Real and Complex Analysis (Rudin)

DIFFERENTIATION 159
21 Iffis a real function on [0, 1] and
y(t) = t + !f(t),
the length of the graph offis, by definition, the total variation of)' on [0, 1]. Show that this length is
finite if and only iff E BV. (See Exercise 13.) Show that it is equal to
fJ1 + [f'(t)]2 dt
iffis absolutely continuous.
n (a) Assume that both f and its maximal function Mf are in IJ(Rt). Prove that then f(x) = 0 a.e.
Em]. Hint: To every other f E IJ(Rk) corresponds a constant c = c(f) > 0 such that
(MfXx) ~ clxl- k
whenever I x I is sufficiently large.
(b) Definef(x) = x-I(log X)-2 if 0 < x < t,f(x) = 0 on the rest of RI. ThenfE LI(RI). Show that
(MfXx) ~ 12x log (2x) 1- I (0 < x < 1/4)
so that J~ (MfXx) dx = ClO.
23 The definition of Lebesgue points, as made in Sec. 7.6, applies to individual integrable functions,
not to the equivalence classes discussed in Sec. 3.10. However, if F E IJ(R,,) is one of these equivalence
classes, one may call a point x E Rt a Lebesgue point of F if there is a complex number, let us call it
(SFXx), such that
lim . -- 1 i If - (SFXx) I dm = 0
.-0 m(B.) Blx ••)
for one (hence for every)f E F.
Define (SF)(x) to be 0 at those points x E Rk that are not Lebesgue points of F.
Prove the following statement: Iff E F, and x is a Lebesgue point off, then x is also a Lebesgue
point of F, andf(x) = (SFXx). Hence SF E F.
Thus S .. selects" a member of F that has a maximal set of Lebesgue points.