Analysis-2-The Riemann Integral, guaranteed and verified 100% Pass

1


The Riemann Integral


When does a Riemann -Stieltjes integral reduce to a Riemann integral? In particular,
when is
𝑏 𝑏
∫𝑎 𝑓 𝑑𝛼 = ∫𝑎 𝑓(𝑥)𝛼′(𝑥) 𝑑𝑥 ?


Theorem: Suppose 𝛼 is increasing and that 𝛼′ exists and is a (bounded)
Riemann integrable function on [𝑎, 𝑏]. Then given a
bounded function, 𝑓 on [𝑎, 𝑏], we have 𝑓 ∈ 𝑅𝛼 [𝑎, 𝑏] if, and only if,
𝑓𝛼′ ∈ 𝑅[𝑎, 𝑏]. In either case,


𝑏 𝑏
∫𝑎 𝑓 𝑑𝛼 = ∫𝑎 𝑓(𝑥)𝛼′(𝑥) 𝑑𝑥


Proof: Let 𝜖 > 0 be given and let’s show that there exists a partition, 𝑃, such
that

|𝑈𝛼 (𝑓, 𝑃 ) − 𝑈(𝑓𝛼 ′ , 𝑃)| ≤ ‖𝑓‖∞ 𝜖 (∗)
and
|𝐿𝛼 (𝑓, 𝑃 ) − 𝐿(𝑓𝛼 ′ , 𝑃)| ≤ ‖𝑓‖∞ 𝜖 (∗∗)


By the triangle inequality this will show that 𝑓 ∈ 𝑅𝛼 [𝑎, 𝑏] if, and

only if, 𝑓𝛼′ ∈ 𝑅[𝑎, 𝑏] and if either exists then:


𝑏 𝑏
∫𝑎 𝑓 𝑑𝛼 = ∫𝑎 𝑓(𝑥)𝛼′(𝑥) 𝑑𝑥.

, 2


First, let’s see why this is true.

Suppose 𝑓 ∈ 𝑅𝛼 [𝑎, 𝑏]. Then there exists a partition 𝑃 such that

𝑈𝛼 (𝑓, 𝑃) − 𝐿𝛼 (𝑓, 𝑃 ) < 𝜖.

Then by the triangle inequality we have:
|𝑈(𝑓𝛼 ′ , 𝑃) − 𝐿(𝑓𝛼 ′ , 𝑃)| ≤ |𝑈(𝑓𝛼 ′ , 𝑃) − 𝑈𝛼 (𝑓, 𝑃)|
+|𝑈𝛼 (𝑓, 𝑃) − 𝐿𝛼 (𝑓, 𝑃)| + |𝐿𝛼 (𝑓, 𝑃) − 𝐿(𝑓𝛼′, 𝑃)|.

Now using inequalities (∗) and (∗∗) we get:
|𝑈(𝑓𝛼 ′ , 𝑃 ) − 𝐿(𝑓𝛼 ′ , 𝑃 )| ≤ ‖𝑓‖∞ 𝜖 + 𝜖 + ‖𝑓‖∞ 𝜖
= (2‖𝑓‖∞ + 1)𝜖.

Since 2‖𝑓‖∞ + 1 is just a constant we have shown that 𝑓𝛼 ′ ∈ 𝑅[𝑎, 𝑏].



A similar argument will show that if 𝑓𝛼 ′ ∈ 𝑅[𝑎, 𝑏] then 𝑓 ∈ 𝑅𝛼 [𝑎, 𝑏].


Notice that if both 𝑓𝛼 ′ ∈ 𝑅[𝑎, 𝑏] and 𝑓 ∈ 𝑅𝛼 [𝑎, 𝑏] then
𝑏
∫𝑎 𝑓𝑑𝛼 = inf 𝑈𝛼 (𝑓, 𝑃)
𝑃
𝑏
∫𝑎 𝑓𝛼′𝑑𝑥 = inf 𝑈 (𝑓𝛼′, 𝑃).
𝑃



But |𝑈𝛼 (𝑓, 𝑃 ) − 𝑈(𝑓𝛼 ′ , 𝑃)| ≤ ‖𝑓‖∞ 𝜖
𝑏 𝑏
⟹ ∫𝑎 𝑓𝑑𝛼 = ∫𝑎 𝑓𝛼′𝑑𝑥.