Total variation, absolute continuity, and the Borel $\sigma $algebra of $C(I)$
1 Total variation
Let $$. A partition of $[a,b]$ is a sequence ${t}_{0},{t}_{1},\mathrm{\dots},{t}_{n}$ such that
$$ 
The total variation of a function $f:[a,b]\to \u2102$ is
$${\mathrm{Var}}_{f}[a,b]=sup\{\sum _{i=1}^{n}f({t}_{i})f({t}_{i1}):{t}_{0},{t}_{1},\mathrm{\dots},{t}_{n}\text{is a partition of}[a,b]\}.$$ 
If $$ then we say that $f$ has bounded variation.
Lemma 1.
If $$, 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.^{1}^{1} 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
$$F(x)={\mathrm{Var}}_{f}[a,x],x\in [a,b].$$ 
Then:

1.
$f(y)f(x)\le F(y)F(x)$ for all $$.

2.
$F$ is a nondecreasing function.

3.
$Ff$ and $F+f$ are nondecreasing functions.

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 ${t}_{0},\mathrm{\dots},{t}_{n}$ is a partition of $[a,x]$ then ${t}_{0},\mathrm{\dots},{t}_{n},y$ is a partition of $[a,y]$, so
$$\sum _{i=1}^{n}f({t}_{i})f({t}_{i1})+f(y)f(x)\le F(y).$$ 
Since this is true for any partition ${t}_{0},\mathrm{\dots},{t}_{n}$ of $[a,x]$,
$$F(x)+f(y)f(x)\le F(y).$$ 
This shows in particular that $F(x)\le F(y)$, and thus that $F$ is nondecreasing.
For $$,
$$f(y)f(x)\le f(y)f(x)\le F(y)F(x),$$ 
thus
$$F(x)f(x)\le F(y)f(y),$$ 
showing that $x\mapsto F(x)f(x)$ is nondecreasing. Likewise,
$$f(x)f(y)\le f(y)f(x)\le F(y)F(x),$$ 
thus
$$f(x)+F(x)\le f(y)+F(y),$$ 
showing that $x\mapsto F(x)+f(x)$ is nondecreasing.
Suppose that $F$ is continuous at ${x}_{0}$ and let $\u03f5>0$. There is some $\delta >0$ such that $$ implies that $$. If $$, then
$$ 
showing that $f$ is continuous at ${x}_{0}$.
Suppose that $f$ is continuous at ${x}_{0}$ and let $\u03f5>0$. Then there is some $\delta >0$ such that $$ implies that $$, and such that ${x}_{0}\delta >a$. Let $$, and let ${t}_{0},\mathrm{\dots},{t}_{n}$ be a partition of $[s,b]$ such that
$$ 
and such that none of ${t}_{0},\mathrm{\dots},{t}_{n}$ is equal to ${x}_{0}$. Say that $$. Then
$${t}_{0},\mathrm{\dots},{t}_{k},{x}_{0},{t}_{k+1},\mathrm{\dots},{t}_{n}$$ 
is a partition of $[s,b]$. For $$ we have $$ and therefore
${\mathrm{Var}}_{f}[s,x]+{\mathrm{Var}}_{f}[x,b]$  $={\mathrm{Var}}_{f}[s,b]$  
$$  
$\le {\displaystyle \sum _{i=1}^{k}}f({t}_{i})f({t}_{i1})+f(x)f({t}_{k})$  
$+f({x}_{0})f(x)$  
$+f({t}_{k+1})f({x}_{0})+{\displaystyle \sum _{i=k+2}^{n}}f({t}_{i})f({t}_{i1})+\u03f5$  
$\le {\mathrm{Var}}_{f}[s,x]+f(x)f({x}_{0})+{\mathrm{Var}}_{f}[{x}_{0},b]+\u03f5$  
$$ 
giving
$$ 
As ${\mathrm{Var}}_{f}[a,b]={\mathrm{Var}}_{f}[a,x]+{\mathrm{Var}}_{f}[x,b]$ and also ${\mathrm{Var}}_{f}[a,b]={\mathrm{Var}}_{f}[a,{x}_{0}]+{\mathrm{Var}}_{f}[{x}_{0},b]$, we have $F(x)+{\mathrm{Var}}_{f}[x,b]=F({x}_{0})+{\mathrm{Var}}_{f}[{x}_{0},b]$, and therefore
$$ 
Thus, if $$ then $$, showing that $F$ is leftcontinuous at ${x}_{0}$. It is straightforward to show in the same way that $F$ is rightcontinuous 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,^{2}^{2} 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 $$ and let $I=[a,b]$. A function $f:I\to \u2102$ is said to be absolutely continuous if for any $\u03f5>0$ there is some $\delta >0$ such that for any $n$ and any collection of pairwise disjoint intervals $({\alpha}_{1},{\beta}_{1}),\mathrm{\dots},({\alpha}_{n},{\beta}_{n})$ satisfying
$$ 
we have
$$ 
It is immediate that if $f$ is absolutely continuous then $f$ is uniformly continuous.
