Math 597. Homework 4 Bryden Cais

(1) (a) For any ϵ > 0 there exists a dyadic cover Q_{j} of E such that μ(E) + ϵ ≥ ∑ μ(Q_{j}). Moreover, we

have

μ(E) ≤ ∑ μ(Q_{j} ∩ E)

by countable subadditivity of μ so that by hypothesis (since the Q_{j} are dyadic) we have

μ(E) ≤ ∑ μ(Q_{j} ∩ E) ≤ c ∑ μ(Q_{j}) ≤ c(μ(E) + ϵ).

It follows that we have, for any ϵ > 0 μ(E) ≤ cμ(E) + ϵ so that

μ(E) (\frac{1 − c}{c}) ≤ ϵ.

If c = 0 we are done since then μ(E) ≤ ϵ for any ϵ > 0. Otherwise, 0 < c < 1 so that, (1 − c)/c is some fixed nonzero constant. Since ϵ > 0 is arbitrary, we must have μ(E) = 0.

(b) Let F = [0, 1] \ E. Then since E is measurable, do is F. Moreover, for any dyadic interval I we have

μ(F ∩ I) = μ(([0, 1] ∩ I) \ (E ∩ I)) = μ(I) − μ(E ∩ I) ≤ |I| − c̃|I| = (1 − c̃)|I|.

Now obviously c̃ ≤ 1 and c̃ > 0 by assumption so that 0 ≤ 1 − c̃ < 1. We now apply part (a) and deduce that μ(F) = 0. It follows that μ(E) = μ([0, 1]) = μ(F) = 1.

(2) Define I_{n}^{k} = [\frac{k}{2^{n}} , \frac{k+1}{2^{n}}] and for any 0 < α < 1 and any closed interval I, let φ_{α}(I) be the cantor set constructed on I with constant 1 − α, so that φ_{α}(I) has measure α|I|. We define E_{j} and α_{j} recursively as follows: let α_{0} = 1/2 and E_{0} = φ_{1/2}([0, 1]), so that E_{0} has measure 1/4. Supposing that α_{j} is defined for 0 ≤ j < n define E_{j} for 0 ≤ j < n as

E_{j} = ⋃_{k=0}^{2^{j}−1} φ_{α_{j}}(I_{j}^{k})

and define

β_{n}^{l} = μ (I_{n}^{l} ∩ ⋃_{0≤j<n} E_{j}) .

We claim that β_{n}^{l} < |I_{n}^{l}|. Indeed, observe that E_{j} is closed in [0, 1] and contains no intervals; this follows directly from the cantor process. It follows that ⋃_{0≤j<n} E_{j} is closed and contains no intervals since a finite union of closed sets that contain no intervals is again closed and contains no intervals. Since I_{n}^{l} is closed in [0, 1], the intersection I_{n}^{l} ∩ ⋃_{0≤j<n} E_{j} is closed in I_{n}^{l} in the subspace topology. Therefore, I_{n}^{l} \ ⋃_{0≤j<n} E_{j} is open in I_{n}^{l} in the subspace topology and nonempty (since ⋃_{0≤j<n} E_{j} contains no intervals). Since any nonempty open set contains an interval, it follows that I_{n}^{l} \ ⋃_{0≤j<n} E_{j} contains an interval and hence that β_{n}^{l} < |I_{n}^{l}|. We therefore set

ϵ_{n}^{l} = |I_{n}^{l}| − β_{n}^{l}.

We have just shown that ϵ_{n}^{l} > 0. We now define

α_{n} = \frac{1}{2^{n}} min_{\begin{matrix}0≤q≤2^{n}−1 \\ 0≤r≤n \end{matrix}}\{ϵ_{r}^{q}, \frac{1}{2^{n+2}} \}.

Observe that α_{n} > 0 since the minimum is taken over a finite set.

