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 $f:[a,b]\to ℂ$ is

$${\mathrm{Var}}_f[a,b]=sup\{\sum _{i=1}^n|f(t_i)-f(t_{i-1})|:t_0,t_1,\mathrm{\dots },t_n\text{ is a partition of }[a,b]\}.$$

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.

Theorem 2.

Suppose that $f:[a,b]\to ℝ$ 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. $F-f$ 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

$$\sum _{i=1}^n|f(t_i)-f(t_{i-1})|+|f(y)-f(x)|\le F(y).$$

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 $ϵ>0$. 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 $ϵ>0$. 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

${\mathrm{Var}}_f[s,x]+{\mathrm{Var}}_f[x,b]$ $={\mathrm{Var}}_f[s,b]$

$\le {\displaystyle \sum _{i=1}^k}|f(t_i)-f(t_{i-1})|+|f(x)-f(t_k)|$

$+|f(x_0)-f(x)|$

$+|f(t_{k+1})-f(x_0)|+{\displaystyle \sum _{i=k+2}^n}|f(t_i)-f(t_{i-1})|+ϵ$

$\le {\mathrm{Var}}_f[s,x]+|f(x)-f(x_0)|+{\mathrm{Var}}_f[x_0,b]+ϵ$

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 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 $f:[a,b]\to ℝ$ is of bounded variation, then Theorem 2 tells us that $F$ and $F+f$ are nondecreasing functions. A monotone function is differentiable almost everywhere, and it follows that $f=(F+f)-F$ is differentiable almost everywhere.

2 Absolute continuity

Let and let $I=[a,b]$. A function $f:I\to ℂ$ is said to be absolutely continuous if for any $ϵ>0$ 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 $f:[a,b]\to ℂ$ is absolutely continuous then $f$ has bounded variation.

Proof.

Because $f$ is absolutely continuous, there is some $\delta >0$ such that if $({\alpha }_1,{\beta }_1),\mathrm{\dots },({\alpha }_n,{\beta }_n)$ are pairwise disjoint and

then

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

$${\mathrm{Var}}_f[a,b]=\sum _{i=1}^N{\mathrm{Var}}_f[x_{i-1},x_i]\le N,$$

showing that $f$ has bounded variation. ∎

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

The following theorem establishes connections between absolute continuity of a function and Lebesgue measure. In the following theorem, we extend $f:[a,b]\to ℝ$ to $ℝ\to ℝ$ 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 $f:I\to ℝ$ 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

   $$f(x)-f(a)={\int }_a^x{f}^{\prime }(t)𝑑\lambda (t),a\le x\le b.$$

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 $ϵ>0$. 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

$$\sum _i|f({\beta }_i)-f({\alpha }_i)|\le ϵ.$$

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

$$\lambda (f(V))=\sum _i(f({\beta }_i)-f({\alpha }_i))=\sum _i|f({\beta }_i)-f({\alpha }_i)|\le ϵ.$$

This is true for all $ϵ>0$, so $\lambda (f(V))=0$. Because $f(E_0)\subset f(V)$, it follows that $f(E_0)\in 𝔐$ (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 $g:I\to ℝ$ 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 $ϵ>0$. 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

$$N=f^{-1}\left(\bigcup _i({\gamma }_i,{\delta }_i)\right)\cap \bigcup _i({\alpha }_i,{\beta }_i)=\bigcup _{i,j}(f^{-1}({\gamma }_i,{\delta }_i)\cap ({\alpha }_i,{\beta }_i))\in 𝔐.$$

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 $ϵ>0$ there is some $N\in 𝔐$ with $g(E)\subset g(N)$ and , from which it follows that $\lambda (g(E))=0$.

Suppose that $E\subset I$ belongs to $𝔐$. Because $E\in 𝔐$, there are $E_0,E_1\in 𝔐$ 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 $𝔐$. On the other hand, because $\lambda (E_0)=0$, $g(E_0)\in 𝔐$. Therefore $g(E)=g(E_0)\cup g(E_1)\in 𝔐$. Define $\mu :𝔐\to [0,\mathrm{\infty })$ by

$$\mu (E)=\lambda (g(E\cap I)),E\in 𝔐.$$

If $E_i$ are countably many pairwise disjoint elements of $𝔐$, then $g(E_i\cap I)$ are pairwise disjoint elements of $𝔐$, 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 Radon-Nikodym theorem there is a unique $h\in L^1(\lambda )$ such that

$$\mu (E)={\int }_{E}h𝑑\lambda ,E\in 𝔐.$$

$h(x)\ge 0$ for $\lambda$-almost all $x\in ℝ$.

