Problems and Solutions in - Mathematics - University of Hawaii

1.3 1998 April 3 1 REAL ANALYSIS
5. Let I be the interval [0, 1], and let C(I), C(I × I) denote the spaces of real valued continuous functions on I and
I × I, respectively, with the usual supremum norm on these spaces. Show that the collection of finite sums of the
form
f(x, y) =
φi(x)ψi(y),
where φi, ψi ∈ C(I) for each i, is dense in C(I × I).
6. Let m be Lebesgue measure on the real line R, and for each Lebesgue measurable subset E of R define
1
µ(E) = dm(x).
1 + x2 E
Show that m is absolutely continuous with respect to µ, and compute the Radon-Nikodym derivative dm/dµ.
i
Solution: Obviously both measures are non-negative. We must first prove m ≪ µ. To this end, suppose m(E) >
0, where E ∈ M, the σ-algebra of Lebesgue measurable sets. Then, if we can show µ(E) > 0, this will establish
that the implication µ(E) = 0 ⇒ m(E) = 0 holds for all E ∈ M; i.e., m ≪ µ.
For n = 1, 2, . . ., define
An =
x ∈ R :
1
n + 1 <
Then Ai ∩ Aj = ∅ for all i = j in N, and, for all n = 1, 2, . . .,
µ(An) ≥ 1
n + 1 m(An).
1
1
≤ .
1 + x2 n
Also, R = ∪An, since 0 < 1
1+x2 ≤ 1 holds for all x ∈ R. Therefore,
µ(E) = µ(E ∩ (∪nAn)) = µ(∪n(An ∩ E)) =
µ(An ∩ E).
The last equality might need a bit of justification: Since f(x) = 1
1+x 2 is continuous, hence measurable, the sets
{An} are measurable. Therefore, the last equality holds by countable additivity of disjoint measurable sets.
Now note that m(E) = m(An∩E) > 0 implies the existence of an n ∈ N such that m(An∩E) > 0. Therefore,
µ(E) ≥ µ(An ∩ E) ≥ 1
n + 1 m(An ∩ E) > 0,
which proves that m ≪ µ. By the Radon-Nikodym theorem (A.1), there is a unique h ∈ L1 (µ) such that
m(E) = h dµ, and f dm = fh dµ ∀ f ∈ L 1 (m).
In particular, if E ∈ M and f(x) = 1
1+x2 χE, then
1
h(x)
µ(E) = dm(x) = dµ(x).
E 1 + x2 E 1 + x2 That is, h(x)
dµ = E E 1+x2 dµ(x) holds for all measurable sets E, which implies16 that, h(x)
1+x2 = 1 holds for
µ-almost every x ∈ R. Therefore,
dm
dµ (x) = h(x) = 1 + x2 .
One final note: h is uniquely defined only up to an equivalence class of functions that are equal to 1+x 2 , µ-a.e. ⊓⊔
16Recall the standard result: if f and g are integrable functions such that E f = E g holds for all measurable sets E, then f = g, µ-a.e.
This is an exam problem, but I can’t remember on which exam it appears. When I come across it again I’ll put a cross reference here.
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