# Total variation, absolute continuity, and the Borel $\sigma$-algebra of $C(I)$

Jordan Bell
March 10, 2015

## 1 Total variation

Let $a. A partition of $[a,b]$ is a sequence $t_{0},t_{1},\ldots,t_{n}$ such that

 $a=t_{0}

The total variation of a function $f:[a,b]\to\mathbb{C}$ is

 $\mathrm{Var}_{f}[a,b]=\sup\left\{\sum_{i=1}^{n}|f(t_{i})-f(t_{i-1})|:\textrm{% t_{0},t_{1},\ldots,t_{n} is a partition of [a,b]}\right\}.$

If $\mathrm{Var}_{f}[a,b]<\infty$ then we say that $f$ has bounded variation.

###### Lemma 1.

If $a\leq c, then

 $\mathrm{Var}_{f}[c,d]=\mathrm{Var}_{f}[c,e]+\mathrm{Var}_{f}[e,d].$

The following theorem establishes properties of functions of bounded variation.11 1 Charalambos D. Aliprantis and Owen Burkinshaw, Principles of Real Analysis, third ed., p. 377, Theorem 39.10.

###### Theorem 2.

Suppose that $f:[a,b]\to\mathbb{R}$ is of bounded variation and define

 $F(x)=\mathrm{Var}_{f}[a,x],\qquad x\in[a,b].$

Then:

1. 1.

$|f(y)-f(x)|\leq F(y)-F(x)$ for all $a\leq x.

2. 2.

$F$ is a nondecreasing function.

3. 3.

$F-f$ and $F+f$ are nondecreasing functions.

4. 4.

For $x_{0}\in[a,b]$, $f$ is continuous at $x_{0}$ if and only if $F$ is continuous at $x_{0}$.

###### Proof.

If $t_{0},\ldots,t_{n}$ is a partition of $[a,x]$ then $t_{0},\ldots,t_{n},y$ is a partition of $[a,y]$, so

 $\sum_{i=1}^{n}|f(t_{i})-f(t_{i-1})|+|f(y)-f(x)|\leq F(y).$

Since this is true for any partition $t_{0},\ldots,t_{n}$ of $[a,x]$,

 $F(x)+|f(y)-f(x)|\leq F(y).$

This shows in particular that $F(x)\leq F(y)$, and thus that $F$ is nondecreasing.

For $a\leq x,

 $f(y)-f(x)\leq|f(y)-f(x)|\leq F(y)-F(x),$

thus

 $F(x)-f(x)\leq F(y)-f(y),$

showing that $x\mapsto F(x)-f(x)$ is nondecreasing. Likewise,

 $f(x)-f(y)\leq|f(y)-f(x)|\leq F(y)-F(x),$

thus

 $f(x)+F(x)\leq f(y)+F(y),$

showing that $x\mapsto F(x)+f(x)$ is nondecreasing.

Suppose that $F$ is continuous at $x_{0}$ and let $\epsilon>0$. There is some $\delta>0$ such that $|x-x_{0}|<\delta$ implies that $|F(x)-F(x_{0})|<\epsilon$. If $|x-x_{0}|<\delta$, then

 $|f(x)-f(x_{0})|\leq|F(x)-F(x_{0})|<\epsilon,$

showing that $f$ is continuous at $x_{0}$.

Suppose that $f$ is continuous at $x_{0}$ and let $\epsilon>0$. Then there is some $\delta>0$ such that $|x-x_{0}|<\delta$ implies that $|f(x)-f(x_{0})|<\epsilon$, and such that $x_{0}-\delta>a$. Let $x_{0}-\delta, and let $t_{0},\ldots,t_{n}$ be a partition of $[s,b]$ such that

 $\mathrm{Var}_{f}[s,b]<\sum_{i=1}^{n}|f(t_{i})-f(t_{i-1})|+\epsilon$

and such that none of $t_{0},\ldots,t_{n}$ is equal to $x_{0}$. Say that $t_{k}. Then

