9.1: L-Integrals and Antiderivatives
- Page ID
- 19218
I. Lebesgue theory makes it possible to strengthen many calculus theorems. We shall start with functions on \(E^{1}, f : E^{1} \rightarrow E.\) (A reader who has omitted the "starred" part of Chapter 8, §7, will have to set \(E=E^{*}\left(E^{n}, C^{n}\right)\) throughout.)
By \(L\)-integrals of such functions, we mean integrals with respect to Lebesgue measure \(m\) in \(E^{1}.\) Notation:
\[L \int_{a}^{b} f=L \int_{a}^{b} f(x) d x=L \int_{[a, b]} f\]
and
\[L \int_{b}^{a} f=-L \int_{a}^{b} f.\]
For Riemann integrals, we replace "\(L\)" by "\(R.\)" We compare such integrals with antiderivatives (Chapter 5, §5), denoted
\[\int_{a}^{b} f,\]
without the "\(L\)" or "\(R.\)" Note that
\[L \int_{[a, b]} f=L \int_{(a, b)} f,\]
etc., since \(m\{a\}=m\{b\}=0\) here.
Let \(f : E^{1} \rightarrow E\) be \(L\)-integrable on \(A=[a, b].\) Set
\[H(x)=L \int_{a}^{x} f, \quad x \in A.\]
Then the following are true.
(i) The function \(f\) is the derivative of \(H\) at any \(p \in A\) at which \(f\) is finite and continuous. (At \(a\) and \(b\), continuity and derivatives may be one-sided from within.)
(ii) The function \(H\) is absolutely continuous on \(A;\) hence \(V_{H}[A]<\infty.\)
- Proof
-
(i) Let \(p \in(a, b], q=f(p) \neq \pm \infty.\) Let \(f\) be left continuous at \(p;\) so, given \(\varepsilon>0,\) we can fix \(c \in (a, p)\) such that
\[|f(x)-q|<\varepsilon \text { for } x \in (c, p).\]
Then
\[\begin{aligned}(\forall x \in(c, p)) &\left|L \int_{x}^{p}(f-q)\right| \leq L \int_{x}^{p}|f-q| \\ & \leq L \int_{x}^{p}(\varepsilon)=\varepsilon \cdot m[x, p]=\varepsilon(p-x). \end{aligned}\]
But
\[\begin{aligned} L \int_{x}^{p}(f-q) &=L \int_{x}^{p} f-L \int_{x}^{p} q \\ L \int_{x}^{p} q &=q(p-x), \quad \text { and } \\ L \int_{x}^{p} f &=L \int_{a}^{p} f-L \int_{a}^{x} f \\ &=H(p)-H(x). \end{aligned}\]
Thus
\[|H(p)-H(x)-q(p-x)| \leq \varepsilon(p-x);\]
i.e.,
\[\left|\frac{H(p)-H(x)}{p-x}-q\right| \leq \varepsilon \quad(c<x<p).\]
Hence
\[f(p)=q=\lim _{x \rightarrow p^{-}} \frac{\Delta H}{\Delta x}=H_{-}^{\prime}(p).\]
If \(f\) is right continuous at \(p \in[a, b),\) a similar formula results for \(H_{+}^{\prime}(p).\) This proves clause (i).
