## 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,

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

## 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.