5.8: Rectifiable Arcs. Absolute Continuity

If a function $f : E^{1} \rightarrow E$ is of bounded variation (§7) on an interval $I=[a, b],$ we can define a real function $v_{f}$ on $I$ by

$$v_{f}(x)=V_{f}[a, x](=\text { total variation of } f \text { on }[a, x]) \text { for } x \in I;$$

$v_{f}$ is called the total variation function, or length function, generated by $f$ on $I$. Note that $v_{f} \uparrow$ on $I.$ (Why?) We now consider the case where $f$ is also relatively continuous on $I,$ so that the set $A=f[I]$ is a rectifiable arc (see §7, Note 1 and Definition 2).

Definition 1

A function $f : E^{1} \rightarrow E$ is (weakly) absolutely continuous on $I=[a, b]$ iff $V_{f}[I]<+\infty$ and $f$ is relatively continuous on $I$.

Theorem $\PageIndex{1}$

The following are equivalent:

(i) $f$ is (weakly) absolutely continuous on $I=[a, b]$;

(ii) $v_{f}$ is finite and relatively continuous on $I ;$ and

(iii) $(\forall \varepsilon>0) \text{ } (\exists \delta>0) \text{ } (\forall x, y \in I | 0 \leq y-x<\delta) \text{ } V_{f}[x, y]<\varepsilon$.

Proof

We shall show that (ii) $\Rightarrow$ (iii) $\Rightarrow$ (i) $\Rightarrow$ (ii).

(ii) $\Rightarrow$ (iii). As $I=[a, b]$ is compact, (ii) implies that $v_{f}$ is uniformly continuous on $I$ (Theorem 4 of Chapter 4, §8). Thus

$$(\forall \varepsilon>0) \text{ } (\exists \delta>0) \text{ } (\forall x, y \in I | 0 \leq y-x<\delta) \quad v_{f}(y)-v_{f}(x)<\varepsilon.$$

However,

$$v_{f}(y)-v_{f}(x)=V_{f}[a, y]-V_{f}[a, x]=V_{f}[x, y]$$

by additivity (Theorem 1 in §7). Thus (iii) follows.

(iii) $\Rightarrow$ (i). By Corollary 3 of §7, $|f(x)-f(y)| \leq V_{f}[x, y].$ Therefore, (iii) implies that

$$(\forall \varepsilon>0) \text{ } (\exists \delta>0) \text{ } (\forall x, y \in I| | x-y |<\delta) \quad|f(x)-f(y)|<\varepsilon,$$

and so $f$ is relatively (even uniformly) continuous on $I$.

Now with $\varepsilon$ and $\delta$ as in (iii), take a partition $P=\left\{t_{0}, \ldots, t_{m}\right\}$ of $I$ so fine that

$$t_{i}-t_{i-1}<\delta, \quad i=1,2, \ldots,\text{ } m.$$

Then $(\forall i) V_{f}\left[t_{i-1}, t_{i}\right]<\varepsilon.$ Adding up these $m$ inequalities and using the additivity of $V_{f},$ we obtain

$$V_{f}[I]=\sum_{i=1}^{m} V_{f}\left[t_{i-1}, t_{i}\right]<m \varepsilon<+\infty.$$

Thus (i) follows, by definition.

That (i) $\Rightarrow$ (ii) is given as the next theorem. $\quad \square$

Theorem $\PageIndex{2}$

If $V_{f}[I]<+\infty$ and if $f$ is relatively continuous at some $p \in I$ (over $I=[a, b]),$ then the same applies to the length function $v_{f}$.

Proof

We consider left continuity first, with $a<p \leq b$.

Let $\varepsilon>0.$ By assumption, there is $\delta>0$ such that

$$|f(x)-f(p)|<\frac{\varepsilon}{2} \text { when }|x-p|<\delta \text { and } x \in[a, p].$$

Fix any such $x.$ Also, $V_{f}[a, p]=\sup _{P} S(f, P)$ over $[a, p].$ Thus

$$V_{f}[a, p]-\frac{\varepsilon}{2}<\sum_{i=1}^{k}\left|\Delta_{i} f\right|$$

for some partition

$$P=\left\{t_{0}=a, \ldots, t_{k-1}, t_{k}=p\right\} \text { of }[a, p]. \text { (Why?)}$$

We may assume $t_{k-1}=x, x$ as above. (If $t_{k-1} \neq x,$ add $x$ to $P. )$ Then

$$\left|\Delta_{k} f\right|=|f(p)-f(x)|<\frac{\varepsilon}{2},$$

and hence

$$V_{f}[a, p]-\frac{\varepsilon}{2}<\sum_{i=1}^{k-1}\left|\Delta_{i} f\right|+\left|\Delta_{k} f\right|<\sum_{i=1}^{k-1}\left|\Delta_{i} f\right|+\frac{\varepsilon}{2} \leq V_{f}\left[a, t_{k-1}\right]+\frac{\varepsilon}{2}.$$