(ii) Let \(\varepsilon>0\) be given. Then Theorem 6 in Chapter 8, §6, yields a \(\delta>0\) such that
\[\left|L \int_{X} f\right| \leq L \int_{X}|f|<\varepsilon\]
whenever
\[m X<\delta \text { and } A \supseteq X, X \in \mathcal{M}.\]
Here we may set
\[X=\bigcup_{i=1}^{r} A_{i} \text { (disjoint)}\]
for some intervals
\[A_{i}=\left(a_{i}, b_{i}\right) \subseteq A\]
so that
\[m X=\sum_{i} m A_{i}=\sum_{i}\left(b_{i}-a_{i}\right)<\delta.\]
Then (1) implies that
\[\varepsilon>L \int_{X}|f|=\sum_{i} L \int_{A_{i}}|f| \geq \sum_{i}\left|L \int_{a_{i}}^{b_{i}} f\right|=\sum_{i}\left|H\left(b_{i}\right)-H\left(a_{i}\right)\right|.\]
Thus
\[\sum_{i}\left|H\left(b_{i}\right)-H\left(a_{i}\right)\right|<\varepsilon\]
whenever
\[\sum_{i}\left(b_{i}-a_{i}\right)<\delta\]
and
\[A \supseteq \bigcup_{i}\left(a_{i}, b_{i}\right) \text { (disjoint).}\]
(This is what we call "absolute continuity in the stronger sense.") By Problem 2 in Chapter 5, §8, this implies "absolute continuity" in the sense of Chapter 5, §8, hence \(V_{H}[A]<\infty . \quad \square\)
Note 1. The converse to (i) fails: the differentiability of \(H\) at \(p\) does not imply the continuity of its derivative \(f\) at \(p\) (Problem 6 in Chapter 5, §2).
Note 2. If \(f\) is continuous on \(A-Q\) (\(Q\) countable), Theorem 1 shows that \(H\) is a primitive (antiderivative): \(H=\int f\) on \(A.\) Recall that "Q countable" implies \(m Q=0,\) but not conversely. Observe that we may always assume \(a, b \in Q.\)
We can now prove a generalized version of the so-called fundamental theorem of calculus, widely used for computing integrals via antiderivatives.
If \(f : E^{1} \rightarrow E\) has a primitive \(F\) on \(A=[a, b],\) and if \(f\) is bounded on \(A-P\) for some \(P\) with \(m P=0,\) then \(f\) is \(L\)-integrable on \(A,\) and
\[L \int_{a}^{x} f=F(x)-F(a) \quad \text {for all } x \in A.\]
- Proof
-
By Definition 1 of Chapter 5, §5, \(F\) is relatively continuous and finite on \(A=[a, b],\) hence bounded on \(A\) (Theorem 2 in Chapter 4, §8).
It is also differentiable, with \(F^{\prime}=f,\) on \(A-Q\) for a countable set \(Q \subseteq A,\) with \(a, b \in Q.\) We fix this \(Q\) along with \(P.\)
As we deal with \(A\) only, we surely may redefine \(F\) and \(f\) on \(-A:\)
\[F(x)=\left\{\begin{array}{ll}{F(a)} & {\text { if } x<a,} \\ {F(b)} & {\text { if } x>b,}\end{array}\right.\]
and \(f=0\) on \(-A.\) Then \(f\) is bounded on \(-P,\) while \(F\) is bounded and
continuous on \(E^{1},\) and \(F^{\prime}=f\) on \(-Q;\) so \(F=\int f\) on \(E^{1}.\)Also, for \(n=1,2, \ldots\) and \(t \in E^{1},\) set
\[f_{n}(t)=n\left[F\left(t+\frac{1}{n}\right)-F(t)\right]=\frac{F(t+1 / n)-F(t)}{1 / n}.\]
Then
\[f_{n} \rightarrow F^{\prime}=f \quad \text {on } -Q;\]
i.e., \(f_{n} \rightarrow f\) (a.e.) on \(E^{1}\) (as \(m Q=0\)).
By (3), each \(f_{n}\) is bounded and continuous (as \(F\) is). Thus by Theorem 1 of Chapter 8, §3, \(F\) and all \(f_{n}\) are \(m\)-measurable on \(A\) (even on \(E^{1}\)). So is \(f\) by Corollary 1 of Chapter 8, §3.