 $t_{0},\ldots,t_{k},x_{0},t_{k+1},\ldots,t_{n}$

is a partition of $[s,b]$. For $t_{k} we have $|x-x_{0}|<\delta$ and therefore

 $\displaystyle\mathrm{Var}_{f}[s,x]+\mathrm{Var}_{f}[x,b]$ $\displaystyle=\mathrm{Var}_{f}[s,b]$ $\displaystyle<\sum_{i=1}^{n}|f(t_{i})-f(t_{i-1})|+\epsilon$ $\displaystyle\leq\sum_{i=1}^{k}|f(t_{i})-f(t_{i-1})|+|f(x)-f(t_{k})|$ $\displaystyle+|f(x_{0})-f(x)|$ $\displaystyle+|f(t_{k+1})-f(x_{0})|+\sum_{i=k+2}^{n}|f(t_{i})-f(t_{i-1})|+\epsilon$ $\displaystyle\leq\mathrm{Var}_{f}[s,x]+|f(x)-f(x_{0})|+\mathrm{Var}_{f}[x_{0},% b]+\epsilon$ $\displaystyle<\mathrm{Var}_{f}[s,x]+\mathrm{Var}_{f}[x_{0},b]+2\epsilon,$

giving

 $\mathrm{Var}_{f}[x,b]-\mathrm{Var}_{f}[x_{0},b]<2\epsilon.$

As $\mathrm{Var}_{f}[a,b]=\mathrm{Var}_{f}[a,x]+\mathrm{Var}_{f}[x,b]$ and also $\mathrm{Var}_{f}[a,b]=\mathrm{Var}_{f}[a,x_{0}]+\mathrm{Var}_{f}[x_{0},b]$, we have $F(x)+\mathrm{Var}_{f}[x,b]=F(x_{0})+\mathrm{Var}_{f}[x_{0},b]$, and therefore

 $F(x_{0})-F(x)<2\epsilon.$

Thus, if $t_{k} then $|F(x_{0})-F(x)|<2\epsilon$, showing that $F$ is left-continuous at $x_{0}$. It is straightforward to show in the same way that $F$ is right-continuous at $x_{0}$, and thus continuous at $x_{0}$. ∎

If $f:[a,b]\to\mathbb{R}$ is of bounded variation, then Theorem 2 tells us that $F$ and $F+f$ are nondecreasing functions. A monotone function is differentiable almost everywhere,22 2 Charalambos D. Aliprantis and Owen Burkinshaw, Principles of Real Analysis, third ed., p. 375, Theorem 39.9. and it follows that $f=(F+f)-F$ is differentiable almost everywhere.

## 2 Absolute continuity

Let $a and let $I=[a,b]$. A function $f:I\to\mathbb{C}$ is said to be absolutely continuous if for any $\epsilon>0$ there is some $\delta>0$ such that for any $n$ and any collection of pairwise disjoint intervals $(\alpha_{1},\beta_{1}),\ldots,(\alpha_{n},\beta_{n})$ satisfying

 $\sum_{i=1}^{n}(\beta_{i}-\alpha_{i})<\delta,$

we have

 $\sum_{i=1}^{n}|f(\beta_{i})-f(\alpha_{i})|<\epsilon.$

It is immediate that if $f$ is absolutely continuous then $f$ is uniformly continuous.

###### Lemma 3.

If $f:[a,b]\to\mathbb{C}$ is absolutely continuous then $f$ has bounded variation.

###### Proof.

Because $f$ is absolutely continuous, there is some $\delta>0$ such that if $(\alpha_{1},\beta_{1}),\ldots,(\alpha_{n},\beta_{n})$ are pairwise disjoint and

 $\sum_{i=1}^{n}(\beta_{i}-\alpha_{i})<\delta,$

then

 $\sum_{i=1}^{n}|f(\beta_{i})-f(\alpha_{i})|<1.$

Let $N$ be an integer that is $>\frac{b-a}{\delta}$ and let $a=x_{0}<\cdots such that $x_{i}-x_{i-1}<\frac{b-a}{N}$ for each $i=1,\ldots,N$. Then

 $\mathrm{Var}_{f}[a,b]=\sum_{i=1}^{N}\mathrm{Var}_{f}[x_{i-1},x_{i}]\leq N,$

showing that $f$ has bounded variation. ∎

Let $\lambda$ be Lebesgue measure on $\mathbb{R}$ and let $\mathfrak{M}$ be the collection of Lebesgue measurable subsets of $\mathbb{R}$.