We have thus defined the sets E_{j} and the constants α_{j} recursively. Let E = ⋃_{j=0}^{∞} E_{j}. We claim that μ(E ∩ I_{n}^{l}) < |I_{n}^{l}|. Indeed, we have

μ(E ∩ I_{n}^{l}) ≤ μ (I_{n}^{l} ∩ ⋃_{0≤j<n} E_{j}) + ∑_{j≥n} μ(E_{j} ∩ I_{n}^{l}) = β_{n}^{l} + ∑_{j≥n} μ(E_{j} ∩ I_{n}^{l}). (1)

by countable subadditivity. We now estimate μ(E_{j} ∩ I_{n}^{l}) for j ≥ n. Since E_{j} is a union of sets of the form φ_{α_{j}}(I_{j}^{m}), each set having measure α_{j}|I_{j}^{m}|, and j ≥ n, we see that I_{n}^{l} ∩ E_{j} is a union of 2^{j−n} sets of the form φ_{α_{j}}(I_{j}^{m}), each set having measure α_{j}|I_{j}^{m}| = \frac{a_{j}}{2^{j}}. Therefore,

μ(E_{j} ∩ I_{n}^{l}) = 2^{−n}α_{j}.

It follows that

(2)  ∑_{j≥n} μ(E_{j} ∩ I_{n}^{l}) = 2^{−n} ∑_{j≥n} α_{j} ≤ 2^{−n} ∑_{j≥n} 2^{−j}ϵ_{n}^{l} = \frac{1}{2^{2n−1}} ϵ_{n}^{l}

since

α_{j} = \frac{1}{2^{j}} min_{\begin{matrix}0≤q≤2^{j}−1 \\ 0≤r≤j \end{matrix}}\{ϵ_{r}^{q}, \frac{1}{2^{j+2}} \} ≤ \frac{1}{2^{j}} ϵ_{n}^{l}

for j ≥ n since, afterall, ϵ_{n}^{l} is one of the terms in the minimum. Combining (2) with (1), we have

μ(E ∩ I_{n}^{l}) ≤ β_{n}^{l} + \frac{ϵ_{n}^{l}}{2^{2n−1}} < |I_{n}^{l}|

by the definition of ϵ_{n}^{l} and β_{n}^{l}. Since any dyadic interval I is a countable disjoint union of intervals of the form I_{n}^{l}, it follows that for any dyadic interval I we have μ(E ∩ I) < |I|.

Observe that E ∩ I_{n}^{l} contains the set φ_{α_{n}}(I_{n}^{l}) of measure α_{n}|I_{n}^{l}| > 0 so by monotonicity of μ, we have 0 < μ(E ∩ I_{n}^{l}). Again, 0 < μ(E ∩ I) for any dyadic interval I follows as above. It is also obvious from this that 0 < μ(E). On the other hand, we have

μ(E) ≤ ∑_{0≤j} μ(E_{j}) ≤ ∑_{\begin{matrix}0≤j \\ 0≤l≤2^{j}−1 \end{matrix}} μ(φ_{α_{j}}(I_{j}^{l})) ≤ ∑_{0≤j} α_{j} ≤ ∑_{0≤j} \frac{1}{2^{2j+2}} < 1,

so we are done.

(3) Let f be the cantor ternary function extended by f(x) = 0 for x < 0 and f(x) = 1 for x > 1. Then f is right-continuous as we saw in class and C is measurable since it is Borel (again from class).

(a) Let ϵ > 0 be arbitrary. Since f is continous there exists a δ > 0 such that |f(x) − f(y)| < ϵ whenever |x − y| < δ. Now we may cover x by the set I = (x − δ/2, x + δ/2), and we have μ_{f}(I) < ϵ. Since ϵ was arbitrary, it follows that μ_{f}({x}) = 0.

(b) From lecture, since f is right continuous we have μ_{f}((0, 1]) = f(1) − f(0) = 1. Since μ_{f}(0) = 0 by part (a), it follows that μ_{f}([0, 1]) = 1.

(c) We may cover [0, 1]\C by sets of the form (α_{j}, β_{j}]. Here (a_{j}, b_{j}) are all the intervals thrown away in the construction of C. From the definition of f and right continuity, we have f(b_{j}) = f(a_{j}) for all j and therefore μ_{f}((a_{j}, b_{j}]) = 0. It follows that μ_{f}([0, 1] \ C) = 0.

(d) Since C is μ_{f} measurable (being Borel), we have

μ_{f}([0, 1]) = μ_{f}([0, 1] ∩ C) + μ_{f}([0, 1] \ C) = μ_{f}(C)

by part (c). Then by part (a), it follows that μ_{f}(C) = 1.