Moreover, by boundedness, \(F\) and \(f_{n}\) are \(L\)-integrable on finite intervals. So is \(f.\) For example, let
\[|f| \leq K<\infty \text { on } A-P;\]
as \(m P=0\),
\[\int_{A}|f| \leq \int_{A}(K)=K \cdot m A<\infty,\]
proving integrability. Now, as
\[F=\int f \text { on any interval }\left[t, t+\frac{1}{n}\right],\]
Corollary 1 in Chapter 5, §4 yields
\[\left(\forall t \in E^{1}\right) \quad\left|F\left(t+\frac{1}{n}\right)-F(t)\right| \leq \sup _{t \in-Q}\left|F^{\prime}(t)\right| \frac{1}{n} \leq \frac{K}{n}.\]
Hence
\[\left|f_{n}(t)\right|=n\left|F\left(t+\frac{1}{n}\right)-F(t)\right| \leq K;\]
i.e., \(\left|f_{n}\right| \leq K\) for all \(n\).
Thus \(f\) and \(f_{n}\) satisfy Theorem 5 of Chapter 8, §6, with \(g=K.\) By Note 1 there,
\[\lim _{n \rightarrow \infty} L \int_{a}^{x} f_{n}=L \int_{a}^{x} f.\]
In the next lemma, we show that also
\[\lim _{n \rightarrow \infty} L \int_{a}^{x} f_{n}=F(x)-F(a),\]
which will complete the proof.\(\quad \square\)
Given a finite continuous \(F : E^{1} \rightarrow E\) and given \(f_{n}\) as in (3), we have
\[\lim _{n \rightarrow \infty} L \int_{a}^{x} f_{n}=F(x)-F(a) \quad \text {for all } x \in E^{1}.\]
- Proof
-
As before, \(F\) and \(f_{n}\) are bounded, continuous, and \(L\)-integrable on any \([a, x]\) or \([x, a].\) Fixing \(a,\) let
\[H(x)=L \int_{a}^{x} F, \quad x \in E^{1}.\]
By Theorem 1 and Note 2, \(H=\int F\) also in the sense of Chapter 5, §5, with \(F=H^{\prime}\) (derivative of \(H\)) on \(E^{1}\).
Hence by Definition 2 the same section,
\[\int_{a}^{x} F=H(x)-H(a)=H(x)-0=L \int_{a}^{x} F;\]
i.e.,
\[L \int_{a}^{x} F=\int_{a}^{x} F,\]
and so
\[\begin{aligned} L \int_{a}^{x} f_{n}(t) d t &=n \int_{a}^{x} F\left(t+\frac{1}{n}\right) d t-n \int_{a}^{x} F(t) d t \\ &=n \int_{a+1 / n}^{b+1 / n} F(t) d t-n \int_{a}^{x} F(t) dt. \end{aligned}\]
(We computed
\[\int F(t+1 / n) d t\]
by Theorem 2 in Chapter 5, §5, with \(g(t)=t+1 / n\).) Thus by additivity,
\[L \int_{a}^{x} f_{n}=n \int_{a+1 / n}^{x+1 / n} F-n \int_{a}^{x} F=n \int_{x}^{x+1 / n} F-n \int_{a}^{a+1 / n} F.\]
But
\[n \int_{x}^{x+1 / n} F=\frac{H\left(x+\frac{1}{n}\right)-H(x)}{\frac{1}{n}} \rightarrow H^{\prime}(x)=F(x).\]
Similarly,
\[\lim _{n \rightarrow \infty} n \int_{a}^{a+1 / n} F=F(a).\]
This combined with (5) proves (4), and hence Theorem 2, too.\(\quad \square\)
We also have the following corollary.
If \(f : E^{1} \rightarrow E^{*}\left(E^{n}, C^{n}\right)\) is \(R\)-integrable on \(A=[a, b],\) then
\[(\forall x \in A) \quad R \int_{a}^{x} f=L \int_{a}^{b} f=F(x)-F(a),\]
provided \(F\) is primitive to \(f\) on \(A.\)
- Proof
-
This follows from Theorem 2 by Definition (c) and Theorem 2 of Chapter 8, §9.
