Real Analysis HW 4 Solutions
Problem Problem 4: Suppose
f is f is a real-valued function on each number c number c.. Is f Is f necessarily necessarily measurable?
R such
that f −1 (c) is measurable for
Solution: No.
Let V Let V be a non-measurab non-measurable le subset of (0, (0, 1) and consider consider the function function f f ((x) = (2 χV − 1). 1). Note Note that that f > 0 on V and f ≤ 0 on R ∼ V . Sincee f e · (2χ V . Sinc f is one-to-one we −1 also have that f that f (c) is either empty or a singleton and therefore measurable. However, by construction we have f have f −1 ((0, ((0, ∞)) = {x : f ( = V . . Therefore f Therefore f is is not measurable. f (x) > 0 > 0 } = V x
Let {f n } be a sequence of measurable functions defined on a measurable set Definee E 0 to be the set of points x in E onve verge rges. s. Is the set E 0 E . Defin E at which {f n (x)} ccon measurable. Problem Problem 9:
Solution: Note
that we may write E 0 as ∞
E 0 = {x : {f n (x)} is Cauchy} =
∞
< 1/k /k}. {x : | f n (x) − f m (x)| < 1
k=1 N =1 n,m≥N
Therefore E 0 is measurable since { x : | f n (x) − f m (x)| < 1 < 1/k /k} is.
Problem Problem 13: A
real valued measurable function is said to be semisimple provided it takes only a countable countable number number of values. alues. Let f Let f be be any measurable function E function E .. Show that there is a sequence sequence of semisimple semisimple functions functions {f n } on E on E that that converge to f uniformly on E .. f uniformly on E Solution: Suppose
for each integer m integer m ≥ 1 we split up R into intervals {I n,m 1/m n,m }n∈Z of size 1/m given by I by I n,m [(n − 1)/m,n/m 1)/m,n/m), ), which in turn partitions E partitions E into into measurable sets {E n,m n,m = [(n n,m }n∈Z −1 given by E by E n,m = f (I n,m n,m = f n,m ). Define the semisimple function φm =
n∈ Z
n−1 · χE n,m n,m m
It follows that for any integer m ≥ 1, if we fix x ∈ E there E there is an n ∈ Z such that φm (x) = (n − 1)/m 1) /m ≤ f ( This means that that for each each x ∈ E and f (x) < n/m and therefore x ∈ E n,m n,m . This any n > 1/ 1 /,, m ≥ 1 we may write f ( f (x) − φm (x) < 1 < 1/m /m.. Therefore if we let > 0, then for any n on E , − φn = |f − φn | < 1 f − < 1/n /n < on E and so φ so φ n → f on E .. f uniformly on E
1
Problem 22: (Dini’s
Theorem) Let { f n } be an increasing sequence of continuous functions on [a, b] which converge pointwise on [a, b] to the continuous function f on [a, b]. Show that the convergence is uniform on [a, b]. Let > 0, and define E n = {x ∈ [a, b] : f (x) − f n (x) < }. Note that since f n and f are continuous and converge pointwise, that E n are relatively open in [a, b] and ∞ n=1 E n = [a, b]. Furthermore, since f n is an increasing sequence then E n ⊆ E n+1 . It follows by Heine Borel that [a, b] is compact and therefore there exists a sub-cover {E nj } jN =1 of [a, b]. However since E n are ascending, Solution:
N
[a, b] =
E nj = E n , ∗
j =1
where n ∗ = max j {n j }. Therefore the convergence is uniform.
Problem 25: Suppose f is
a function that is continuous on a closed set F of real numbers. Show that f has a continuous extension to all of R. This is a special case of the forthcoming Tietze Extension Theorem. Solution: We
may assume F is non-empty. Since F is closed we see that R ∼ F is open, and therefore may be written as a countable disjoint union of open intervals R ∼ F = ∞ ˜ k=1 (ak , bk ). On each of these intervals, we extend f continuously to f as a linear function in the following way: suppose x ∈ (ak , bk ), then
• if a k , bk are finite we define ˜ = f (bk ) − f (ak ) (x − ak ) + f (ak ), f (x) bk − ak
• if b k = ∞ define • if a k = −∞ define
˜ = f (ak ), f (x) ˜ = f (bk ). f (x)
˜ now continuous on R since it matches f on the endpoints of the Clearly the extended f is intervals and is therefore continuous at every point. Problem 27: Show
that the conclusion of Egoroff’s Theorem can fail if we drop the assumption that the domain has finite measure. Solution: Consider
sequence f n (x) = χ[n,∞) (x). Clearly f n → 0 pointwise on R. Suppose that there existed a set F such that m(R ∼ F ) < and f n → 0 uniformly on F . Since f n is an indicator function, this means we can find an N such that f N = 0 on F . This implies F ⊆ ( −∞, N ) and so [N, ∞) ⊆ R ∼ F , giving m([N, ∞)) ≤ m(R ∼ F ) < , 2
which is a contradiction. Problem 29:
Prove the extension of Lusin’s Theorem to the case that E has infinite
measure. Solution: Suppose
f : E → R is measurable. We split up E into disjoint finite measure sets {E n }n∈Z given by E n = E ∩ [n, n + 1). By Lusin’s theorem for finite measure sets, we know that there exist closed sets F n and continuous functions gn : F n → R such that m(E n ∼ F n ) < /2|n|+1 and f = g n on F n . Define F = k∈Z F n and g(x) =
gn (x)χF n (x).
k ∈Z
Note that g is continuous on F . We now claim that F is closed. To see this consider {xn } ⊆ F , converging to an x ∈ E . Since x ∈ E k for some k, we see that for large enough N , {xn }n≥N belongs to F k−1 ∪ F k which is a closed set and therefore x ∈ F k−1 ∪ F k ⊆ F . Since F is closed, by problem (25) we may extend g continuously to f = g on F and m(E ∼ F ) = m
(E n ∼ F n )
n∈Z
=
R.
It now follows that
m(E n ∼ F n ) < .
k ∈Z
3