The following theorem establishes connections between absolute continuity of a function and Lebesgue measure.33 3 Walter Rudin, Real and Complex Analysis, third ed., p. 146, Theorem 7.18. In the following theorem, we extend $f:[a,b]\to\mathbb{R}$ to $\mathbb{R}\to\mathbb{R}$ by defining $f(x)=f(b)$ for $x>b$ and $f(x)=f(a)$ for $x. In particular, for any $x>b$, $f^{\prime}(x)$ exists and is equal to $0$, and for any $x, $f^{\prime}(x)$ exists and is equal to $0$.

###### Theorem 4.

Suppose that $I=[a,b]$ and that $f:I\to\mathbb{R}$ is continuous and nondecreasing. Then the following statements are equivalent.

1. 1.

$f$ is absolutely continuous.

2. 2.

If $E\subset I$ and $\lambda(E)=0$ then $\lambda(f(E))=0$. (In words: $f$ has the Luzin property.)

3. 3.

$f$ is differentiable $\lambda$-almost everywhere on $I$, $f^{\prime}\in L^{1}(\lambda)$, and

 $f(x)-f(a)=\int_{a}^{x}f^{\prime}(t)d\lambda(t),\qquad a\leq x\leq b.$
###### Proof.

Assume that $f$ is absolutely continuous and let $E\subset I$ with $\lambda(E)=0$. Let $E_{0}=E\setminus\{a,b\}$; to prove that $\lambda(f(E))=0$ it suffices to prove that $\lambda(f(E_{0}))=0$. Let $\epsilon>0$. As $f$ is absolutely continuous, there is some $\delta>0$ such that for any $n$ and any collection of pairwise disjoint intervals $(\alpha_{1},\beta_{1}),\ldots,(\alpha_{n},\beta_{n})$ satisfying

 $\sum_{i=1}^{n}(\beta_{i}-\alpha_{i})<\delta,$

we have

 $\sum_{i=1}^{n}|f(\beta_{i})-f(\alpha_{i})|<\epsilon.$

There is an open set $V$ such that $E_{0}\subset V\subset I$ and such that $\lambda(V)<\delta$. (Lebesgue measure is outer regular.) There are countably many pairwise disjoint intervals $(\alpha_{i},\beta_{i})$ such that $V=\bigcup_{i}(\alpha_{i},\beta_{i})$. Then

 $\sum_{i}(\beta_{i}-\alpha_{i})=\lambda(V)<\delta,$

so for any $n$,

 $\sum_{i=1}^{n}(\beta_{i}-\alpha_{i})<\delta,$

and because $f$ is absolutely continuous it follows that

 $\sum_{i=1}^{n}|f(\beta_{i})-f(\alpha_{i})|<\epsilon.$

This is true for all $n$, so

 $\sum_{i}|f(\beta_{i})-f(\alpha_{i})|\leq\epsilon.$

Because $f$ is continuous and nondecreasing, $f(\alpha_{i},\beta_{i})=(f(\alpha_{i}),f(\beta_{i}))$ for each $i$. Therefore

 $f(V)=f\left(\bigcup_{i}(\alpha_{i},\beta_{i})\right)=\bigcup_{i}f(\alpha_{i},% \beta_{i})=\bigcup_{i}(f(\alpha_{i}),f(\beta_{i})),$

which gives

 $\lambda(f(V))=\sum_{i}(f(\beta_{i})-f(\alpha_{i}))=\sum_{i}|f(\beta_{i})-f(% \alpha_{i})|\leq\epsilon.$

This is true for all $\epsilon>0$, so $\lambda(f(V))=0$. Because $f(E_{0})\subset f(V)$, it follows that $f(E_{0})\in\mathfrak{M}$ (Lebesgue measure is complete) and that $\lambda(f(E_{0}))=0$.

