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HW11

1. Sets of measure zero cannot contain intervals. If a set A has measure zero, then any collection of intervals whose union contains A must have total length/measure equal to 0. However, if A contained an interval (c,d), then the total length would be at least d-c > 0, a contradiction. 2. If a function f is Riemann integrable over [a,b] and positive over an interval (c,d) contained in [a,b], then the integral of f over [a,b] is positive. This is because f must be continuous almost everywhere, and sets of measure zero cannot contain intervals. 3. If G is non

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93 views2 pages

HW11

1. Sets of measure zero cannot contain intervals. If a set A has measure zero, then any collection of intervals whose union contains A must have total length/measure equal to 0. However, if A contained an interval (c,d), then the total length would be at least d-c > 0, a contradiction. 2. If a function f is Riemann integrable over [a,b] and positive over an interval (c,d) contained in [a,b], then the integral of f over [a,b] is positive. This is because f must be continuous almost everywhere, and sets of measure zero cannot contain intervals. 3. If G is non

Uploaded by

Bobby Tastr
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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HW 11

1. Prove that sets of measure zero contain no intervals.


A has measure zero inf{
P
i=1 (bi ai ) : A i=1 (ai , bi )} = 0 where inf is taken
over all collections of countable number of intervals which their union contains A.
Suppose there is some interval (c, d) A
Then for any collection of intervals {(ai , bi )} i=1 containing A,


Pi=1 (ai , bi ) A (c, d)

i=1 (bi ai ) > (d c) > 0 for all collection of intervals {(ai , bi )}i=1 containing A,
then measure of A would be greater than or equal to d c > 0 which is a contradiction
to A has measure zero.
Therefore A contains no open interval, but this also implies that A cannot contain any
interval.

2. Let f : [a, b] R be Riemann integrable and nonnegative, with f positive over


Rb
(c, d) [a, b]. Then a f dx > 0.
By the Lebesgue theorem, f must be continuous almost everywhere (discontinuous at
a set of measure zero).
Since by the previous problem, sets of measure zero contain no intervals,
p (c, d) such that f is continuous at p.
We have f (p) > 0.
By continuity of f , , 0 < < min{b p, p a}, such that if |x p| 6 , then
|f (x) f (p)| < f (p)
2
,
which implies that if |x p| 6 , then f (x) > f (p) 2
R p+ R p+ f (p)
p
f > p 2
= f (p) > 0 where we use the monotonicity property of the integral.
R p
Since f (x) > 0 for all x [a, b], again by monotonicity property, a f > 0 and
Rb
p+
f >0
combining
Rb R pthe three
R p+parts,R b
a
f = a f + p f + p+ f > f (p) > 0

3. Let G : [a, b] R be nondecreasing and continuous over [a, b] and differentiable


over (a, b). Suppose g is an extension of G0 to [a, b] that is Riemann integrable. Let
f : [G(a), G(b)] R be Riemann integrable and have a primitive F , where F :
[G(a), G(b)] R is continuous over [G(a), G(b)] and differentiable over (G(a), G(b)).
Then Z G(b) Z b
f (y)dy = f (G(x))g(x)dx
G(a) a

Let P = [p0 , , pl ] be the partition of [G(a), G(b)] associated with the primitive F .
Let Q = [q0 , , qm ] be the partition of [a, b] associated with the primitive G.
Since G is not strictly increasing, the inverse of G may not be well defined.
Still, we can choose any of the points that maps to p1 , , pl1 to be G1 (p1 ), , G1 (pl1 ),
and let G1 (p0 ) = a and G1 (pl ) = b.
Then since G is nondecreasing, G1 (p0 ) 6 G1 (p1 ) 6 6 G1 (pl ).
Let G1 (P ) = [G1 (p0 ), , G1 (pl )] and R = Q G1 (P ) = [r0 , , rn ].
Then we have that F (G) is differentiable over (ri1 , ri ) (i = 1, , n).

1
By the Chain Rule we have
F (G)0 (x) = F 0 (G(x))G0 (x) = f (G(x))g(x) for every x (ri1 , ri )
So F (G) is a primitive of f (G)g over [a, b]
The rest of the proof is the same as the increasing case. (Proposition 3.1 in the notes).
R G(b) Rb
4. Find an example such that G(a) f (y)dy = a f (G(x))g(x)dx fails.

Let 
x for x > 0
f (x) =
0 for x<0
and G(x) = x2 then g(x) = 2x
R G(b)a = 1 andR b1 = 1
Let
1
R1 R0 Rb
G(a)
f (y)dy = 1 f (x)dx = 0 6= 2
= 0
2x3 dx + 1
0dx = a
f (G(x))g(x)dx

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