1
(2.2.35): Let f be an additive function.
x
1
(i) Clearly f (nx) = nf (x) and therefore f n = n f (x) for all n Z+ and
x R. Hence, for any m, n Z+ , f m = m f (1), and so, if f is continuous,
n
n
f (x) = f (1)x.
(ii) Suppose that f is bounded on some non
1
1.1.11: Given C and
> 0, set
C( ) =
I C : sup f inf f
I
I
Assume that f is Riemann integrable, and let
exists a > 0 such that
U (f ; C ) L(f ; C ) <
.
> 0 be given. Then there
2
whenever C < . Thus, if C < , then
vol(I )
I C ( )
sup f inf f
I C ( )
I
1
2.1.19:
(i) The preservation of unions is obvious, as is the fact that 1 preserves
dierences. To see that may not preserve dierences, consider the case when
E contains at least two elements and is constant. In general, (B \ A)
(B ) \ (A). When is one-t
1
(3.3.17):
(i) Since
1
f (t) R (dt) f (x)
[x,x+ )
1
2
f (t) f (x) R (dt)
[x,x+ )
1
2
(x,x+ )
f (t) f (x) R (dt),
it is clear that
x Leb(f ) =
1
[x,x+ )
1
(x,x]
f (t) R (dt) f (x)
as
0.
f (t) R (dt) f (x)
as
0.
Similarly,
x Leb(f ) =
(ii) Assume that f