8.4.E: Problems on Integration of Elementary Functions
( \newcommand{\kernel}{\mathrm{null}\,}\)
Verify Note 2.
Prove Corollary 1(iv)−( vii ).
Prove that ∫Af=0 if mA=0 or f=0 on A. Disprove the converse by examples.
Find a primitive F for f=CR in our example. Show that
∫[0,1]fdm=F(1)−F(0).
Fill in the proof details in Theorem 2.
[Hint: Use comparison test for series.]
⇒5. Show that if f and g are elementary and nonnegative with f≥g on A, then
∫Af≥∫Ag≥0.
[Hint: As in Theorem 2, let
f=∑iaiCAi and g=∑ibiCAi.
Then f≥g≥0 implies ai≥bi≥0.]
⇒6. Prove that if f and g are elementary and (extended) real on A, then
∫A(f±g)=∫Af±∫Ag,
provided
(i) ∫Af or ∫Ag is finite, or
(ii) ∫Af,fAg, and ∫Af±∫Ag are all orthodox.
[Outline: As in Theorem 2, let
f=∑iaiCAi and g=∑ibiCAi,
so
f±g=ai±bi on Ai.
Now, if
|∫Af|<∞,
then by Problem 14 in Chapter 4, §13, and formula (4),∑aimAi converges absolutely; so its termwise addition to any other series does not affect the absolute convergence or divergence of the latter, i.e., the finiteness or infiniteness of its positive and negative parts. For example,
∑i(ai±bi)+mAi=∞
iff
∑b+imAi=∞.
Thus if
∫Ag=±∞,
then
∫A(f±g)=∫Ag=±∞=∫Af±∫Ag.
If both
∫Af,∫Ag≠±∞,
Theorem 2( ii) applies. In the orthodox infinite case, a similar proof works on noting that either the positive or the negative parts of both series are finite if
∫Af±∫Ag
is orthodox, too. (Verify!)]
Show that if f is elementary and nonnegative on A and
∫Af>p∈E∗,
then there is an elementary and nonnegative map g on A such that
∫Af≥∫Ag>p,
g=0 on A(f=0), and
f>g on A−A(f=0).
[Hints: Let
B=A(f=∞)
and
C=A−B;
so gn is elementary and nonnegative on A and
gn=n on B
and
gn=(1−1n)f on C;
so gn is elementary and nonnegative on A and
f>gn on A−A(f=0).(Why?)
By Theorem 1 and Corollary 1(iv)(vii),
∫Agn=∫Bgn+∫Cgn=∫B(n)+∫C(1−1n)f=n⋅mB+(1−1n)∫Cf.
Deduce that
limn→∞∫Agn=∫Bf+∫Cf=∫Af>p;
so
(∃n)∫Agn>p.
Take g=gn for that n.]
Show that if E=E∗, Theorem 1( i ) holds also if ∫Af is infinite but orthodox.
(i) Prove that if f is elementary and integrable on A, so is −f, and
∫A(−f)=−∫Af.
(ii) Show that this holds also if f is elementary and (extended) real and ∫Af is orthodox.