Problems and Solutions in - Mathematics - University of Hawaii

1.1 1991 November 21 1 REAL ANALYSIS
Solution: For n = 1, 2, . . ., define
Clearly,
and each An is measurable (why?). 3 Next, define
An = {x ∈ X : 1/n ≤ |f(x)| < n}.
A1 ⊆ A2 ⊆ · · · ↑ A
∞
n=1
A0 = {x ∈ X : f(x) = 0} and A∞ = {x ∈ X : |f(x)| = ∞}.
Then X = A0 ∪ A ∪ A∞ is a disjoint union, and
|f| = |f| +
X
A0
A
|f| +
A∞
An
|f| =
A
|f|. (2)
The first term in the middle expression is zero since f is zero on A0, and the third term is zero since f ∈ L1 (µ)
implies µ(A∞) = 0. To prove the result, then, we must find a measurable set E such that
|f| < ɛ, and
A\E
µ(E) < ∞.
Define fn = |f|χAn . Then {fn} is a sequence of non-negative measurable functions and, for each x ∈ X,
limn→∞ fn(x) = |f(x)|χA(x). Since An ⊆ An+1, we have 0 ≤ f1(x) ≤ f2(x) ≤ · · · , so the monotone
convergence theorem4 implies
X fn →
|f|, and, by (2),
A
lim
n→∞
An
Therefore, there is some N > 0 for which
Finally, note that 1/N ≤ |f| < N on AN, so
µ(AN) ≤ N
|f| dµ =
X\AN
AN
A
|f| dµ =
|f| dµ < ɛ.
|f| dµ ≤ N
X
X
|f| dµ.
|f| dµ < ∞.
Therefore, the set E = AN meets the given criteria. ⊓⊔
4. 5 If (X, Σ, µ) is a measure space, f is a non-negative measurable function, and ν(E) =
measure.
E
f dµ, show that ν is a
Solution: Clearly µ(E) = 0 ⇒ ν(E) = 0. Therefore, ν(∅) = µ(∅) = 0. In particular ν is not identically infinity,
so we need only check countable additivity. Let {E1, E2, . . .} be a countable collection of disjoint measurable sets.
3 Answer: f is measurable and x → |x| is continuous, so g = |f| is measurable. Therefore, An = g −1 ([1/n, n)) is measurable (theorem A.2).
4 Alternatively, we could have cited the dominated convergence theorem here since fn(x) ≤ |f(x)| (x ∈ X; n = 1, 2, . . .).
5 See also: Rudin [8], chapter 1. Thanks to Matt Chasse for pointing out a mistake in my original solution to this problem. I believe the solution
given here is correct, but the skeptical reader is encouraged to consult Rudin.
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