Suppose that $x\in ℝ$ 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, $f^{\prime }(x)=h(x)-1$ for $\lambda$-almost all $x\in ℝ$, and it follows that $f^{\prime }\in L^1(\lambda )$ and

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

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

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

Let $ϵ>0$ 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 that there is some $\delta >0$ such that for $E\in 𝔐$, implies that . This gives us

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

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

The following lemma establishes properties of the total variation of absolutely continuous functions.

Lemma 5.

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

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

is absolutely continuous.

Proof.

Let $ϵ>0$. 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 $(t_{i,j-1},t_{i,j})$, $1\le i\le n$, $1\le j\le m_i$, are disjoint intervals whose total length is , hence

It follows that

$$\sum _{i=1}^n|F({\beta }_i)-F({\alpha }_i)|=\sum _{i=1}^n{\mathrm{Var}}_f[{\alpha }_i,{\beta }_i]\le ϵ,$$

which shows that $F$ is absolutely continuous. ∎

We now prove the fundamental theorem of calculus for absolutely continuous functions.

Theorem 6.

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

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

Proof.

Define $F:I\to ℝ$ by

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

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

$$f_1=\frac{F+f}{2},f_2=\frac{F-f}{2},$$

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

$$f_1(x)-f_1(a)={\int }_a^x{f}_1^{\prime }(t)𝑑\lambda (t),a\le x\le b$$

and

$$f_2(x)-f_2(a)={\int }_a^x{f}_2^{\prime }(t)𝑑\lambda (t),a\le x\le b.$$

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

$$f(x)-f(a)={\int }_a^x{f}^{\prime }(t)𝑑\lambda (t),a\le x\le b,$$

proving the claim. ∎

3 Borel sets

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

$AC(I)\in {ℬ}_{C(I)}$.

Proof.

If $X,Y$ are Polish spaces, $f:X\to Y$ is continuous, $A\in {ℬ}_X$, and $f|A$ is injective, then $f(A)\in {ℬ}_Y$. Let $X=ℂ\times L^1(I)$, which is a Banach space with the norm

$${\parallel (A,g)\parallel }_X=|A|+{\int }_a^b|g|𝑑\lambda ,(A,g)\in X.$$

Furthermore, $ℂ$ 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 $\mathrm{\Phi }:X\to C(I)$ by

$$\mathrm{\Phi }(A,g)(x)=A+{\int }_a^{x}g(t)𝑑\lambda (t),(A,g)\in X,x\in I.$$

For $(A_1,g_1),(A_2,g_2)\in X$,

${\parallel \mathrm{\Phi }(A_1,g_1)-\mathrm{\Phi }(A_2,g_2)\parallel }_{C(I)}$ $={\parallel (A_1-A_2)+{\displaystyle {\int }_a^x}(g_1(t)-g_2(t))𝑑\lambda (t)\parallel }_{C(I)}$

$=|A_1-A_2|+\underset{x\in I}{sup}\left|{\displaystyle {\int }_a^x}(g_1(t)-g_2(t))𝑑\lambda (t)\right|$

$\le |A_1-A_2|+{\displaystyle {\int }_a^b}|g_1(t)-g_2(t)|𝑑\lambda (t)$

$={\parallel (A_1,g_1)-(A_2,g_2)\parallel }_X,$

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

Let $(A,g)\in X$ and $ϵ>0$. Because $g\in L^1(I)$, there is some $\delta >0$ such that if then . If $({\alpha }_1,{\beta }_1),\mathrm{\dots },({\alpha }_n,{\beta }_n)$ are disjoint intervals whose total length is , then, with $E={\bigcup }_{i=1}^n({\alpha }_i,{\beta }_i)$,

${\displaystyle \sum _{i=1}^n}|\mathrm{\Phi }(A,g)({\beta }_i)-\mathrm{\Phi }(A,g)({\alpha }_i)|$ $={\displaystyle \sum _{i=1}^n}\left|{\displaystyle {\int }_{{\alpha }_i}^{{\beta }_i}}g(t)𝑑\lambda (t)\right|$

$\le {\displaystyle \sum _{i=1}^n}{\displaystyle {\int }_{{\alpha }_i}^{{\beta }_i}}|g(t)|𝑑\lambda (t)$

$={\displaystyle {\int }_E}|g|𝑑\lambda$

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

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

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

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

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 $G:I\to ℂ$ 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 theorem 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 $\mathrm{\Phi }(X)\in {ℬ}_{C(I)}$. ∎