MATH 501: REAL ANALYSIS Homework 12 Due to 12/10/14

MATH 501: REAL ANALYSIS
Homework 12
Due to 12/10/14
Student’s name:
(1) Let X = Y be the set of positive integers and µ = ν the counting
measure. State the Fubini and Tonelli theorems explicitly for
this case.
Solution: When µ = ν is the counting measure, the integral
with respect to these two measure is equivalent to the sum of
series.
Fubini Theorem (discrete case): If the series {xm,n } is summable, i.e., m,n xm,n < ∞, then for every m, n xm,n < ∞
and m,n xm,n = m n xm,n .
Tonelli Theorem (discrete case): If for a series {xm,n }, for
every m, n xm,n < ∞ and m,n xm,n = m n xm,n < ∞,
Then {xm,n } is summable, i.e., m,n xm,n < ∞.
Where m,n xm,n < ∞ is defined as supF { (m,n)∈F xm,n , F
is the finite subset of N × N}.
(2) Let X = Y be the set of positive integers and µ = ν the counting
measure. Let also

−x

if x = y
2 − 2
−x
f (x, y) = −2 + 2
if x = y + 1

0
otherwise
Show that the Fubini theorem fails for f and explain why.
Solution: Assuming the Fubini theorem doesn’t fail. Calculating h(x) = f (x, ·)dν, h(1) = 2− 12 and h(x) = 0 otherwise
where x ∈ N. Calculating g(y) = f (·, y)µ = 2−y − 2−y+1 =
−2−y . Obviously h(x)dµ = 32 , g(y)dν < 0 contradicts.
1
2
MATH 501: REAL ANALYSIS
(3) Let h and g be integrable functions on (X, A, µ) and (X, B, ν)
respectively. Define f (x, y) = h(x)g(y). Show that f is integrable on (X × Y, A × B, µ × ν) and that
f d(µ × ν) =
X×Y
h dµ
X
g dν.
Y
Solution: First suppose h, g are nonnegative almost everywhere. Using Tonelli theorem, we could prove that f is integrable (X × Y, A × B, µ × ν) and that X×Y f d(µ × ν) =
h dµ Y g dν.. In the general case, we could write f = (h+ −
X
−
h )(g + −g − ) = h+ g + +h− g − −(h+ g − +h− g + ). Applying Tonelli
theorem again, we get X×Y h+ g + d(µ×ν) = X h+ dµ Y g + dν,
h− g − d(µ×ν) = X h− dµ Y g − dν, X×Y h+ g − d(µ×ν) =
X×Y
h+ dµ Y g − dν, X×Y h− g + d(µ × ν) = X h− dµ Y g + dν.
X
Therefore X×Y f d(µ × ν) = X h dµ Y g dν.
(4) Let f (r, α) = r2 sin α be a function defined on the unit disk
D = {(r, α) : r ≤ 1, 0 ≤ α < 2π} in R2 . Show that f is
integrable with respect to area and compute its integral.
Solution: f is bounded in D hence integrable. Observing
the fact in calculus dxdy = rdrdα, D f dxdy = D f (r, α)rdrdα =
r3 sin αdrdα. As is showed in (3), D f drdα = [0,1] r3 dr [0,2π] sin αdα =
D
1
∗ 0 = 0.
4
(5) Let a function
x+y
x2 + y 2
be defined on the unit square S = [0, 1] × [0, 1] in R2 . Is this
function integrable with respect to area? Justify your answer.
Solution:
Considering a bigger area D which is the 41
part of the disk with radius 2, obviously S ⊂ D. S f dxdy ≤
f dxdy. Under the polar coordinate system, f can be written
D
θ
into f (r, θ) = sin θ+cos
and D = {(γ, θ) : 0 ≤ γ ≤ 2, 0 ≤ θ ≤ π2 }.
γ
Then D f (x, y)dxdy = D f (γ, θ)γdγdθ = D sin θ + cos θdγdθ
is bounded. f is integrable with respect to area.
f (x, y) =

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