Caution. Formulas (2) and (6) may fail if \(f\) is unbounded, or if \(F\) is not a primitive in the sense of Definition 1 of Chapter 5, §5: We need \(F^{\prime}=f\) on \(A-Q, Q\) countable (\(m Q=0\) is not enough!). Even \(R\)-integrability (which makes \(f\) bounded and a.e. continuous) does not suffice if
\[F \neq \int f.\]
For examples, see Problems 2-5.
If \(f\) is relatively continuous and finite on \(A=[a, b]\) and has a bounded derivative on \(A-Q\) (\(Q\) countable), then \(f^{\prime}\) is \(L\)-integrable on \(A\) and
\[L \int_{a}^{x} f^{\prime}=f(x)-f(a) \quad \text { for } x \in A.\]
This is simply Theorem 2 with \(F, f, P\) replaced by \(f, f^{\prime}, Q,\) respectively
If in Theorem 2 the primitive
\[F=\int f\]
is exact on some \(B \subseteq A,\) then
\[f(x)=\frac{d}{d x} L \int_{a}^{x} f, \quad x \in B.\]
(Recall that \(\frac{d}{d x} F(x)\) is classical notation for \(F^{\prime}(x)\).)
- Proof
-
By (2), this holds on \(B \subseteq A\) if \(F^{\prime}=f\) there.\(\quad \square\)
II. Note that under the assumptions of Theorem 2,
\[L \int_{a}^{x} f=F(x)-F(a)=\int_{a}^{x} f.\]
Thus all laws governing the primitive \(\int f\) apply to \(L \int f.\) For example, Theorem 2 of Chapter 5, §5, yields the following corollary.
Let \(g : E^{1} \rightarrow E^{1}\) be relatively continuous on \(A=[a, b]\) and have a bounded derivative on \(A-Q\) (\(Q\) countable).
Suppose that \(f : E^{1} \rightarrow E\) (real or not) has a primitive on \(g[A],\) exact on \(g[A-Q],\) and that \(f\) is bounded on \(g[A-Q]\).
Then \(f\) is \(L\)-integrable on \(g[A],\) the function
\[(f \circ g) g^{\prime}\]
is \(L\)-integrable on \(A,\) and
\[L \int_{a}^{b} f(g(x)) g^{\prime}(x) d x=L \int_{p}^{q} f(y) d y,\]
where \(p=g(a)\) and \(q=g(b)\).
For this and other applications of primitives, see Problem 9. However, often a direct approach is stronger (though not simpler), as we illustrate next.
Suppose \(f : E^{1} \rightarrow E^{1}\) is \(\geq 0\) and monotonically decreasing on \(A=[a, b].\) Then, if \(g : E^{1} \rightarrow E^{1}\) is \(L\)-integrable on \(A,\) so also is \(f g,\) and
\[L \int_{a}^{b} f g=f(a) \cdot L \int_{a}^{c} g \quad \text { for some } c \in A.\]
- Proof
The \(L\)-integrability of \(f g\) follows by Theorem 3 in Chapter 8, §6, as \(f\) is monotone and bounded, hence even \(R\)-integrable (Corollary 3 in Chapter 8, §9).