Lemma 3.
If $f:[a,b]\to \u2102$ is absolutely continuous then $f$ has bounded variation.
Proof.
Because $f$ is absolutely continuous, there is some $\delta >0$ such that if $({\alpha}_{1},{\beta}_{1}),\mathrm{\dots},({\alpha}_{n},{\beta}_{n})$ are pairwise disjoint and
$$ 
then
$$ 
Let $N$ be an integer that is $>\frac{ba}{\delta}$ and let $$ such that $$ for each $i=1,\mathrm{\dots},N$. Then
$${\mathrm{Var}}_{f}[a,b]=\sum _{i=1}^{N}{\mathrm{Var}}_{f}[{x}_{i1},{x}_{i}]\le N,$$ 
showing that $f$ has bounded variation. ∎
Let $\lambda $ be Lebesgue measure on $\mathbb{R}$ and let $\U0001d510$ be the collection of Lebesgue measurable subsets of $\mathbb{R}$.
The following theorem establishes connections between absolute continuity of a function and Lebesgue measure.^{3}^{3} 3 Walter Rudin, Real and Complex Analysis, third ed., p. 146, Theorem 7.18. In the following theorem, we extend $f:[a,b]\to \mathbb{R}$ to $\mathbb{R}\to \mathbb{R}$ by defining $f(x)=f(b)$ for $x>b$ and $f(x)=f(a)$ for $$. In particular, for any $x>b$, ${f}^{\prime}(x)$ exists and is equal to $0$, and for any $$, ${f}^{\prime}(x)$ 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.
$f$ is absolutely continuous.

2.
If $E\subset I$ and $\lambda (E)=0$ then $\lambda (f(E))=0$. (In words: $f$ has the Luzin property.)

3.
$f$ is differentiable $\lambda $almost everywhere on $I$, ${f}^{\prime}\in {L}^{1}(\lambda )$, and
$$f(x)f(a)={\int}_{a}^{x}{f}^{\prime}(t)\mathit{d}\lambda (t),a\le x\le b.$$
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 $\u03f5>0$. As $f$ is absolutely continuous, there is some $\delta >0$ such that for any $n$ and any collection of pairwise disjoint intervals $({\alpha}_{1},{\beta}_{1}),\mathrm{\dots},({\alpha}_{n},{\beta}_{n})$ satisfying
$$ 
we have
$$ 
There is an open set $V$ such that ${E}_{0}\subset V\subset I$ and such that $$. (Lebesgue measure is outer regular.) There are countably many pairwise disjoint intervals $({\alpha}_{i},{\beta}_{i})$ such that $V={\bigcup}_{i}({\alpha}_{i},{\beta}_{i})$. Then
$$ 
so for any $n$,
$$ 
and because $f$ is absolutely continuous it follows that
$$ 
This is true for all $n$, so
$$\sum _{i}f({\beta}_{i})f({\alpha}_{i})\le \u03f5.$$ 
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
$$\lambda (f(V))=\sum _{i}(f({\beta}_{i})f({\alpha}_{i}))=\sum _{i}f({\beta}_{i})f({\alpha}_{i})\le \u03f5.$$ 
This is true for all $\u03f5>0$, so $\lambda (f(V))=0$. Because $f({E}_{0})\subset f(V)$, it follows that $f({E}_{0})\in \U0001d510$ (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
$$g(x)=x+f(x),x\in I.$$ 
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 $\u03f5>0$. There are countably many pairwise disjoint intervals $({\alpha}_{i},{\beta}_{i})$ such that $E\subset {\bigcup}_{i}({\alpha}_{i},{\beta}_{i})$ and $$, and because $\lambda (f(E))=0$, there are countably many pairwise disjoint intervals $({\gamma}_{i},{\delta}_{i})$ such that $f(E)\subset {\bigcup}_{i}({\gamma}_{i},{\delta}_{i})$ and $$. Let
$$N={f}^{1}\left(\bigcup _{i}({\gamma}_{i},{\delta}_{i})\right)\cap \bigcup _{i}({\alpha}_{i},{\beta}_{i})=\bigcup _{i,j}({f}^{1}({\gamma}_{i},{\delta}_{i})\cap ({\alpha}_{i},{\beta}_{i}))\in \U0001d510.$$ 
We check that
$$\lambda (g(N))=\lambda (N)+\lambda (f(N)),$$ 
and because
$$ 
we have
$$ 
Finally, $E\subset N$ so $g(E)\subset g(N)$. Therefore, for every $\u03f5>0$ there is some $N\in \U0001d510$ with $g(E)\subset g(N)$ and $$, from which it follows that $\lambda (g(E))=0$.
