Problems and Solutions in - Mathematics - University of Hawaii

1.7 2007 November 16 1 REAL ANALYSIS
where f − fn1 → 0, by assumption, and supx∈A |fn(x) − g(x)| → 0 since fn → g uniformly on A. Since ɛ > 0
was arbitrary, it follows that n
−n |f − g| dm = 0 and, for functions f, g ∈ L1 ([−n, n], m), this implies that f = g
a.e. on [−n, n].
It remains to show that f, g ∈ L1 ([−n, n], m). It’s clear that f ∈ L1 since f1 ≤ f − fn1 + fn1 < ∞. To
a.e.
prove g ∈ L1 ([−n, n], m) note that fn −−→ g implies limn |fn(x)| = |g(x)| for almost all x, so by Fatou’s lemma,
g1 = |g| dm = lim |fn| dm ≤ lim |fn| dm = lim fn1 = f1 < ∞.
The last equality holds because, by the triangle inequality, |fn1 − f1| ≤ fn − f1 → 0. ⊓⊔
3. Let K be a compact subset in R 3 and let f(x) = dist(x, K).
(a) Prove that f is a continuous function and that f(x) = 0 if and only if x ∈ K.
n (b) Let g = max(1 − f, 0) and prove that limn→∞ g exists and is equal to m3(K).
Solution: (a) Define dist(x, K) = f(x) = infk∈K |x − k|. Clearly, for all k ∈ K, f(x) ≤ |x − k|. Therefore, by
the triangle inequality, for any x, y ∈ R 3 ,
and so, taking the infimum over k ∈ K on the right,
Similarly,
f(x) ≤ |x − y| + |y − k|, ∀k ∈ K,
f(x) ≤ |x − y| + f(y). (26)
f(y) ≤ |x − y| + f(x). (27)
Obviously, for any given x ∈ R 3 , f(x) is finite. Therefore, (26) and (27) together imply that
Whence f is (Lipschitz) continuous.
|f(x) − f(y)| ≤ |x − y|, ∀ x, y ∈ R 3 .
Now, if x ∈ K, then it’s clear that f(x) = 0. Suppose x /∈ K; that is, x is in the complement K c of K. Since K is
closed, K c is open and we can find an ɛ-neighborhood about x fully contained in K c , in which case f(x) > ɛ. We
have thus proved that f(x) = 0 if and only if x ∈ K. ⊓⊔
(b) First observe that f(x) = 0 for all x ∈ K, and f(x) > 0 for all x /∈ K. Define K1 to be a closed and bounded
set containing K on which f(x) ≤ 1. That is, K1 is the set of points that are a distance of not more than 1 unit
from the set K. In particular K ⊂ K1. Notice that g = max(1 − f, 0) = (1 − f)χK1 . Also, if x ∈ K1 \ K,
then 0 ≤ 1 − f(x) < 1, so g n → 0 on the set K1 \ K, while on the set K, g n = 1 for all n ∈ N. Therefore,
g n → χK. Finally, note that g n ≤ χK1 ∈ L 1 (R 3 ) so the dominated convergence theorem can be applied to yield
limn→∞
g n = χK = m3(K). ⊓⊔
4. Let E be a Borel subset of R 2 .
(a) Explain what this means.
(b) Suppose that for every real number t the set Et = {(x, y) ∈ E | x = t} is finite. Prove that E is a Lebesgue
null set.
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