+})\right\|$ $\displaystyle=$ $\displaystyle\left\|\int_{t}^{t+\eta}(f-f(t^{+}))+\int_{t+\eta}^{t+r}(f-f(t^{+% }))\right\|$ $\displaystyle\leq$ $\displaystyle\eta\sup_{t\leq s\leq t+\eta}\left\|f(s)-f(t^{+})\right\|$ $\displaystyle+(r-\eta)\sup_{t+\eta\leq s\leq t+r}\left\|f(s)-f(t^{+})\right\|$ $\displaystyle\leq$ $\displaystyle 2\left\|f\right\|_{\infty}\eta+(r-\eta)\epsilon.$

This is true for all $0<\eta, so

 $\left\|\int_{a}^{t+r}f-\int_{a}^{t}f-\int_{t}^{t+r}f(t^{+})\right\|\leq r\epsilon,$

i.e.

 $\left\|\frac{g(t+r)-g(t)}{r}-f(t^{+})\right\|\leq\epsilon.$

This shows that

 $g_{+}^{\prime}(t)=f(t^{+}).$

Similarly,

 $g_{-}^{\prime}(t)=f(t^{-}).$

Because $f$ is regulated, Lemma 4 shows that there is a countable set $D\subset I$ such that $t\in I\setminus D$ implies that $f$ is continuous at $t$. Therefore, if $t\in I\setminus D$ then $f(t^{+})=f(t^{-})=f(t)$, so $g_{+}^{\prime}(t)=g_{-}^{\prime}(t)$, which means that if $t\in I\setminus D$ then $g$ is differentiable at $t$, with $g^{\prime}(t)=f(t)$. To prove that $g$ is a primitive of $f$ on $I$ it suffices now to show that $g$ is continuous. For $\epsilon>0$ and $t\in I$, let $\delta=\frac{\epsilon}{\left\|f\right\|_{\infty}}$, and then for $|s-t|<\delta$ we have by Lemma 8 that

 $\left\|g(s)-g(t)\right\|=\left\|\int_{a}^{s}f-\int_{a}^{t}f\right\|=\left\|% \int_{s}^{t}f\right\|\leq|t-s|\left\|f\right\|_{\infty}<\delta\left\|f\right\|% _{\infty}=\epsilon,$

showing that $g$ is continuous at $t$, completing the proof. ∎

Suppose that $X$ is a Banach space and that $f:[a,b]\to X$ is a primitive of a regulated function $h:[a,b]\to X$. Because $h$ is regulated, by Theorem 13 the function $g:[a,b]\to X$ defined by $g(t)=\int_{a}^{t}f$ is a primitive of $f$ on $[a,b]$. Then applying Lemma 12, there is some $c\in X$ such that $f(t)-g(t)=c$ for all $t\in[a,b]$. But $f(a)-g(a)=f(a)$, so $c=f(a)$. Hence, for all $t\in[a,b]$,

 $f(t)=f(a)+\int_{a}^{t}h.$

But

 $\int_{a}^{t}h=\int_{a}^{a+\eta_{1}}h+\int_{a+\eta_{1}}^{t-\eta_{2}}h+\int_{t-% \eta_{2}}^{t}h=\int_{a}^{a+\eta_{1}}h+\int_{a+\eta_{1}}^{t-\eta_{2}}f^{\prime}% +\int_{t-\eta_{2}}^{t}h$

and

 $\left\|\int_{a}^{a+\eta_{1}}h\right\|\leq\eta_{1}\left\|h\right\|_{\infty},% \qquad\left\|\int_{t-\eta_{2}}^{t}h\right\|\leq\eta_{2}\left\|h\right\|_{% \infty},$

hence as $\eta_{1}\to 0^{+}$ and $\eta_{2}\to 0^{+}$,

 $\int_{a+\eta_{1}}^{t-\eta_{2}}f^{\prime}\to\int_{a}^{t}h,$

and so it makes sense to write

 $\int_{a}^{t}f^{\prime}=\int_{a}^{t}h,$

and thus for all $t\in[a,b]$,

 $f(t)=f(a)+\int_{a}^{t}f^{\prime}.$