Suppose that $E\subset I$ belongs to $\U0001d510$. Because $E\in \U0001d510$, there are ${E}_{0},{E}_{1}\in \U0001d510$ such that $E={E}_{0}\cup {E}_{1}$, $\lambda ({E}_{0})=0$, and ${E}_{1}$ is a countable union of closed sets (namely, an ${F}_{\sigma}$set). On the one hand, as ${E}_{1}\subset I$, ${E}_{1}$ is a countable union of compact sets, and because $g$ is continuous, $g({E}_{1})$ is a countable union of compact sets, and in particular belongs to $\U0001d510$. On the other hand, because $\lambda ({E}_{0})=0$, $g({E}_{0})\in \U0001d510$. Therefore $g(E)=g({E}_{0})\cup g({E}_{1})\in \U0001d510$. Define $\mu :\U0001d510\to [0,\mathrm{\infty})$ by
$$\mu (E)=\lambda (g(E\cap I)),E\in \U0001d510.$$ 
If ${E}_{i}$ are countably many pairwise disjoint elements of $\U0001d510$, then $g({E}_{i}\cap I)$ are pairwise disjoint elements of $\U0001d510$, hence
$\mu \left({\displaystyle \bigcup _{i}}{E}_{i}\right)$  $=\lambda \left(g\left(\left({\displaystyle \bigcup _{i}}{E}_{i}\right)\cap I\right)\right)$  
$=\lambda \left({\displaystyle \bigcup _{i}}g({E}_{i}\cap I)\right)$  
$={\displaystyle \sum _{i}}\lambda (g({E}_{i}\cap I))$  
$={\displaystyle \sum _{i}}\mu ({E}_{i}),$ 
showing that $\mu $ is a measure. If $\lambda (E)=0$, then $\lambda (E\cap I)=0$ so $\lambda (g(E\cap I))=0$, i.e. $\mu (E)=0$. This shows that $\mu $ is absolutely continuous with respect to $\lambda $. Therefore by the RadonNikodym theorem^{4}^{4} 4 Walter Rudin, Real and Complex Analysis, third ed., p. 121, Theorem 6.10. there is a unique $h\in {L}^{1}(\lambda )$ such that
$$\mu (E)={\int}_{E}h\mathit{d}\lambda ,E\in \U0001d510.$$ 
$h(x)\ge 0$ for $\lambda $almost all $x\in \mathbb{R}$.
