Math 597. Homework 10 Bryden Cais

(1) Let B_{n} denote the unit ball in R^{n} and let C_{n} be the volume of B_{n}. Let μ be the usual Lebesgue measure on R^{n}. Then by Prop. III.15.1 we have

∫_{B_{n}} ∥x∥^{p} = p ∫_{0}^{∞} t^{p−1}μ({x ∈ B_{n} : ∥x∥ > t}) dt

= p ∫_{0}^{1} t^{p−1}μ({x ∈ B_{n} : ∥x∥ > t}) dt

= p ∫_{0}^{∞} t^{p−1}(C_{n} − C_{n}t^{n}) dt

= pC_{n} ∫_{0}^{∞}(t^{p−1} − t^{p+n−1}) dt

= pC_{n} \{\frac{1}{p} − \frac{1}{p + n}\}

= \frac{nC_{n}}{p + n}.

Similarly, we have

∫_{R^{n}−B_{n}} ∥x∥^{p} = p ∫_{0}^{∞} t^{p−1}μ({x ∈ R^{n} − B_{n} : ∥x∥ > t}) dt,

but for all t, we have μ({x ∈ R^{n} − B_{n} : ∥x∥ > t}) = +∞ so that the integral is infinite.

(2) Let f, g be integrable on [a, b] and set F = ∫_{a}^{x} f dμ, G = − ∫_{x}^{b} g dμ.

(a) Since F, G are integrals of f, g, they are of bounded variation and absolutely continuous. Moreover, F′, G′ exist almost everywhere and are equal a.e. to f, g respectively. It follows that F G is of bounded variation and absolutely continuous (a result from class). Therefore, (F G)′ exists a.e. and we have (F G)′ = F′g + G′f a.e. Moreover, we have

(F G)′ = (F G)(a) + ∫_{a}^{x}(F G)′ dμ
= (F G)(b) + ∫_{b}^{x}(F G)′ dμ.

Using F (a) = G(b) = 0 we see that

∫_{a}^{x}(F G)′ dμ + ∫_{x}^{b}(F G)′ dμ = ∫_{a}^{b}(F G)′ dμ = 0.

Using the fact that (F G)′ = F′g + G′f a.e. it follows that

∫_{a}^{b} F g dμ = − ∫_{a}^{b} Gf dμ.

Of course, this is formally just integration by parts for differentiable functions F, G.

(b) No, the proposed formula is false without further restrictions. For example, let

F (x) = \begin{cases}0 & if x ∈ [0, 1/2] \\ 1 & if x ∈ (1/2, 1] \end{cases} ,

and

G(x) = \begin{cases}1 & if x ∈ [0, 1/2] \\ 0 & if x ∈ (1/2, 1] \end{cases} .

Then we have F (0) = G(1) = 0 and

∫_{0}^{1} F dμ_{G} = F (1/2)μ_{G}({1/2}) = 0 · (−1) = 0,

while

− ∫_{0}^{1} G dμ_{F} = G(1/2)μ_{F} ({1/2}) = −1 · 1 = −1.

(3) Let f(x) = x sin (\frac{1}{x^{α}} ) for α > 0. We consider f on the interval [0, 1] and show that it is of bounded variation for α < 1. It is clear that f has a single extremum in any interval [\frac{1}{((j+1)π)^{1/α}} , \frac{1}{(jπ)^{1/α}}] for j ≥ 1. Let m_{2n} for n ≥ 0 denote the local maxima values of f and m_{2n+1} for n ≥ 0 denote the values of f at the local minima. We easily have the estimates

\frac{1}{((j + 1)π)^{α}} ≤ |m_{j−1}| ≤ \frac{1}{(jπ)^{α}}

for 1 ≤ j. Moreover, since f is piecewise monotone on the intervals between any consecutive extreme points, we have

V_{f}[0, 1] = ∑_{n≥0}(m_{2n} − m_{2n+1})

so that

∑_{j≥2} \frac{1}{(jπ)^{1/α}} ≤ V_{f}[0, 1] ≤ ∑_{j≥1} \frac{1}{(jπ)^{1/α}} .

It follows that f is of bounded variation if and only if the sum ∑_{j>0} \frac{1}{j^{1/α}} converges. By the integral test, this is the case if and only if α < 1, as claimed.

(4) Let μ be a finite signed Borel measure on [0, 1].

(a) Let E′ = {x ∈ [0, 1] : μ({x}) ≠ 0}. Since μ is finite, so are μ^{+} and μ^{−} and hence so is |μ|. Clearly, μ(x) ≠ 0 implies |μ|(x) > 0. By monotonicity and finite additivity of |μ| we then have

|μ|([0, 1]) ≥ sup{ ∑_{x∈S⊂E′} |μ|(x) : S finite} = ∑_{x∈E′} |μ|(x).

It follows from HW 1 that E′ is countable and that ∑_{x∈E′} |μ|(x) is finite. since

|μ|(x) = μ_{1}(x) + μ_{2}(x) = |μ_{1}(x)| + |μ_{2}(x)| ≥ |μ_{1}(x) − μ_{2}(x)| = |μ(x)|,

we see that ∑_{x∈E′} |μ(x)| < ∞.

(b) It follows that μ_{pp}(E) := ∑_{x∈E} μ({x}) is a finite Borel measure on [0, 1] and that μ_{pp}⊥μ_{Leb}. Indeed, we have |μ_{pp}|(E^{′c}) = 0 and μ_{Leb}(E′ = 0). By Thm. III.18.2 we can write μ = μ_{0} + μ_{1} with μ_{0}⊥μ_{Leb} and μ_{1} ≪ μ_{Leb}. Thus let F be such that |μ_{0}|(F^{c}) = μ_{Leb}(F) = 0. We claim that μ_{pp}(F^{c}) = 0. Indeed, by definition,

μ_{pp}(F^{c}) = ∑_{x∈F^{c}} μ({x}).

Since μ = μ_{0} + μ_{1} and μ_{1} ≪ μ_{Leb}, we see that μ_{1}({x}) = 0 for any point x so that μ({x}) = μ_{0}({x}). Therefore, for any x ∈ F^{c} we have |μ_{0}|({x}) ≤ |μ_{0}|(F^{c}) = 0 where we have used the monotonicity of |μ_{0}|. Thus, for all x ∈ F^{c} we have μ({x}) = 0 and therefore μ_{pp}(F^{c}) = 0 as claimed. Thus μ_{3} = μ_{0} −μ_{pp} satisfies μ_{3}(F^{c}) = 0 and of course we still have μ_{Leb}(F) = 0 so that μ_{3}⊥μ_{Leb}. Now by our above discussion, we have seen that for any point x we have μ({x}) = μ_{0}({x}). Similarly, by definition we have μ_{pp}({x}) = ∑_{x∈{x}} μ(x) = μ({x}) = μ_{0}({x}) so that μ_{3}({x}) = 0 for all points x.