However,

$$V_{f}[a, p]=v_{f}(p)$$

and

$$V_{f}\left[a, t_{k-1}\right]=V_{f}[a, x]=v_{f}(x).$$

Thus (1) yields

$$\left|v_{f}(p)-v_{f}(x)\right|=V_{f}[a, p]-V_{f}[a, x]<\varepsilon \text { for } x \in[a, p] \text { with }|x-p|<\delta.$$

This shows that $v_{f}$ is left continuous at $p$.

Right continuity is proved similarly on noting that

$$v_{f}(x)-v_{f}(p)=V_{f}[p, b]-V_{f}[x, b] \text { for } p \leq x<b. \text { (Why?)}$$

Thus $v_{f}$ is, indeed, relatively continuous at $p.$ Observe that $v_{f}$ is also of bounded variation on $I,$ being monotone and finite (see Theorem 3(ii) of §7).

This completes the proof of both Theorem 2 and Theorem 1. $\quad \square$

We also have the following.

Corollary $\PageIndex{1}$

If $f$ is real and absolutely continuous on $I=[a, b]$ (weakly), so are the nondecreasing functions $g$ and $h(f=g-h)$ defined in Theorem 3 of §7.

Indeed, the function $g$ as defined there is simply $v_{f}.$ Thus it is relatively continuous and finite on $I$ by Theorem 1. Hence so also is $h=f-g.$ Both are of bounded variation (monotone!) and hence absolutely continuous (weakly).

Note 1. The proof of Theorem 1 shows that (weak) absolute continuity implies uniform continuity. The converse fails, however (see Problem 1(iv) in §7).

We now apply our theory to antiderivatives (integrals).

Corollary $\PageIndex{2}$

If $F=\int f$ on $I=[a, b]$ and if $f$ is bounded $\left(|f| \leq K \in E^{1}\right)$ on $I-Q$ ($Q$ countable), then $F$ is weakly absolutely continuous on $I.$

(Actually, even the stronger variety of absolute continuity follows. See Chapter 7, §11, Problem 17).

Proof

By definition, $F=\int f$ is finite and relatively continuous on $I,$ so we only have to show that $V_{F}[I]<+\infty.$ This, however, easily follows by Problem 3 of §7 on noting that $F^{\prime}=f$ on $I-S$ ($S$ countable). Details are left to the reader. $\quad \square$

Our next theorem expresses arc length in the form of an integral.

Theorem $\PageIndex{3}$

If $f : E^{1} \rightarrow E$ is continuously differentiable on $I=[a, b]$ (§6), then $v_{f}=\int\left|f^{\prime}\right|$ on $I$ and

$$V_{f}[a, b]=\int_{a}^{b}\left|f^{\prime}\right|.$$

Proof

Let $a<p<x \leq b, \Delta x=x-p,$ and

$$\Delta v_{f}=v_{f}(x)-v_{f}(p)=V_{f}[p, x] . \quad \text {(Why?)}$$

As a first step, we shall show that

$$\frac{\Delta v_{f}}{\Delta x} \leq \sup _{[p, x]}\left|f^{\prime}\right|.$$

For any partition $P=\left\{p=t_{0}, \ldots, t_{m}=x\right\}$ of $[p, x],$ we have

$$S(f, P)=\sum_{i=1}^{m}\left|\Delta_{i} f\right| \leq \sum_{i=1}^{m} \sup _{\left[t_{i-1}, t_{i}\right]}\left|f^{\prime}\right|\left(t_{i}-t_{i-1}\right) \leq \sup _{[p, x]}\left|f^{\prime}\right| \Delta x.$$

Since this holds for any partition $P,$ we have

$$V_{f}[p, x] \leq \sup _{[p, x]}\left|f^{\prime}\right| \Delta x,$$

which implies (2).

On the other hand,

$$\Delta v_{f}=V_{f}[p, x] \geq|f(x)-f(p)|=|\Delta f|.$$

Combining, we get

$$\left|\frac{\Delta f}{\Delta x}\right| \leq \frac{\Delta v_{f}}{\Delta x} \leq \sup _{[p, x]}\left|f^{\prime}\right|<+\infty$$

since $f^{\prime}$ is relatively continuous on $[a, b],$ hence also uniformly continuous and bounded. (Here we assumed $a<p<x \leq b$. However, (3) holds also if $a \leq x<p<b,$ with $\Delta v_{f}=-V[x, p]$ and $\Delta x<0.$ Verify!)

