## Chapter 8

## Q. 8.T.5

The following statements are equivalent:

(i) f ∈ \mathcal{R}(a, b) with integral over [a, b] equal to A.

(ii) For any ε > 0, there is a δ > 0 such that, if P is any partition satisfying \left\|P\right\| < δ and α is any mark on P, then

|S(f, P, α) − A| < ε,

that is,

\underset{\left\|P\right\|→0 }{\lim} S (f, P, α) = A.

## Step-by-Step

## Verified Solution

Assume (i) is true. Then

A = U (f) = L (f )

and, if ε > 0 is given, we know from Darboux’s theorem that there is a δ > 0 such that, if P is any partition satisfying \left\|P\right\| < δ, then

A − L (f, P) < ε, U (f, P) − A < ε.

Since, for every mark α on P, we have

L (f, P) ≤ S (f, P, α) ≤ U (f, P) ,

it follows that

|S (f, P, α) − A| < ε,

as required in (ii).

Now assume the validity of (ii). Let ε > 0 be given and suppose that δ > 0 is chosen so that

\left\|P\right\| < δ ⇒ |S (f, P, α) − A| < ε/2

for every mark α on P. Take any partition P = \left\{x_{0}, x_{1}, x_{2}, …, x_{n}\right\} satisfying \left\|P\right\|< δ. From the definitions of m_{i} and M_{i}, we are assured of the existence of α_{i} and β_{i} in \left[x_{i}, x_{i+1}\right] such that

m_{i} > f (α_{i}) − \frac{ε}{2 (b − a)}, M_{i} < f (β_{i}) + \frac{ε}{2 (b − a)}.

This implies that α = (α_{0}, α_{1}, . . . , α_{n}) and β = (β_{0}, β_{1}, . . . , β_{n}) are marks on P and that

L (f, P) > S (f, P, α) − ε /2, U (f, P) < S (f, P, β) + ε /2.

Since \left\|P\right\| < δ, we get

L (f, P) > A − ε, U (f, P) < A + ε,

and since L (f, P) ≤ L (f ) and U (f, P) ≥ U (f ) , we obtain

L (f ) > A − ε, U (f ) < A + ε.

Letting ε → 0, and recalling that L(f ) ≤ U (f ), we conclude that

U (f) = L (f) = A.