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.