Master Dissertation

We want to show by induction that for x ∈ [a + m − 1, a + m],
|f(x)| ≤ |f(a)| + m. (6.8)
For m = 1, let x ∈ [a, a + 1] then we may write x = a + k for some
k ∈ [0, 1]. Then
|f(a + k)| = |f(a + k) − f(a) + f(a)| ≤ |f(a + k) − f(a)| + |f(a)|
≤ |f(a)| + 1.
In the (n + 1)th step, let x ∈ [a + n, a + n + 1]. Then there exists a
kn ∈ [0, 1] such that x = a + n + kn and
|f(x)| = |f(a + n + kn)| = |f(a + n + kn) − f(a + n) + f(a + n)|
≤ |f(a + n + kn) − f(a + n)| + |f(a + n)|
< |f(a)| + n + 1
by the induction hypothesis. This concludes the proof by induction.
Clearly if x satisfies (6.8), then it is also satisfied by all y ∈ [a, a + m] and
we have proven that f is bounded on the interval [a, b].
Corollary 6.5. Let f(x) = ln r(e −x ). Then there exists a constant x0 such
that f is integrable on every interval [a, b] where x0 ≤ a.
Proof. This follows directly from the definition of integrability since f is
measurable and bounded on [a, b] by Lemma 6.4.
Lemma 6.6. Let a ≥ x0 from Lemma 6.4. Then f(x) = ln r(e−x ) can be
represented for all x ≥ a as
f(x) = g(x) +
x
a
h(s)ds, (6.9)
where g and h are measurable and bounded on every interval [a, b] and
g(x) → g0 (|g0| ≤ ∞) and h(x) → 0 for x → ∞.
Proof. Let x ≥ a we may write f(x) on the form (6.9) by letting
g(x) :=
x+1
FIXME: maybe write out the above
By (6.4) h(x) → 0 for x → ∞.
x
f(x) − f(s)ds +
a+1
h(x) := f(x + 1) − f(x)
44
a
f(s)ds