Integration Theory / Mathematical Analysis (5B1452/MA429)

Solutions to homework assignment # 1

1. Sets A ∩ B, A − B, and B − A are disjoint. Hence, since μ is additive,

μ(A) + μ(B) = (μ(A − B) + μ(A ∩ B)) + (μ(A ∩ B) + μ(B − A))
= μ(A ∩ B) + (μ(A − B) + μ(A ∩ B) + μ(B − A)) = μ(A ∩ B) + μ(A ∪ B) .

2. a. If set E is countable or finite, then we can cover it by one-point sets {a_{j}} and obtain 0 ≤ μ^{∗}(E) ≤ ∑_{j} λ{a_{j}} = 0. On the other hand, if E is uncountable, μ^{∗}(E) ≤ λ(X) = 1 and for any cover A_{j} one of the sets, say A_{p}, has to coincide with X (if all A_{j} are one-point or empty, then E is countable), hence ∑_{j} λ(A_{j}) ≥ λ(A_{p}) = λ(X) = 1.

Thus we conclude, that μ^{∗}(E) = 0 if E is countable and μ^{∗}(E) = 1 if E is uncountable. b. If set E is countable, then for countable A both A ∩ E and A − E are countable and

μ^{∗}(A) = 0 = 0 + 0 = μ^{∗}(A ∩ E) + μ^{∗}(A − E) .

For uncountable A set A ∩ E is countable while A − E is uncountable and

μ^{∗}(A) = 1 = 0 + 1 = μ^{∗}(A ∩ E) + μ^{∗}(A − E) .

We conclude that countable sets are μ^{∗} measurable, and hence their complements also are. On the other hand, if both sets E and E^{c} are uncountable, taking A = X we obtain

μ^{∗}(A) = μ^{∗}(X) = 1 ≠ 1 + 1 = μ^{∗}(E) + μ^{∗}(E^{c}) = μ^{∗}(A ∩ E) + μ^{∗}(A − E) ,

so E is not μ^{∗}-measurable.

Therefore A = {E : E or E^{c} is countable}, and μ is zero for countable sets and one for uncountable ones.

3. Take arbitrary set E, then for any A one has

μ^{∗}(A) = μ^{∗}(A ∩ E) + μ^{∗}(A − E) ,

so E is μ^{∗}-measurable. Thus all sets are μ^{∗}-measurable, and μ^{∗} is a measure.

4. Denote σ := s_{1} … s_{n} (the binary number). Numbers, where the string S occurs as the digits number (kn + 1), …, (kn + n), are exactly those in the intervals [σ/2^{kn+n}, (σ + 1)σ/2^{kn+n}). Simlarly to construction in class, one sees by induction in l that the set E_{l} of numbers, where the string S does not occur as the digits number (kn + 1), …, (kn + n) for 1 ≤ k ≤ l, can be covered by (2^{n} − 1)^{l} intervals of length 2^{−ln}. Therefore, the Lebesgue measure of E_{l} satisfies μ(E_{l}) ≤ (2^{n} − 1)^{l} · 2^{−ln} = [(2^{n} − 1)/2^{n}]^{l}. As l tends to infinity, the latter tends to zero, because 2^{n} − 1)/2^{n} < 1. Thus, intersection ∩_{l}E_{l} has measure zero. But it contains all the numbers which do not have S in their binary expansion, so latter is a nullset. 5.a. It is sufficient to show that μ^{∗} is a metric outer measure, i.e. that for any sets A and B, separated by an interval of positive length, μ_{f}^{∗}(A) + μ_{f}^{∗}(B) = μ_{f}^{∗}(A ∪ B), or, equivalently, “≤”, since the opposite inequality always holds. This definition was given in class, and it is weaker, than one in the book.

Take such sets A and B, then we can chose a point x so that A ⊂ (−∞, x) and B ⊂ (x, +∞) (or vice versa). Take any covering of A ∪ B by a collection of open intervals {I_{j}}. For a given I_{j} denote by I′_{j} and I′′_{j} the intervals I_{j} ∩ (−∞, x) and I_{j} ∩ (x, +∞) correspondingly. Clearly, λ(I_{j}) = λ(I′_{j}) + λ(I′′_{j}), and collections {I′_{j}} and {I′_{j}} cover A and B. Therefore

μ_{f}^{∗}(A) + μ_{f}^{∗}(B) ≤ ∑_{j} λ(I′_{j}) + ∑_{j} λ(I′′_{j})
= ∑_{j}(λ(I′_{j}) + λ(I′′_{j})) = ∑_{j} λ(I_{j}) ,

and taking infimum over all covers {I′_{j}} we arrive at desired inequality μ_{f}^{∗}(A) + μ_{f}^{∗}(B) ≤ μ_{f}^{∗}(A ∪ B). b. We will not assume the continuity of f. Take any cover of [a, b] by open intervals. By the Heine-Borel lemma the closed interval is compact, hence we can chose a finite subcover {I_{j}}_{j=1...n}. We can through away any interval non-intersecting [a, b], and among the remaining ones there will be two intervals I_{j} and I_{i} which intersect. Replacing them by their union I. If I = (x, y), we can assume that I_{i} = (x, x′) and I_{j} = (y′, y), with y′ < x′, and hence

λ(I) = f(y) − f(x) ≤ f(y) − f(y′) + f(x′) − f(x) = λ(I_{j}) + λ(I_{i}) .

Therefore joining two intervals only decreases ∑_{j} λ(I_{j}). Repeat this operation untill only one interval J = (c, d) remains. Clearly, c < a, b < d, thus (function f is increasing!)

lim_{x→b+} f(x) − lim_{x→a−} f(x) ≤ f(d) − f(c) = λ(J) ≤ ∑_{j} λ(I_{j}) .

Since we have started with an arbitrary cover, this reasoning shows, that lim_{x→b+} f(x) − lim_{x→a−} f(x) ≤ μ_{f}[a, b]. On the other hand, covering by an interval (a − ε, b + ε), and then tending ε to zero, shows that the reverse inequality also holds. Thus,

μ_{f}[a, b] = lim_{x→b+} f(x) − lim_{x→a−} f(x) .

Particularly, applying this to one-point closed inetrval [a, a] we get

μ_{f}{a} = lim_{x→a+} f(x) − lim_{x→a−} f(x) .

Since all Borel sets are Lebesgue-Stieltjes measurable,

μ_{f}(a, b) = μ_{f}[a, b] − μ_{f}{a} − μ_{f}{b}
= ( lim_{x→b+} f(x) − lim_{x→a−} f(x)) − ( lim_{x→a+} f(x) − lim_{x→a−} f(x)) − ( lim_{x→b+} f(x) − lim_{x→b−} f(x))
= lim_{x→b−} f(x) − lim_{x→a+} f(x) .

c. By the definition, λ(I) = 1 if 0 ∈ I and λ(I) = 0 if 0 ∉ I. Any set E, not containing 0 can be covered by intervals (−∞, 0) and (0, +∞) (which do not contain 0), hence μ_{f}^{∗}(E) = 0. If E contains 0, then any covering contains some interval I_{p}, containing 0, and hence μ_{f}^{∗}(E) ≥ 1. Covering E by the interval (−∞, +∞) we obtain the reverse inequality. So μ_{f}^{∗}(E) = 1 if 0 ∈ E and μ_{f}^{∗}(E) = 0 if 0 ∉ E. It is easy to check that all sets are measurable.