Assume that for all $E\subset I$ with $\lambda(E)=0$, $\lambda(f(E))=0$. Define $g:I\to\mathbb{R}$ by

 $g(x)=x+f(x),\qquad x\in I.$

Because $f$ is continuous and nondecreasing, $g$ is continuous and strictly increasing. Thus if $(\alpha,\beta)\subset I$ then $g(\alpha,\beta)=(g(\alpha),g(\beta))$ and so

 $\lambda(g(\alpha,\beta))=g(\beta)-g(\alpha)=\beta+f(\beta)-(\alpha+f(\alpha))=% \beta-\alpha+f(\beta)-f(\alpha),$

showing that if $J\subset I$ is an interval then $\lambda(g(J))=\lambda(J)+\lambda(f(J))$. Suppose that $E\subset I$ and $\lambda(E)=0$, and let $\epsilon>0$. There are countably many pairwise disjoint intervals $(\alpha_{i},\beta_{i})$ such that $E\subset\bigcup_{i}(\alpha_{i},\beta_{i})$ and $\sum_{i}(\beta_{i}-\alpha_{i})<\epsilon$, and because $\lambda(f(E))=0$, there are countably many pairwise disjoint intervals $(\gamma_{i},\delta_{i})$ such that $f(E)\subset\bigcup_{i}(\gamma_{i},\delta_{i})$ and $\sum_{i}(\delta_{i}-\gamma_{i})<\epsilon$. Let

 $N=f^{-1}\left(\bigcup_{i}(\gamma_{i},\delta_{i})\right)\cap\bigcup_{i}(\alpha_% {i},\beta_{i})=\bigcup_{i,j}(f^{-1}(\gamma_{i},\delta_{i})\cap(\alpha_{i},% \beta_{i}))\in\mathfrak{M}.$

We check that

 $\lambda(g(N))=\lambda(N)+\lambda(f(N)),$

and because

 $\lambda(N)+\lambda(f(N))\leq\sum_{i}(\beta_{i}-\alpha_{i})+\sum_{i}(\delta_{i}% -\gamma_{i})<2\epsilon$

we have

 $\lambda(g(N))<2\epsilon.$

Finally, $E\subset N$ so $g(E)\subset g(N)$. Therefore, for every $\epsilon>0$ there is some $N\in\mathfrak{M}$ with $g(E)\subset g(N)$ and $\lambda(g(N))<\epsilon$, from which it follows that $\lambda(g(E))=0$.

Suppose that $E\subset I$ belongs to $\mathfrak{M}$. Because $E\in\mathfrak{M}$, there are $E_{0},E_{1}\in\mathfrak{M}$ such that $E=E_{0}\cup E_{1}$, $\lambda(E_{0})=0$, and $E_{1}$ is a countable union of closed sets (namely, an $F_{\sigma}$-set). On the one hand, as $E_{1}\subset I$, $E_{1}$ is a countable union of compact sets, and because $g$ is continuous, $g(E_{1})$ is a countable union of compact sets, and in particular belongs to $\mathfrak{M}$. On the other hand, because $\lambda(E_{0})=0$, $g(E_{0})\in\mathfrak{M}$. Therefore $g(E)=g(E_{0})\cup g(E_{1})\in\mathfrak{M}$. Define $\mu:\mathfrak{M}\to[0,\infty)$ by

 $\mu(E)=\lambda(g(E\cap I)),\qquad E\in\mathfrak{M}.$

If $E_{i}$ are countably many pairwise disjoint elements of $\mathfrak{M}$, then $g(E_{i}\cap I)$ are pairwise disjoint elements of $\mathfrak{M}$, hence

 $\displaystyle\mu\left(\bigcup_{i}E_{i}\right)$ $\displaystyle=\lambda\left(g\left(\left(\bigcup_{i}E_{i}\right)\cap I\right)\right)$ $\displaystyle=\lambda\left(\bigcup_{i}g(E_{i}\cap I)\right)$ $\displaystyle=\sum_{i}\lambda(g(E_{i}\cap I))$ $\displaystyle=\sum_{i}\mu(E_{i}),$

showing that $\mu$ is a measure. If $\lambda(E)=0$, then $\lambda(E\cap I)=0$ so $\lambda(g(E\cap I))=0$, i.e. $\mu(E)=0$. This shows that $\mu$ is absolutely continuous with respect to $\lambda$. Therefore by the Radon-Nikodym theorem44 4 Walter Rudin, Real and Complex Analysis, third ed., p. 121, Theorem 6.10. there is a unique $h\in L^{1}(\lambda)$ such that

 $\mu(E)=\int_{E}hd\lambda,\qquad E\in\mathfrak{M}.$

$h(x)\geq 0$ for $\lambda$-almost all $x\in\mathbb{R}$.

