Math 597. Homework 9 Bryden Cais

(1) Let μ be counting measure and ν Lebesgue measure on (say) the unit interval. Obviously we have ν ≪ μ since the only set E with μ(E) = 0 is the empty set E = ∅, which has Lebesgue measure 0. However, the unit interval is not σ-finite, so that we may not apply the Radon-Nikodym Theorem. Indeed, there is no measurable function f with ν(E) = ∫_{E} f dμ. To see this, let x be a point. Then we would have 0 = ν(x) = ∫_{x} f dμ = f(x). It follows that f is identically 0, but then we would have 1 = ν([0, 1]) = ∫_{[0,1]} f dμ = 0, a contradiction.

(2) Let ν, μ be σ-finite measures on the measure space (X, A) with ν ≪ μ and set λ = ν + μ. Then ν ≪ λ so we may set f = \frac{dν}{dλ}.

(a) Let U_{i} be a countable covering of X by measurable sets with ν(U_{i}) finite for each i. Let E = {x ∈ X : f(x) ≥ 1}. Then we have

ν(E ∩ U_{i}) = ∫_{E∩U_{i}} f(x) dλ ≥ ∫_{E∩U_{i}} dλ = λ(E ∩ U_{i}) = μ(E ∩ U_{i}) + ν(E ∩ U_{i}),

so that we have ν(E ∩ U_{i}) ≥ ν(E ∩ U_{i}) + μ(E ∩ U_{i}). Since ν(E ∩ U_{i}) is finite, it follows that we have μ(E ∩U_{i}) = 0 for each i and hence μ(E) = 0. Thus, f < 1 μ-almost everywhere. Similarly, for δ > 0 let F_{δ} = {x : f(x) < −δ}. Then ν(F_{δ}) = ∫_{F_{δ}} f dλ < −δλ(F_{δ}) < 0 if λ(F_{δ}) > 0. Thus, since ν is positive,we conclude that λ(F_{δ}) = 0 for all δ > 0 and consequently that μ(F_{δ}) = 0 for all δ > 0. It follows that μ(∪_{n>0}F_{1/n}) = 0 so that f ≥ 0 μ-almost everywhere.

(b) It follows from the definition of an integral as a supremum over all simple functions bounded above by the integrand on the set of integration that we have for any λ-measurable set E and any integrable function h on E

∫_{E} h dλ = ∫_{E} h dμ + ∫_{E} h dν.

Moreover, we have that

∫_{E} hf dλ = ∫_{E} h dν.

Combining these two relations gives

∫_{E} hf dμ + ∫_{E} hf dν = ∫_{E} h dν

from which we obtain

∫_{E} hf dμ = ∫_{E} h(1 − f) dν.

This gives f dμ = (1 − f) dν from which the proposed expression for \frac{dν}{dμ} follows.

(3) Let (X, A, μ) be a finite measure space, f a nonnegative μ-measurable function, B a subalgebra of

A and ν = μ|_{B}. Define

f_{n}(x) = min{f(x), n}

and observe that f_{n} → f pointwise as n → ∞ and that the sequence {f_{n}} is monotone. Let

λ_{n}(E) = ∫_{E} f_{n} dμ.

Then since X is finite and E ⊂ X and f_{n} is bounded above (and below) for each n we see that λ_{n} is a finite measure on (X, B). Moreover, if μ(E) = 0 then we have λ_{n}(E) = 0 so that λ_{n} ≪ μ for each n, and since ν is just a restriction of μ, we also have λ_{n} ≪ ν. Since ν is finite (and also λ_{n} is finite by construction) we may apply the Radon-Nikodym Theorem to obtain a sequence of functions g_{n} with

∫_{E} g_{n} dν = λ_{n}(E) = ∫_{E} f_{n} dμ

for each n. Moreover, since the sequence {f_{n}} is monotone, we have that the sequence {g_{n}} is monotone almost everywhere. We then define h_{n}(x) = max{g_{1}(x), g_{2}(x), …, g_{n}(x)} and observe that since {g_{n}} is monotone a.e. we have h_{n} = g_{n} a.e. and {h_{n}} monotone. The limit lim_{n→∞} h_{n} therefore exists and we call it h. From our remarks above, we now have

∫_{E} f dμ = ∫_{E} lim f_{n} dμ
= lim ∫_{E} f_{n} dμ

by the Monotone Convergence Theorem,

= lim ∫_{E} g_{n} dν
= lim ∫_{E} h_{n} dν
= ∫_{E} lim h_{n} dν

again by MCT,

= ∫_{E} h dν

for any E ∈ B, as required.

(4) (a) Let μ = λ − ν be a signed measure on a space X with λ, ν positive. As usual, we have the decomposition X = X^{+} ∪ X^{−} for μ. By definition, we have

μ^{+}(E) = μ(E ∩ X^{+}) = (λ − ν)(E ∩ X^{+}) = λ(E ∩ X^{+}) − ν(E ∩ X^{+}) ≤ λ(E ∩ X^{+})

since ν is positive. But since E ∩ X^{+} ⊂ E, we have λ(E ∩ X^{+}) ≤ λ(E) so that μ^{+} ≤ λ. Similarly, we have

−μ^{−}(E) = μ(E ∩ X^{−}) = (λ − ν)(E ∩ X^{−}) = λ(E ∩ X^{−}) − ν(E ∩ X^{−}) ≥ −ν(E ∩ X^{−})

since λ is positive. This gives −μ^{−}(E) ≥ −ν(E) so that ν ≥ μ^{−}.

(b) Observe that for any signed measures μ_{1}, μ_{2} we may write μ_{1} + μ_{2} = (μ_{1}^{+} + μ_{2}^{+}) − (μ_{1}^{−} + μ_{2}^{−}). Since μ_{1}^{+} +μ_{2}^{+} and μ_{1}^{−} +μ_{2}^{−} are positive, we may apply part (a) to obtain (μ_{1} +μ_{2})^{+} ≤ μ_{1}^{+} +μ_{2}^{+} and (μ_{1} + μ_{2})^{−} ≤ μ_{1}^{−} + μ_{2}^{−}. Adding these inequalities together gives

(μ_{1} + μ_{2})^{+} + (μ_{1} + μ_{2})^{−} ≤ (μ_{1}^{+} + μ_{1}^{−}) + (μ_{2}^{+} + μ_{2}^{−}).

Recalling that |μ| = μ^{−} + μ^{+} we obtain |μ_{1} + μ_{2}| ≤ |μ_{1}| + |μ_{2}| as required.