Chapter 7: The Riemann Integral When the derivative is introduced ...

The fact that |f| is integrable is new information — the proof is an exercise. To do it, you
should verify that, if m i , M i are the inf and sup of f on [x i−1 , x i ] and m ′ i , M i ′ are the same for |f|,
then M i − m i ≥ M i ′ − m′ i . To get the second assertion in (3b), use −|f| ≤ f ≤ |f| on [a.b].
Remarks on the proposition:
(0) Part of what is to be proved here is that f is integrable on [a, b] iff it is integrable on both
[a, c] and [c, b]. But if we define ∫ a
a f = 0 and ∫ a
b f = − ∫ b
a
f, then we don’t need to add the
“If” part of this statement as long as f is integrable on the intervals determined by a, b, c.
(1) The proof is an exercise.
(2) Here is a proof of the k < 0 case: Let ε > 0 be given, and pick a partition P of [a, b] for which
U(f, P ) − ∫ b
a f < ε/(2|k|) and ∫ b
a
f − L(f, P ) < ε/(2|k|). Then because, on each subinterval
[x i−1 , x i ] in P we have
sup{kf(x) : x i−1 ≤ x ≤ x i } = k · inf{f(x) : x i−1 ≤ x ≤ x i }
and similarly with sup and inf reversed, it follows that
U(kf, P ) − L(kf, P ) = kL(f, P ) − kU(f, P ) = |k|(U(f, P ) − L(f, P ))
( ∫ b ∫ b
)
≤ |k| U(f, P ) − f + f − L(f, P ) ) < ε 2 + ε 2 = ε ,
so kf is integrable, and
U(kf, P ) − k
so k ∫ b
a f = ∫ b
a kf.//
∫ b
a
(
f = k L(f, P ) −
∫ b
a
a
a
) (∫ b
)
f = |k| f − L(f, P ) < ε
a
2 < ε ,
(3) Because f ≤ g, for any subinterval [x i−1 , x i ] of any partition P of [a, b] we have
sup{f(x) : x i−1 ≤ x ≤ x i } ≤ sup{g(x) : x i−1 ≤ x ≤ x i }
so U(f, P ) ≤ U(g, P ), so ∫ b
a f = inf P U(f, P ) ≤ inf P U(g, P ) = ∫ b
a g.
[At this point students are ready to do the twelfth problem set.]
Theorem. (Fundamental Theorem of Calculus) (1) If f is the derivative of F on [a, b] and
f is integrable, then ∫ b
a
f = F (b) − F (a).
(2) Let g : [a, b] → R be integrable, and define G(t) = ∫ t
a
g(x) dx. Then G : [a, b] → R is
continuous. If g is continuous at c in [a, b], then G is differentiable at c and G ′ (c) = g(c).
Proof. (1) For any subinterval [x i−1 , x i ] in any partition P of [a, b], by the Mean Value Theorem
∃ x ∗ i ∈ [x i−1 , x i ] s.t. F (x i ) − F (x i−1 ) = f(x ∗ i )(x i − x i−1 ), so ∑ n
i=1 (F (x i) − F (x i−1 )) is between
L(f, P ) and U(f, P ). But the sum is telescoping, to F (b) − F (a), independent of the partition P ,
so F (b) − F (a) is between L(f) and U(f), which are both equal to ∫ b
a f.