Suppose that $x\in\mathbb{R}$ and let $E=[a,x]$. Then $g(E)=[g(a),g(x)]$, and

 $\mu(E)=\int_{E}h(t)d\lambda(t)=\int_{a}^{x}h(t)d\lambda(t).$

On the other hand,

 $\mu(E)=\lambda(g(E))=\lambda([g(a),g(x)])=g(x)-g(a)=x+f(x)-(a+f(a)).$

Hence

 $f(x)-f(a)=\int_{a}^{x}h(t)d\lambda(t)-(x-a),$

i.e.,

 $f(x)-f(a)=\int_{a}^{x}(h(t)-1)d\lambda(t).$

By the Lebesgue differentiation theorem,55 5 Walter Rudin, Real and Complex Analysis, third ed., p. 141, Theorem 7.11. $f^{\prime}(x)=h(x)-1$ for $\lambda$-almost all $x\in\mathbb{R}$, and it follows that $f^{\prime}\in L^{1}(\lambda)$ and

 $f(x)-f(a)=\int_{a}^{x}f^{\prime}(t)d\lambda(t),\qquad x\in I.$

Assume that $f$ is differentiable $\lambda$-almost everywhere in $I$, $f^{\prime}\in L^{1}(\lambda)$, and

 $f(x)-f(a)=\int_{a}^{x}f^{\prime}(t)d\lambda(t),\qquad x\in I.$

Let $\epsilon>0$ and let $(\alpha_{1},\beta_{1}),\ldots,(\alpha_{n},\beta_{n})$ be pairwise disjoint intervals satisfying

 $\sum_{i=1}^{n}(\beta_{i}-\alpha_{i})<\delta.$

Because $f$ is nondecreasing, for $\lambda$-almost all $x\in I$, $f^{\prime}(x)\geq 0$, and hence the measure $\mu$ defined by $d\mu=f^{\prime}d\lambda$ is absolutely continuous with respect to $\lambda$. It follows66 6 Walter Rudin, Real and Complex Analysis, third ed., p. 124, Theorem 6.11. that there is some $\delta>0$ such that for $E\in\mathfrak{M}$, $\lambda(E)<\delta$ implies that $\mu(E)<\epsilon$. This gives us

 $\mu\left(\bigcup_{i=1}^{n}(\alpha_{i},\beta_{i})\right)<\epsilon,$

and as

 $\mu(\alpha_{i},\beta_{i})=\int_{\alpha_{i}}^{\beta_{i}}f^{\prime}(t)d\lambda(t% )=f(\beta_{i})-f(\alpha_{i}),$

we get

 $\sum_{i=1}^{n}f(\beta_{i})-f(\alpha_{i})<\epsilon.$

This shows that $f$ is absolutely continuous, completing the proof. ∎

The following lemma establishes properties of the total variation of absolutely continuous functions.77 7 Walter Rudin, Real and Complex Analysis, third ed., p. 147, Theorem 7.19.

###### Lemma 5.

Suppose that $I=[a,b]$ and that $f:I\to\mathbb{R}$ is absolutely continuous. Then the function $F:I\to\mathbb{R}$ defined by

 $F(x)=\mathrm{Var}_{f}[a,x],\qquad x\in I$

is absolutely continuous.

###### Proof.

Let $\epsilon>0$. Because $f$ is absolutely continuous, there is some $\delta>0$ such that if $(a_{1},b_{1}),\ldots,(a_{m},b_{m})$ are disjoint intervals with $\sum_{k=1}^{m}(b_{k}-a_{k})<\delta$, then

 $\sum_{k=1}^{m}|f(b_{k})-f(a_{k})|<\epsilon.$

Suppose that $(\alpha_{1},\beta_{1}),\ldots,(\alpha_{n},\beta_{n})$ are disjoint intervals with $\sum_{i=1}^{n}(\beta_{i}-\alpha_{i})<\delta$. If $\alpha_{i}=t_{i,0}<\cdots for $i=1,\ldots,n$, then $(t_{i,j-1},t_{i,j})$, $1\leq i\leq n$, $1\leq j\leq m_{i}$, are disjoint intervals whose total length is $<\delta$, hence