Suppose that $x\in \mathbb{R}$ and let $E=[a,x]$. Then $g(E)=[g(a),g(x)]$, and
$$\mu (E)={\int}_{E}h(t)\mathit{d}\lambda (t)={\int}_{a}^{x}h(t)\mathit{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)\mathit{d}\lambda (t)(xa),$$ 
i.e.,
$$f(x)f(a)={\int}_{a}^{x}(h(t)1)\mathit{d}\lambda (t).$$ 
By the Lebesgue differentiation theorem,^{5}^{5} 5 Walter Rudin, Real and Complex Analysis, third ed., p. 141, Theorem 7.11. ${f}^{\prime}(x)=h(x)1$ for $\lambda $almost all $x\in \mathbb{R}$, and it follows that ${f}^{\prime}\in {L}^{1}(\lambda )$ and
$$f(x)f(a)={\int}_{a}^{x}{f}^{\prime}(t)\mathit{d}\lambda (t),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)\mathit{d}\lambda (t),x\in I.$$ 
Let $\u03f5>0$ and let $({\alpha}_{1},{\beta}_{1}),\mathrm{\dots},({\alpha}_{n},{\beta}_{n})$ be pairwise disjoint intervals satisfying
$$ 
Because $f$ is nondecreasing, for $\lambda $almost all $x\in I$, ${f}^{\prime}(x)\ge 0$, and hence the measure $\mu $ defined by $d\mu ={f}^{\prime}d\lambda $ is absolutely continuous with respect to $\lambda $. It follows^{6}^{6} 6 Walter Rudin, Real and Complex Analysis, third ed., p. 124, Theorem 6.11. that there is some $\delta >0$ such that for $E\in \U0001d510$, $$ implies that $$. This gives us
$$ 
and as
$$\mu ({\alpha}_{i},{\beta}_{i})={\int}_{{\alpha}_{i}}^{{\beta}_{i}}{f}^{\prime}(t)\mathit{d}\lambda (t)=f({\beta}_{i})f({\alpha}_{i}),$$ 
we get
$$ 
This shows that $f$ is absolutely continuous, completing the proof. ∎
The following lemma establishes properties of the total variation of absolutely continuous functions.^{7}^{7} 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
$$F(x)={\mathrm{Var}}_{f}[a,x],x\in I$$ 
is absolutely continuous.
Proof.
Let $\u03f5>0$. Because $f$ is absolutely continuous, there is some $\delta >0$ such that if $({a}_{1},{b}_{1}),\mathrm{\dots},({a}_{m},{b}_{m})$ are disjoint intervals with $$, then
$$ 
Suppose that $({\alpha}_{1},{\beta}_{1}),\mathrm{\dots},({\alpha}_{n},{\beta}_{n})$ are disjoint intervals with $$. If $$ for $i=1,\mathrm{\dots},n$, then $({t}_{i,j1},{t}_{i,j})$, $1\le i\le n$, $1\le j\le {m}_{i}$, are disjoint intervals whose total length is $$, hence
$$ 
It follows that
$$\sum _{i=1}^{n}F({\beta}_{i})F({\alpha}_{i})=\sum _{i=1}^{n}{\mathrm{Var}}_{f}[{\alpha}_{i},{\beta}_{i}]\le \u03f5,$$ 
which shows that $F$ is absolutely continuous. ∎
We now prove the fundamental theorem of calculus for absolutely continuous functions.^{8}^{8} 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
$$f(x)f(a)={\int}_{a}^{x}{f}^{\prime}(t)\mathit{d}\lambda (t),x\in I.$$ 
Proof.
Define $F:I\to \mathbb{R}$ by
$$F(x)={\mathrm{Var}}_{f}[a,x],x\in I.$$ 
By Lemma 3, $f$ has bounded variation, and then using Theorem 2, $Ff$ and $F+f$ are nondecreasing. Furthermore, by Lemma 5, $F$ is absolutely continuous, so $Ff$ and $F+f$ are absolutely continuous. Let
$${f}_{1}=\frac{F+f}{2},{f}_{2}=\frac{Ff}{2},$$ 
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
$${f}_{1}(x){f}_{1}(a)={\int}_{a}^{x}{f}_{1}^{\prime}(t)\mathit{d}\lambda (t),a\le x\le b$$ 
and
$${f}_{2}(x){f}_{2}(a)={\int}_{a}^{x}{f}_{2}^{\prime}(t)\mathit{d}\lambda (t),a\le x\le b.$$ 
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
$$f(x)f(a)={\int}_{a}^{x}{f}^{\prime}(t)\mathit{d}\lambda (t),a\le x\le b,$$ 
proving the claim. ∎
3 Borel sets
Let $I=[a,b]$. Denote by $C(I)$ the set of continuous functions $I\to \u2102$, which with the norm
$${\parallel f\parallel}_{C(I)}=\underset{x\in I}{sup}f(x),f\in C(I),$$ 
is a Banach space. Denote by $AC(I)$ the set of absolutely continuous functions $I\to \u2102$. Let ${\mathcal{B}}_{C(I)}$ 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.
$AC(I)\in {\mathcal{B}}_{C(I)}$.
Proof.
