Download Quiz 2 1. Let m be the counting measure on the set of all integers Z

Quiz 2
1. Let m be the counting measure on the set of all integers Z. If µ is any
R measure on
Z, then there is a non-negative function f on Z such that µ(E) = E f (x)dm(x)
for any subset E of Z. Can this be true for any measure m.
Proof:
Let f (x)P
= µ({x}) forRany x ∈ Z. Now for any subset E of Z, µ(E) =
P
x∈E µ({x}) =
x∈E f (x) = E f (x)dm(x).
R
Let m = δ0 . Suppose µ = f dm for some non-negative function f . Then
µ(E) = 0 if 0 6∈ E and µ(E) = f (0) if 0 ∈ E. Thus, µ = f (0)m and hence
δ1 + δ0 is not of this form.
In general the result is true for m if and only if m({x}) 6= 0 for any x ∈ Z.
2. Let (X, B, µ) be a measure space and f be an integrable function on X. Show
that {x ∈ X | f (x) 6= 0} is σ-finite, that is, is a countable union of sets of finite
measure.
Proof: Let
R Et = R{x ∈ X | |fR (x)| > t} for any t > 0. Then for t > 0,
tµ(Et ) = Et tdµ ≤ Et |f (x)| ≤ X |f (x)| < ∞. This implies that Et has finite
measure. Since {x ∈ X | f (x) 6= 0} = ∪∞
n=1 E 1 , we get the result.
n
3. Is the MCT true for decreasing sequence of functions that converge. That
is, find a measure space (X, B, µ) and a decreasing
of non-negative
R sequence
R
measurable functions (fn ) such that fn → f but fn 6→ f .
Proof: Let m be the Lebesgue-measure on R. Define a sequence of functions
(fn ) by fn (x) = 0 if x < n and fn (x) = 1 if x ≥ n. Then fn (x) → 0 for all x ∈ R.
It can easily RverifiedRthat (fn ) is a decreasing sequence of
functions
R non-negative
R
on R. Now, fn = [n,∞) fn = m([n, ∞)) = ∞. Thus, fn 6→ lim fn = 0.
4. Let X be a uncountable set and B be the σ-algebra of all subsets E of X such
that E or E c is countable. Let µ be the measure on B defined by µ(E) = 0 if
E is countable and µ(E) = 1 if E c is countable. Find all measurable functions
on X and its integral with respect to µ.
Proof: For b ∈ [−∞, ∞], let Xb = {x ∈ X | f (x) ≥ b} and B = {b ∈ [−∞, ∞] |
Xb is uncountable }. Let α = supb∈B b. We will show that f = α a.e. If
α = ∞, then an ∈ B such that an < an+1 and an → ∞. Now Xan+1 ⊂ Xan and
∩Xan = X∞ . Since µ is a probability measure, µ(Xan ) = 1 → µ(X∞ ). Thus,
f = ∞ a.e.
If α = −∞, then Xb is countable for any b > −∞ and hence {x | f (x) 6=
−∞} = ∪n≥1 X−n is countable. Thus, f = −∞ a.e.
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Assume that |α| < ∞. Now for any n ≥ 1, there exists a bn ∈ B such that
α − n1 < bn . This implies that Xbn ⊂ Xα− 1 is uncountable. Now {x | f (x) <
n
α} = ∪n≥1 {x | f (x) < α − n1 } and hence {x | f (x) < a} is countable. Now
{x | f (x) > α} = ∪n≥1 {x | f (x) > α + n1 } and hence since α + n1 > α,
{x | f (x) > α + n1 } is countable. This implies that {x | f (x) > a} is countable.
Thus, {x | f (x) 6= α} is countable. This proves that f (x) = α a.e.
Thus, measurable functions are constant α a.e. and hence its integral with
respect to µ is that constant α.
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