Question 8.T.10: Let f : [a, b] → [c, d] be Riemann integrable. If φ : [c, d]...

Let ε > 0 be given. We shall prove the existence of a partition P ∈ \mathcal{P} (a, b) such that

U (φ ◦ f ) − L (φ ◦ f ) < ε.

Since φ is continuous on the compact interval [c, d] , it is bounded and uniformly continuous. Consequently there is a real constant K such that

|φ(t)| ≤ K  for all t ∈ [c, d] ,        (8.12)

and if we set ε^{\prime} =\frac{ε}{2K + (b − a)}, we know that there is a δ > 0 such that

s, t ∈ [c, d] , |t − s| < δ ⇒ |φ (t) − φ (s)| < ε^{\prime}.        (8.13)

On the other hand, since f ∈ \mathcal{R}(a, b) , there is a partition P = \left\{x_{0}, x_{1}, …, x_{n}\right\} such that

U (f, P ) − L (f, P ) < ε^{\prime}δ.        (8.14)

Let

M_{i} = \sup \left\{f (x) : x ∈ [x_{i}, x_{i+1}]\right\}

M^{∗}_{i} = \sup \left\{φ (f (x)) : x ∈ [x_{i}, x_{i+1}]\right\}

m_{i} = \inf \left\{f (x) : x ∈ [x_{i}, x_{i+1}]\right\}

m^{∗}_{i} = \inf \left\{φ (f (x)) : x ∈ [x_{i}, x_{i+1}]\right\} .

We then have

U (φ \circ f ) − L (φ \circ f) = \sum\limits_{i=0}^{n−1}{(M^{∗}_{ i} − m^{∗}_{i} ) (x_{i+1} − x_{i})}

= \sum\limits_{i∈J_{1}}{(M^{∗}_{ i} − m^{∗}_{i} ) (x_{i+1} − x_{i})}

+ \sum\limits_{i∈J_{2}}{(M^{∗}_{ i} − m^{∗}_{i} ) (x_{i+1} − x_{i})} ,        (8.15)

where

J_{1} = \left\{i ∈ \left\{0, 1, . . . , n − 1\right\} : M_{i} − m{i} < δ\right\}

J_{2} = \left\{i ∈ \left\{0, 1, . . . , n − 1\right\} : M_{i} − m_{i} ≥ δ\right\} .

If i ∈ J_{1}, then

|f (x) − f (y)| < δ  for all x, y ∈ [x_{i}, x_{i+1}] ,

and therefore, using (8.13),

|φ (f (x)) − φ (f (y))| < ε^{\prime}  for all x, y ∈ [x_{i}, x_{i+1}] ,

which implies that M^{∗}_{i} − m^{∗}_{i} ≤ ε^{\prime}. Therefore

\sum\limits_{i∈J_{1}}{(M^{∗}_{ i} − m^{∗}_{i} ) (x_{i+1} − x_{i})} ≤ \sum\limits_{i∈J_{1}}{ε^{\prime} (x_{i+1} − x_{i}) ≤ ε^{\prime} (b − a)} .

Now from (8.14) we see that

ε^{\prime}δ >\sum\limits_{i=0}^{n−1}{(M_{i} − m_{i}) (x_{i+1} − x_{i})}

≥ \sum\limits_{i∈J_{2}}{(M_{i} − m_{i}) (x_{i+1} − x_{i})}

≥ δ \sum\limits_{i∈J_{2}}{(x_{i+1} − x_{i})} ,

hence

\sum\limits_{i∈J_{2}}{(x_{i+1} − x_{i}) < ε^{\prime}} .

Since M^{∗}_{ i} − m^{∗}_{i} ≤ 2K, we must have

\sum\limits_{i∈J_{2}}{(M^{∗}_{ i} − m^{∗}_{i} ) (x_{i+1} − x_{i})} < 2Kε^{\prime}.

Using these inequalities in (8.15) yields

U (φ \circ f ) − L (φ \circ f ) < ε^{\prime} (b − a) + 2Kε^{\prime} = ε.