Using this and Lemma 1 of the same section, fix for each \(n\) a \(\mathcal{C}\)-partition
\[\mathcal{P}_{n}=\left\{A_{n i}\right\} \quad\left(i=1,2, \ldots, q_{n}\right)\]
of \(A\) so that
\[(\forall n) \quad \frac{1}{n}>\overline{S}\left(f, \mathcal{P}_{n}\right)-\underline{S}\left(f, \mathcal{P}_{n}\right)=\sum_{i=1}^{q_{n}} w_{n i} m A_{n i},\]
where we have set
\[w_{n i}=\sup f\left[A_{n i}\right]-\inf f\left[A_{n i}\right].\]
Consider any such \(\mathcal{P}=\left\{A_{i}\right\}, i=1, \ldots, q\) (we drop the "\(n\)" for brevity). If \(A_{i}=\left[a_{i-1}, a_{i}\right],\) then since \(f \downarrow\),
\[w_{i}=f\left(a_{i-1}\right)-f\left(a_{i}\right) \geq\left|f(x)-f\left(a_{i-1}\right)\right|, \quad x \in A_{i}.\]
Under Lebesgue measure (Problem 8 of Chapter 8, §9), we may set
\[A_{i}=\left[a_{i-1}, a_{i}\right] \quad(\forall i)\]
and still get
\[\begin{aligned} L \int_{A} f g &=\sum_{i=1}^{q} f\left(a_{i-1}\right) L \int_{A_{i}} g(x) d x \\ &+\sum_{i=1}^{q} L \int_{A_{i}}\left[f(x)-f\left(a_{i-1}\right)\right] g(x) d x. \end{aligned}\]
(Verify!) Here \(a_{0}=a\) and \(a_{q}=b\).
Now, set
\[G(x)=L \int_{a}^{x} g\]
and rewrite the first sum (call it \(r\) or \(r_{n}\)) as
\[\begin{aligned} r &=\sum_{i=1}^{q} f\left(a_{i-1}\right)\left[G\left(a_{i}\right)-G\left(a_{i-1}\right)\right] \\ &=\sum_{i=1}^{q-1} G\left(a_{i}\right)\left[f\left(a_{i-1}\right)-f\left(a_{i}\right)\right]+G(b) f\left(a_{q-1}\right), \end{aligned}\]
or
\[r=\sum_{i=1}^{q-1} G\left(a_{i}\right) w_{i}+G(b) f\left(a_{q-1}\right),\]
because \(f\left(a_{i-1}\right)-f\left(a_{i}\right)=w_{i}\) and \(G(a)=0\).
Now, by Theorem 1 (with \(H, f\) replaced by \(G, g\)), \(G\) is continuous on \(A=\) \([a, b];\) so \(G\) attains a largest value \(K\) and a least value \(k\) on \(A.\)
As \(f \downarrow\) and \(f \geq 0\) on \(A,\) we have
\[w_{i} \geq 0 \text { and } f\left(a_{q-1}\right) \geq 0.\]
Thus, replacing \(G(b)\) and \(G\left(a_{i}\right)\) by \(K(\text { or } k)\) in \((13)\) and noting that
\[\sum_{i=1}^{q-1} w_{i}=f(a)-f\left(a_{q-1}\right),\]
we obtain
\[k f(a) \leq r \leq K f(a);\]
more fully, with \(k=\min G[A]\) and \(K=\max G[A]\),
\[(\forall n) \quad k f(a) \leq r_{n} \leq K f(a).\]
Next, let \(s\) (or rather \(s_{n}\) be the second sum in (12). Noting that
\[w_{i} \geq\left|f(x)-f\left(a_{i-1}\right)\right|,\]
suppose first that \(|g| \leq B\) (bounded) on \(A\).
