Problem 1
Part a
Suppose first that H(x) = h(x). Let ǫ > 0 and let δ be such that sup|y−x|<δ f (y) ≤
H(x) + ǫ and inf |y−x|<δ f (y) ≥ h(x) − ǫ. Then because H(x) − h(x) = 0, we
have that sup|y−x|<δ f (y) − inf |y−x|<δ f (y) < 2ǫ. Note that f (x) is in between
these two quantities, so both quantities are within 2ǫ of f (x). Taking ǫ to be
sufficiently small proves the result.
Conversely, suppose that f is continuous at x and that ǫ > 0. Then there
exists a δ > 0 such that |f (y) − f (x)| < ǫ for all y such that |x − y| < δ.
Then by the triangle inequality, we have that sup|x−y|≤δ f (y) < f (x) + ǫ, and
inf |x−y|≤δ > f (x) − ǫ. This shows that the quantities H(x) and h(x) both
approach f (x).
Part b
Consider the function G from Theorem 2.28 in Folland. The value of this function at x is the limit as k → ∞ of Mk , where Mk is the supremum of f on
the interval pk , where pk is the element of the partition Pk that contains x.
Note that for any δ > 0, there exists a k0 such that pk is contained in the
δ-neighborhood of x for any k > k0 . Therefore, Mk is smaller than the supremum of f (y) over all y in the δ-neighborhood of x, giving that G(x) ≤ H(x)
everywhere. For the reverse inequality, note that, for any interval pk containing
x in its interior, there is a δ such that the δ-neighborhood of x is contained
in the interval pk . This implies that if δ is sufficiently small, we have that the
supremum of f over pk is larger than the supremum of f over δ. Apply this
argument to all of the x that are not left endpoints of some interval of any
of the partitions pk . For such x, we therefore have that H(x) ≤ G(x) by the
monotonicity of the supremum. Because only countably many points occur as
the endpoint of an interval in the partition, this proves that H(x) ≤ G(x) at all
except countably many points. The argument for g and h is basically the same.
This implies that H and h are Lebesgue
measurable
because
G andR g are
R
R
R
Lebesgue measurable. Furthermore, RE H dm = E RG dm and E h dm = E g dm
for any measurable set E. So we get [a,b] H dm = [a,b] G dm, and the same for
h and g.
Now, suppose that f is continuous at almost every x. Then
R h(x) = H(x)
at all except for countably many x. Then the above integrals [a,b] H dm and
R
[a,b] h dm are equal, and therefore the integrals of G and g are equal, and we
can apply Theorem 2.28a.
suppose that f is Riemann integrable. Then we have that
R
R Conversely,
H
dm for every Lebesgue measurable set E. By Theorem 2.23,
h
dm
=
E
E
this implies that h = H almost everywhere.
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Problem 2
Rx
Suppose that f ∈ L1 (m). Let F (x) = −∞ f (t) dt.
Ry
Let ǫ > 0, Consider F (y). If y > x, then F (y) − F (x) = x f (t) dt and if
Rx
Rx
x > y, then F (x)−F (y) = y f (t) dt. Therefore, we need to show that y f (t) dt
Ry
and x f (t) dt can be made arbitrarily close to 0 if y is taken to be sufficiently
close to x.
R
Recall from problem 14 of section 2.2 that λ defined by λ(E) = E |f (t)| dm
is a finite measure. We have that λ({x}) = 0 because |f |χ{x} = 0 almost
everywhere. Furthermore, λ([−∞, x]) is finite because λ is a finite measure.
So by continuity of measure, for any ǫ > 0, there must exist a y0 such that
λ([y0 , x]) < ǫ. Therefore, by monotonicity of measure we have the same inequalRx
ity for any y0 < y < x. By the triangle inequality, we have y f (t) dm < ǫ for
all y such that y0 < y < x. A similar argument works for y > x. This shows
that the function F is continuous at x.
Problem 3
Suppose that |fn | ≤ g, where g is in L1 , and that fn → f in measure. I will show
part b, which immediately implies part a by the triangle inequality. If fn → f
in measure, then for any δ, ǫ > 0, there exists N such that µ({|fn (x) − f (x)|)} >
ǫ) < δfor any n > N . For such an ǫ, δ, and n, we have
Z
Z
Z
|fn (x) − f (x)| +
|fn (x) − f (x)|,
|fn (x) − f (x)| =
X
Sǫc
Sǫ
where Sǫ is the set of points where |fn (x) − f (x)| > ǫ. Note that we have
the bound
|fn (x) − f (x)| < 2g(x) on Sǫ , so the first integral is bounded
R above
R
by 2 Sǫ g(x). Note that Sǫ has measure no more than δ. Therefore, Sǫ g(x)
R
is bounded above by supµ(E)≤δ E g(x). If this supremum does not go to zero
as δ → 0, (say the limit is η) then there exist Sj with µ-summable measures
that have λ-measure at least η. Let S = lim sup Sj . On the one hand, S
has µ-measure 0, because it is contained in sets of arbitrarily small µ-measure.
This implies that the λ-measure of S is zero as well, because it is given as the
integral of a measurable function on a set of measure zero. On the other hand,
continuity of measure (together with the fact that λR is a finite measure) implies
that λ(S) ≥ η. This is a contradiction. Therefore, Sǫ g(x) → 0 as n → ∞.
Therefore, we only need to deal with the remaining term: the integral over
the places where |fn (x) − f (x)| < ǫ. We want to use this bound, but not
directly.
R Let r > 0 and pick a measurable subset E of X of finite measure for
which E c g(x) < r. (We can do this because the support of any L1 function
is σ-finite). On E c ∩ Sǫc , apply the bound |fn (x) − f (x)| < 2g(x). On E ∩ Sǫc ,
apply the bound |fn (x) − f (x)| < ǫ. This shows that the integral of |fn − f |
over Sǫc is no more than 2r + ǫµ(E). Because ǫ is arbitrary, this can be made
2
smaller than 3r, and because r is arbitrary, this shows that the integral can be
made arbitrarily small.
Problem 4
Suppose that |fn | ≤ g for all n, where g is in L1 (µ). As in Egoroff’s theorem,
I’m going to ignore the null set on which fn does not converge to f . We want
to show that for any ǫ > 0 there exits a set E such that µ(E) < ǫ and fn → f
uniformly on E c . As in the proof of Egoroff’s theorem, we want to keep track
of where |fn − f | is large. For any δ > 0 define Eǫ,N to be the set where |fn − f |
is at least ǫ for some n ≥ N . Consider the sets E1/j,N for various n. Note
that |fn − f | ≤ 2g a.e for every n, so E1/j,N is contained in the set of points
where g(x) ≤ 2j for every N . Note that this set has finite measure (otherwise
g would not be integrable). Therefore, we can select anTNj such that |E1/j,Nj |
has measure no more than 2ǫj by using the fact that N E1/j,N has measure
zero and using continuity of measure. Now, take the union over all j to get the
desired set. Clearly fn → f uniformly on E c . Note that you can replace the
statement that g ∈ L1 with the much weaker statement that the measure of the
set of points on which g > ǫ is finite for any ǫ > 0.
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