If $X,Y$ are Polish spaces, $f:X\to Y$ is continuous, $A\in {\mathcal{B}}_{X}$, and $fA$ is injective, then $f(A)\in {\mathcal{B}}_{Y}$.^{9}^{9} 9 Alexander Kechris, Classical Descriptive Set Theory, p. 89, Theorem 15.1. Let $X=\u2102\times {L}^{1}(I)$, which is a Banach space with the norm
$${\parallel (A,g)\parallel}_{X}=A+{\int}_{a}^{b}g\mathit{d}\lambda ,(A,g)\in X.$$ 
Furthermore, $\u2102$ 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 $\mathrm{\Phi}:X\to C(I)$ by
$$\mathrm{\Phi}(A,g)(x)=A+{\int}_{a}^{x}g(t)\mathit{d}\lambda (t),(A,g)\in X,x\in I.$$ 
For $({A}_{1},{g}_{1}),({A}_{2},{g}_{2})\in X$,
${\parallel \mathrm{\Phi}({A}_{1},{g}_{1})\mathrm{\Phi}({A}_{2},{g}_{2})\parallel}_{C(I)}$  $={\parallel ({A}_{1}{A}_{2})+{\displaystyle {\int}_{a}^{x}}({g}_{1}(t){g}_{2}(t))\mathit{d}\lambda (t)\parallel}_{C(I)}$  
$={A}_{1}{A}_{2}+\underset{x\in I}{sup}\left{\displaystyle {\int}_{a}^{x}}({g}_{1}(t){g}_{2}(t))\mathit{d}\lambda (t)\right$  
$\le {A}_{1}{A}_{2}+{\displaystyle {\int}_{a}^{b}}{g}_{1}(t){g}_{2}(t)\mathit{d}\lambda (t)$  
$={\parallel ({A}_{1},{g}_{1})({A}_{2},{g}_{2})\parallel}_{X},$ 
which shows that $\mathrm{\Phi}:X\to C(I)$ is continuous.
Let $(A,g)\in X$ and $\u03f5>0$. Because $g\in {L}^{1}(I)$, there is some $\delta >0$ such that if $$ then $$.^{10}^{10} 10 Walter Rudin, Real and Complex Analysis, third ed., p. 32, exercise 1.12. If $({\alpha}_{1},{\beta}_{1}),\mathrm{\dots},({\alpha}_{n},{\beta}_{n})$ are disjoint intervals whose total length is $$, then, with $E={\bigcup}_{i=1}^{n}({\alpha}_{i},{\beta}_{i})$,
$\sum _{i=1}^{n}}\mathrm{\Phi}(A,g)({\beta}_{i})\mathrm{\Phi}(A,g)({\alpha}_{i})$  $={\displaystyle \sum _{i=1}^{n}}\left{\displaystyle {\int}_{{\alpha}_{i}}^{{\beta}_{i}}}g(t)\mathit{d}\lambda (t)\right$  
$\le {\displaystyle \sum _{i=1}^{n}}{\displaystyle {\int}_{{\alpha}_{i}}^{{\beta}_{i}}}g(t)\mathit{d}\lambda (t)$  
$={\displaystyle {\int}_{E}}g\mathit{d}\lambda $  
$$ 
showing that $\mathrm{\Phi}(A,g)$ 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
$$f(x)f(a)={\int}_{a}^{x}{f}^{\prime}(t)\mathit{d}\lambda (t),x\in I.$$ 
Then $(f(a),{f}^{\prime})\in X$, and the above gives us, for all $x\in I$,
$$\mathrm{\Phi}(f(a),{f}^{\prime})(x)=f(a)+{\int}_{a}^{x}{f}^{\prime}(t)\mathit{d}\lambda (t)=f(x),$$ 
thus $\mathrm{\Phi}(f(a),{f}^{\prime})=f$. Therefore
$$\mathrm{\Phi}(X)=AC(I).$$ 
If $\mathrm{\Phi}({A}_{1},{g}_{1})=\mathrm{\Phi}({A}_{2},{g}_{2})$, then $\mathrm{\Phi}({A}_{1},{g}_{1})(a)=\mathrm{\Phi}({A}_{2},{g}_{2})(a)$ gives ${A}_{1}={A}_{2}$. Using this, and defining $G:I\to \u2102$ by $G={\int}_{a}^{x}({g}_{1}(t){g}_{2}(t))\mathit{d}\lambda (t)$, 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 theorem^{11}^{11} 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 $\mathrm{\Phi}:X\to C(I)$ is injective.
Therefore $\mathrm{\Phi}(X)\in {\mathcal{B}}_{C(I)}$. ∎