Now

$$| | f^{\prime}(p)|-| f^{\prime}(x)| | \leq\left|f^{\prime}(p)-f^{\prime}(x)\right| \rightarrow 0 \quad \text { as } x \rightarrow p,$$

so, taking limits as $x \rightarrow p,$ we obtain

$$\lim _{x \rightarrow p} \frac{\Delta v_{f}}{\Delta x}=\left|f^{\prime}(p)\right|.$$

Thus $v_{f}$ is differentiable at each $p$ in $(a, b),$ with $v_{f}^{\prime}(p)=\left|f^{\prime}(p)\right|.$ Also, $v_{f}$ is relatively continuous and finite on $[a, b]$ (by Theorem 1). Hence $v_{f}=\int\left|f^{\prime}\right|$ on $[a, b],$ and we obtain

$$\int_{a}^{b}\left|f^{\prime}\right|=v_{f}(b)-v_{f}(a)=V_{f}[a, b], \text { as asserted.} \quad \square$$

Note 2. If the range space $E$ is $E^{n}$ (*or $C^{n}$), $f$ has $n$ components

$$f_{1}, f_{2}, \ldots, f_{n}.$$

By Theorem 5 in §1, $f^{\prime}=\left(f_{1}^{\prime}, f_{2}^{\prime}, \ldots, f_{n}^{\prime}\right),$ so

$$\left|f^{\prime}\right|=\sqrt{\sum_{k=1}^{n}\left|f_{k}^{\prime}\right|^{2}},$$

and we get

$$V_{f}[a, b]=\int_{a}^{b} \sqrt{\sum_{k=1}^{n}\left|f_{k}^{\prime}\right|^{2}}=\int_{a}^{b} \sqrt{\sum_{k=1}^{n}\left|f_{k}^{\prime}(t)\right|^{2}} d t \quad\text {(classical notation).}$$

In particular, for complex functions, we have (see Chapter 4, §3, Note 5)

$$V_{f}[a, b]=\int_{a}^{b} \sqrt{f_{\mathrm{re}}^{\prime}(t)^{2}+f_{\mathrm{im}}^{\prime}(t)^{2}} d t.$$

In practice, formula (5) is used when a curve is given parametrically by

$$x_{k}=f_{k}(t), \quad k=1,2, \ldots, \text{ }n,$$

with the $f_{k}$ differentiable on $[a, b].$ Curves in $E^{2}$ are often given in nonparametric form as

$$y=F(x), \quad F : E^{1} \rightarrow E^{1}.$$

Here $F[I]$ is $n o t$ the desired curve but simply a set in $E^{1}.$ To apply (5) here, we first replace "$y=F(x)$" by suitable parametric equations,

$$x=f_{1}(t) \text { and } y=f_{2}(t);$$

i.e., we introduce a function $f : E^{1} \rightarrow E,$ with $f=\left(f_{1}, f_{2}\right).$ An obvious (but not the only) way of achieving it is to set

$$x=f_{1}(t)=t \text { and } y=f_{2}(t)=F(t)$$

so that $f_{1}^{\prime}=1$ and $f_{2}^{\prime}=F^{\prime}.$ Then formula (5) may be written as

$$V_{f}[a, b]=\int_{a}^{b} \sqrt{1+F^{\prime}(x)^{2}} d x, \quad f(x)=(x, F(x)).$$

Example

Find the length of the circle

$$x^{2}+y^{2}=r^{2}.$$

Here it is convenient to use the parametric equations

$$x=r \cos t \text { and } y=r \sin t,$$

i.e., to define $f : E^{1} \rightarrow E^{2}$ by

$$f(t)=(r \cos t, r \sin t),$$

or, in complex notation,

$$f(t)=r e^{t i}.$$

Then the circle is obtained by letting $t$ vary through $[0,2 \pi].$ Thus (5) yields

$$V_{f}[0,2 \pi]=\int_{a}^{b} r \sqrt{\cos ^{2} t+\sin ^{2} t} d t=r \int_{a}^{b} 1 d t=r\left.t\right|_{0} ^{2 \pi}=2 r \pi.$$

Note that $f$ describes the same circle $A=f[I]$ over $I=[0,4 \pi].$ More generally, we could let $t$ vary through any interval $[a, b]$ with $b-a \geq 2 \pi.$ However, the length, $V_{f}[a, b],$ would change (depending on $b-a)$. This is because the circle $A=f[I]$ is not a simple arc (see §7, Note 1), so $\ell A$ depends on $f$ and $I,$ and one must be careful in selecting both appropriately.