 $\sum_{i=1}^{n}\sum_{j=1}^{m_{i}}|f(t_{i,j})-f(t_{i,j-1})|<\epsilon.$

It follows that

 $\sum_{i=1}^{n}|F(\beta_{i})-F(\alpha_{i})|=\sum_{i=1}^{n}\mathrm{Var}_{f}[% \alpha_{i},\beta_{i}]\leq\epsilon,$

which shows that $F$ is absolutely continuous. ∎

We now prove the fundamental theorem of calculus for absolutely continuous functions.88 8 Walter Rudin, Real and Complex Analysis, third ed., p. 148, Theorem 7.20.

###### Theorem 6.

Suppose that $I=[a,b]$ and that $f:I\to\mathbb{R}$ is absolutely continuous. Then $f$ is differentiable at almost all $x$ in $I$, $f^{\prime}\in L^{1}(\lambda)$, and

 $f(x)-f(a)=\int_{a}^{x}f^{\prime}(t)d\lambda(t),\qquad x\in I.$
###### Proof.

Define $F:I\to\mathbb{R}$ by

 $F(x)=\mathrm{Var}_{f}[a,x],\qquad x\in I.$

By Lemma 3, $f$ has bounded variation, and then using Theorem 2, $F-f$ and $F+f$ are nondecreasing. Furthermore, by Lemma 5, $F$ is absolutely continuous, so $F-f$ and $F+f$ are absolutely continuous. Let

 $f_{1}=\frac{F+f}{2},\qquad f_{2}=\frac{F-f}{2},$

which are thus nondecreasing and absolutely continuous. Applying Theorem 4, we get that $f_{1},f_{2}$ are differentiable at almost all $x\in I$, $f_{1}^{\prime},f_{2}^{\prime}\in L^{1}(\lambda)$, and

 $f_{1}(x)-f_{1}(a)=\int_{a}^{x}f_{1}^{\prime}(t)d\lambda(t),\qquad a\leq x\leq b$

and

 $f_{2}(x)-f_{2}(a)=\int_{a}^{x}f_{2}^{\prime}(t)d\lambda(t),\qquad a\leq x\leq b.$

Because $f=f_{1}-f_{2}$, $f$ is differentiable at almost all $x\in I$, $f^{\prime}=f_{1}^{\prime}-f_{2}^{\prime}\in L^{1}(\lambda)$, and

 $f(x)-f(a)=\int_{a}^{x}f^{\prime}(t)d\lambda(t),\qquad a\leq x\leq b,$

proving the claim. ∎

## 3 Borel sets

Let $I=[a,b]$. Denote by $C(I)$ the set of continuous functions $I\to\mathbb{C}$, which with the norm

 $\left\|f\right\|_{C(I)}=\sup_{x\in I}|f(x)|,\qquad f\in C(I),$

is a Banach space. Denote by $AC(I)$ the set of absolutely continuous functions $I\to\mathbb{C}$. Let $\mathscr{B}_{C(I)}$ be the Borel $\sigma$-algebra of $C(I)$. We have $AC(I)\subset C(I)$, and in the following theorem we prove that $AC(I)$ is a Borel set in $C(I)$.

###### Theorem 7.

$AC(I)\in\mathscr{B}_{C(I)}$.

###### Proof.

If $X,Y$ are Polish spaces, $f:X\to Y$ is continuous, $A\in\mathscr{B}_{X}$, and $f|A$ is injective, then $f(A)\in\mathscr{B}_{Y}$.99 9 Alexander Kechris, Classical Descriptive Set Theory, p. 89, Theorem 15.1. Let $X=\mathbb{C}\times L^{1}(I)$, which is a Banach space with the norm

 $\left\|(A,g)\right\|_{X}=|A|+\int_{a}^{b}|g|d\lambda,\qquad(A,g)\in X.$

Furthermore, $\mathbb{C}$ and $L^{1}(I)$ are separable and thus so is $X$, so $X$ is indeed a Polish space. The Banach space $C(I)$ is separable and thus is a Polish space. Define $\Phi:X\to C(I)$ by

