Math 597. Homework 2 Bryden Cais

(1) Suppose that Q is of type G_{δ}, i.e. that we can write Q = ⋂_{i=1}^{∞} U_{i} where each U_{i} is open. Observe that since Q is dense in R and Q ⊂ U_{i} for all i, each U_{i} is dense in R, so that Q is a countable intersection of dense, open sets. For any α ∈ Q, let S_{α} = (−∞, α) ∪ (α, ∞). Clearly, S_{α} is open and dense in R. Then Q \ {α} = Q ∩ S_{α} is a countable intersection of dense open sets, and hence

⋂_{α∈Q} Q \ {α} = ∅

is a countable intersection of dense open sets, and by Baire’s Theorem, is dense. This is absurd, so that Q is not of type G_{δ}.

(2) Recall that Axiom (iv) in the definition of a measure is that μ(E) < ∞ for some set E. We show that the axiom μ(∅) = 0 is equivalent to this. One direction is trivial, as we just set E = ∅. In the other direction, suppose that μ(E) < ∞. Since E ∩ ∅ = ∅, by Axiom (iii) (countable additivity) we have

μ(E) = μ(E ∪ ∅) = μ(E) + μ(∅)

from which it follows that μ(∅) = 0.

(3) We have lim sup E_{j} := ⋂_{n=1}^{∞} ⋃_{j=n}^{∞} E_{j} by definition. Suppose that ∑_{j=1}^{∞} μ(E_{j}) < ∞. Observe, for each n we have lim sup E_{j} ⊂ ⋃_{j=n}^{∞} E_{j} so that, by the monotonicity of μ, we have

μ(lim sup E_{j}) ≤ μ( ⋃_{j=n}^{∞} E_{j}) = ∑_{j=n}^{∞} μ(E_{j}).

Now since ∑_{j=1}^{∞} μ(E_{j}) < ∞, for any ϵ > 0 there exists some N > 0 such that ∑_{j=N}^{∞} μ(E_{j}) < ϵ. It follows that for any ϵ > 0 we have an N > 0 with

μ(lim sup E_{j}) ≤ μ( ⋃_{j=N}^{∞} E_{j}) < ϵ.

Since this holds for any ϵ > 0 we have μ(lim sup E_{j}) = 0.

(4) Let μ be a finitely additive outer measure and {E_{j}} and countable collection of pairwise disjoint measurable sets. Since ⋃_{j=1}^{N} E_{j} ⊂ ⋃_{j=1}^{∞} E_{j} for every N > 0, by monotonicity of μ, we have μ(⋃_{j=1}^{N} E_{j}) ≤ μ(⋃_{j=1}^{∞} E_{j}). Thus, by finite additivity and pairwise disjointness of the E_{j}, we have

∑_{j=1}^{N} μ(E_{j}) ≤ μ( ⋃_{j=1}^{∞} E_{j}).

Since this inequality is independent of N, it holds in the limit, i.e.

∑_{j=1}^{∞} μ(E_{j}) ≤ μ( ⋃_{j=1}^{∞} E_{j}).

Countable subadditivity of μ gives the reverse inequality, so we must have

∑_{j=1}^{∞} μ(E_{j}) = μ( ⋃_{j=1}^{∞} E_{j}).

Thus, μ is countably additive.

(5) Recall that

H_{α,ϵ}(E) = inf {∑ diam(Q_{j})^{α} : E ⊂ ⋃ Q_{j}, diam(Q_{j}) < ϵ} ,

and that without loss of generality we may suppose that Q_{j} ⊂ E for all j (which we assume from now on). Observe that if E ⊂ ⋃ Q_{j} then f(E) ⊂ ⋃ f(Q_{j}), and that since

||f(x) − f(y)|| ≤ C||x − y||^{η}

we have diam(f(Q_{j}))^{α} ≤ C^{α} diam(Q_{j})^{αη}. It follows that for every covering {Q_{j}} of E with

diam(Q_{j}) < ϵ

we have the covering {f(Q_{j})} of f(E) with

diam(f(Q_{j})) < Cϵ^{η}

such that

∑ diam(f(Q_{j}))^{α} ≤ ∑ C^{α} diam(Q_{j})^{αη}.

Thus, for every element x of the set

{∑ diam(Q_{j})^{αη} : E ⊂ ⋃ Q_{j}, diam(Q_{j}) < ϵ}

we have an element y of the set

{∑ diam(f(Q′_{j}))^{α} : f(E) ⊂ ⋃ Q′_{j}, diam(Q′_{j}) < Cϵ^{η}}

with y ≤ C^{α}x (just let Q′_{j} = f(Q_{j})). Therefore, by the properties of the infimum, we have

H_{α,Cϵ^{η}}(f(E)) ≤ C^{α}H_{ηα,ϵ}(E).

Since this holds for all positive ϵ, we have (since C, η > 0)

H_{α}(f(E)) = lim_{ϵ→0} H_{α,ϵ}(f(E)) = lim_{ϵ→0} H_{α,Cϵ^{η}}(f(E)) ≤ C^{α} lim_{ϵ→0} H_{ηα,ϵ}(E) = C^{α}H_{ηα}(E),

so that

H_{α}(f(E)) ≤ C^{α}H_{ηα}(E).