Then for all \(n\),
\[\left|s_{n}\right| \leq \sum_{i=1}^{q_{n}} L \int_{A_{n i}}\left(w_{n i} B\right)=B \sum_{i=1}^{q_{n}} w_{n i} m A_{n i}<\frac{B}{n} \rightarrow 0 \quad \ \text{(by (11)).}\]
But by (12),
\[L \int_{A} f g=r_{n}+s_{n} \quad(\forall n).\]
As \(s_{n} \rightarrow 0\),
\[L \int_{A} f g=\lim _{n \rightarrow \infty} r_{n},\]
and so by (14),
\[k f(a) \leq L \int_{A} f g \leq K f(a).\]
By continuity, \(f(a) G(x)\) takes on the intermediate value \(L \int_{A} f g\) at some \(c \in A;\) so
\[L \int_{A} f g=f(a) G(c)=f(a) L \int_{a}^{c} g,\]
since
\[G(x)=L \int_{a}^{x} f.\]
Thus all is proved for a bounded \(g.\)
The passage to an unbounded \(g\) is achieved by the so-called truncation method described in Problems 12 and 13. (Verify!)\(\quad \square\)
Let \(f : E^{1} \rightarrow E^{1}\) be monotone on \(A=[a, b].\) Then if \(g : E^{1} \rightarrow E^{1}\) is \(L\)-integrable on \(A,\) so also is \(f g,\) and
\[L \int_{a}^{b} f g=f(a) L \int_{a}^{c} g+f(b) L \int_{c}^{b} g \quad \text {for some } c \in A.\]
- Proof
-
If, say, \(f \downarrow\) on \(A,\) set
\[h(x)=f(x)-f(b).\]
Then \(h \geq 0\) and \(h \downarrow\) on \(A;\) so by Lemma 2,
\[\int_{a}^{b} g h=h(a) L \int_{a}^{c} g \quad \text {for some } c \in A.\]
As
\[h(a)=f(a)-f(b),\]
this easily implies (15).
If \(f \uparrow,\) apply this result to \(-f\) to obtain (15) again.\(\quad \square\)
Note 3. We may restate (15) as
\[(\exists c \in A) \quad L \int_{a}^{b} f g=p L \int_{a}^{c} g+q L \int_{c}^{b} g,\]
provided either
(i) \(f \uparrow\) and \(p \leq f(a+) \leq f(b-) \leq q,\) or
(ii) \(f \downarrow\) and \(p \geq f(a+) \geq f(b-) \geq q\).
This statement slightly strengthens (15).
To prove clause (i), redefine
\[f(a)=p \text { and } f(b)=q.\]
Then still \(f \uparrow;\) so (15) applies and yields the desired result. Similarly for (ii). For a continuous \(g,\) see also Problem 13(ii') in Chapter 8, §9, based on Stieltjes theory.
III. We now give a useful analogue to the notion of a primitive.
A map \(F : E^{1} \rightarrow E\) is called an \(L\)-primitive or an indefinite \(L\)-integral of \(f : E^{1} \rightarrow E,\) on \(A=[a, b]\) iff \(f\) is \(L\)-integrable on \(A\) and
\[F(x)=c+L \int_{a}^{x} f\]
for all \(x \in A\) and some fixed finite \(c \in E\).
Notation:
\[F=L \int f \quad\left(\text {not } F=\int f\right)\]
or
\[F(x)=L \int f(x) d x \quad \text {on } A.\]
By (16), all \(L\)-primitives of \(f\) on \(A\) differ by finite constants only.
If \(E=E^{*}\left(E^{n}, C^{n}\right),\) one can use this concept to lift the boundedness restriction on \(f\) in Theorem 2 and the corollaries of this section. The proof will be given in §2. However, for comparison, we state the main theorems already now.
Let
\[F=L \int f \quad \text {on } A=[a, b]\]
for some \(f : E^{1} \rightarrow E^{*}\left(E^{n}, C^{n}\right)\).
Then \(F\) is differentiable, with
\[F^{\prime}=f \quad \text {a.e. on } A.\]
In classical notation,
\[f(x)=\frac{d}{d x} L \int_{a}^{x} f(t) d t \quad \text {for almost all } x \in A.\]
A proof was sketched in Problem 6 of Chapter 8, §12. (It is brief but requires more "starred" material than used in §2.)
Let \(F : E^{1} \rightarrow E^{n}\left(C^{n}\right)\) be differentiable on \(A=[a, b]\) (at \(a\) and \(b\) differentiability may be one sided). Let \(F^{\prime}=f\) be \(L\)-integrable on \(A\).
Then
\[L \int_{a}^{x} f=F(x)-F(a) \quad \text {for all } x \in A.\]