 $\Phi(A,g)(x)=A+\int_{a}^{x}g(t)d\lambda(t),\qquad(A,g)\in X,\qquad x\in I.$

For $(A_{1},g_{1}),(A_{2},g_{2})\in X$,

 $\displaystyle\left\|\Phi(A_{1},g_{1})-\Phi(A_{2},g_{2})\right\|_{C(I)}$ $\displaystyle=\left\|(A_{1}-A_{2})+\int_{a}^{x}(g_{1}(t)-g_{2}(t))d\lambda(t)% \right\|_{C(I)}$ $\displaystyle=|A_{1}-A_{2}|+\sup_{x\in I}\left|\int_{a}^{x}(g_{1}(t)-g_{2}(t))% d\lambda(t)\right|$ $\displaystyle\leq|A_{1}-A_{2}|+\int_{a}^{b}|g_{1}(t)-g_{2}(t)|d\lambda(t)$ $\displaystyle=\left\|(A_{1},g_{1})-(A_{2},g_{2})\right\|_{X},$

which shows that $\Phi:X\to C(I)$ is continuous.

Let $(A,g)\in X$ and $\epsilon>0$. Because $g\in L^{1}(I)$, there is some $\delta>0$ such that if $\lambda(E)<\delta$ then $\int_{E}|g|d\lambda<\epsilon$.1010 10 Walter Rudin, Real and Complex Analysis, third ed., p. 32, exercise 1.12. If $(\alpha_{1},\beta_{1}),\ldots,(\alpha_{n},\beta_{n})$ are disjoint intervals whose total length is $<\delta$, then, with $E=\bigcup_{i=1}^{n}(\alpha_{i},\beta_{i})$,

 $\displaystyle\sum_{i=1}^{n}|\Phi(A,g)(\beta_{i})-\Phi(A,g)(\alpha_{i})|$ $\displaystyle=\sum_{i=1}^{n}\left|\int_{\alpha_{i}}^{\beta_{i}}g(t)d\lambda(t)\right|$ $\displaystyle\leq\sum_{i=1}^{n}\int_{\alpha_{i}}^{\beta_{i}}|g(t)|d\lambda(t)$ $\displaystyle=\int_{E}|g|d\lambda$ $\displaystyle<\epsilon,$

showing that $\Phi(A,g)$ is absolutely continuous. On the other hand, let $f\in AC(I)$. From Theorem 6, $f$ is differentiable at almost all $x\in I$, $f^{\prime}\in L^{1}(I)$, and

 $f(x)-f(a)=\int_{a}^{x}f^{\prime}(t)d\lambda(t),\qquad x\in I.$

Then $(f(a),f^{\prime})\in X$, and the above gives us, for all $x\in I$,

 $\Phi(f(a),f^{\prime})(x)=f(a)+\int_{a}^{x}f^{\prime}(t)d\lambda(t)=f(x),$

thus $\Phi(f(a),f^{\prime})=f$. Therefore

 $\Phi(X)=AC(I).$

If $\Phi(A_{1},g_{1})=\Phi(A_{2},g_{2})$, then $\Phi(A_{1},g_{1})(a)=\Phi(A_{2},g_{2})(a)$ gives $A_{1}=A_{2}$. Using this, and defining $G:I\to\mathbb{C}$ by $G=\int_{a}^{x}(g_{1}(t)-g_{2}(t))d\lambda(t)$, we have $G(x)=0$ for all $x\in I$. Then $G^{\prime}(x)=0$ for all $x\in I$, and by the Lebesgue differentiation theorem1111 11 Walter Rudin, Real and Complex Analysis, third ed., p. 141, Theorem 7.11. we have $G^{\prime}(x)=g_{1}(x)-g_{2}(x)$ for almost all $x\in I$. That is, $g_{1}(x)=g_{2}(x)$ for almost all $x\in I$, and thus in $L^{1}(I)$ we have $g_{1}=g_{2}$. Therefore $\Phi:X\to C(I)$ is injective.

Therefore $\Phi(X)\in\mathscr{B}_{C(I)}$. ∎