Math 597. Homework 3 Bryden Cais

(1) Let Q consist of X, ∅, and all singletons of X. Define the set functions λ_{i} : Q → R by

λ_{1}(X) = ∞  λ_{1}(∅) = 0  λ_{1}(E) = 1

for all E ≠ X, ∅ in Q, and

λ_{2}(X) = 1  λ_{2}(E) = 0

for all E ≠ X in Q. Then we claim that for any set U ⊂ X,

μ_{λ_{1}}^{∗}(U) = \begin{cases}#U & if #U < ∞ \\ ∞ & otherwise \end{cases} ,

μ_{λ_{2}}^{∗}(U) = \begin{cases}0 & if U is countable \\ 1 & otherwise \end{cases} .

Indeed, if U is countable we have

μ_{λ_{1}}^{∗}(U) = inf {∑ λ_{1}(Q_{j}) : U ⊂ ⋃ Q_{j}} = ∑_{x∈U} λ_{1}(x) = ∑_{x∈U} 1,

which is clearly equal to #U if #U is finite and ∞ otherwise. On the other hand, if U is not countable, then the only way to cover U by a countable union of sets in Q is by X itself (possibly with additional terms, but at any rate, X must be part of the cover) so that μ_{λ_{1}}^{∗}(U) = ∞, as claimed.

Again, for countable U we have

μ_{λ_{2}}^{∗}(U) = inf {∑ λ_{2}(Q_{j}) : U ⊂ ⋃ Q_{j}} = ∑_{x∈U} λ_{2}(x) = ∑_{x∈U} 0 = 0,

while for uncountable U, we must have X as an element of our cover Q_{j} so that

μ_{λ_{2}}^{∗}(U) = 1

(since we can take the cover consisting only of X).

Let us now determine the σ-algebra of μ_{λ_{i}}^{∗} for i = 1, 2. We have already seen in class that for i = 1, every subset of X is measurable. Indeed, for any U ⊂ X we have two cases: if μ_{λ_{1}}^{∗}(A) < ∞ then certainly μ_{λ_{1}}^{∗}(A ∩ E), μ_{λ_{1}}^{∗}(A \ E) < ∞ by monotonicity, and

μ_{λ_{1}}^{∗}(A \ E) + μ_{λ_{1}}^{∗}(A ∩ E) = #(A \ E) + #(A ∩ E) = #A ≤ μ_{λ_{1}}^{∗}(A).

On the other hand, if μ_{λ_{1}}^{∗}(A) = ∞ then there is nothing to prove. Now we claim that the σ-algebra of μ_{λ_{2}}^{∗} measurable sets is generated (under countable unions and complemantations) by all countable subsets of X. For it is clear that every set that is obtainable in this way is either countable, or its complement in X is countable. (One must check a few trivial things: for example, obviously countable unions of countable sets are countable. The key is that (X \ E_{1}) ∪ (E_{2}) = X \ ((X \ E_{1}) ∩ E_{2}) which is the complement of a countable set if E_{1}, E_{2} are countable.) Observe that for any A ⊂ X,if E or X \ E is countable then

μ_{λ_{2}}^{∗}(A \ E) + μ_{λ_{2}}^{∗}(A ∩ E) = μ_{λ_{2}}^{∗}(A).

On the other hand, if E and X \ E are uncountable then for any A ⊂ X, since μ_{λ_{2}}^{∗}(A) ≤ 1 we have

μ_{λ_{2}}^{∗}(X \ E) + μ_{λ_{2}}^{∗}(X ∩ E) = 1 + 1 > μ_{λ_{2}}^{∗}(X)

so that E is not measurable. Another way to characterize this σ-algebra (which ic plain from the description above) is as the collection of all subsets of X that are either countable or have countable complement in X.

(2) It is clear that μ_{λ}^{∗}(U) = λ(U) when U ∈ Q, so that we need only determine μ_{λ}^{∗}(U) when U is a set consisting of one element. But the only way to cover a one element set of X is by a set in Q of measure at least 1. On the other hand, every element of X is contained in one of the two-element sets in Q so that any (nonempty) singleton has measure 1. We claim that no subsets of X other than X, ∅ are measurable. For let E ⊂ X consist of one element and let A ⊃ E be any two element set. Then μ_{λ}^{∗}(A ∩ E) + μ_{λ}^{∗}(A \ E) = 1 + 1 > μ_{λ}^{∗}(A) = 1. If E is a two element set, let A ≠ E be any two element set having nontrivial intersection with E. Then, again, μ_{λ}^{∗}(A ∩ E) + μ_{λ}^{∗}(A \ E) = 1 + 1 > μ_{λ}^{∗}(A) = 1. Clearly X, ∅ are measurable as for any A we have A ∩ X = A and A \ X = ∅ while A ∩ ∅ = ∅ and A \ ∅ = A.

(3) Let Q be the collection of squares in R^{2} together with the empty set and set λ(Q) = diam Q. Assume that μ_{λ}^{∗}(Q) = λ(Q). Define Q_{j} as in the problem set for 1 ≤ j ≤ 4.

(a) Since ⋃_{j=1}^{4} Q_{j} is covered by the square [0, 3] × [0, 3], we have

μ_{λ}^{∗} ( ⋃_{j=1}^{4} Q_{j}) ≤ diam([0, 3] × [0, 3]) = 3\sqrt{2}.

On the other hand, μ_{λ}^{∗}(Q_{j}) = \sqrt{2} for each j by assumption so that

μ_{λ}^{∗} ( ⋃_{j=1}^{4} Q_{j}) ≤ 3\sqrt{2} < 4\sqrt{2} = ∑_{j=1}^{4} μ_{λ}^{∗}(Q_{j}).

(b) Observe that μ_{λ}^{∗} is translation invariant. Indeed, if Q_{j} is any covering of E with ∑_{j} λ(Q_{j}) = α then x + Q_{j} is a covering of x + A with λ(x + Q_{j}) = λ(Q_{j}) and hence ∑_{j} λ(x + Q_{j}) = α. Conversely, we may replace x by −x to obtain a covering of E from a covering of x + E. This gives a bijection of sets

{∑_{j} λ(Q_{j}) : E ⊂ ⋃ Q_{j}} ↔ {∑_{j} λ(Q′_{j}) : x + E ⊂ ⋃ Q′_{j}}

from which it follows that μ_{λ}^{∗}(E) = μ_{λ}^{∗}(x + E). Now if Q_{1} were measurable, then Q_{2} would be also, since it is just a translation of Q_{1}. For suppose not, and let A be such that

μ_{λ}^{∗}(A \ Q_{2}) + μ_{λ}^{∗}(A ∩ Q_{2}) ≠ μ_{λ}^{∗}(A).

Then A′ = A − (2, 0) satisfies

μ_{λ}^{∗}(A \ Q_{2}) + μ_{λ}^{∗}(A ∩ Q_{2}) = μ_{λ}^{∗}(A′ \ Q_{1}) + μ_{λ}^{∗}(A′ ∩ Q_{1})

and μ_{λ}^{∗}(A) = μ_{λ}^{∗}(A′), so that Q_{1} is not measurable. Thus, if Q_{1} is measurable, then Q_{i} is measurable for 1 ≤ i ≤ 4 and also ⋃_{i=1}^{4} Q_{i} is measurable. But part (a) contradicts this, as countable additivity of measurable sets holds for μ_{λ}^{∗}. It follows that Q_{1} is not measurable.