Chapter 3

Densities and derivatives

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Problem 3.1

Let μ\mu be the measure defined for all subsets of XX by μA=+∞\mu A = +\infty if A≠∅A \neq \emptyset, and μ(∅)=0\mu (\emptyset ) = 0. Show that μf=μ(2f)\mu f = \mu (2 f) for all nonnegative functions ff.

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Problem 3.2

Let λ\lambda denote the measure on R\mathbb {R} with λA\lambda A equal to the number of points in AA (possibly infinite), and let μA=∞\mu A = \infty for all nonempty AA. Show that λ\lambda has no density with respect to μ\mu, even though both measures have the same negligible sets.

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Problem 3.3

Suppose Δ1\Delta_{1} and Δ2\Delta_{2} are functions in M+(X,A)\mathcal{M}^{+}(\mathcal{X}, \mathcal{A}) for which μ(fΔ1)=μ(fΔ2)\mu \left(f \Delta_{1}\right) = \mu \left(f \Delta_{2}\right) for all ff in M+(X,A)\mathcal{M}^{+}(X, \mathcal{A}), for some measure μ\mu.

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(i)

Show that Δ1=Δ2\Delta_{1} = \Delta_{2} a.e. [ μ\mu ] if μΔ1<∞\mu \Delta_{1} < \infty. Hint: Consider the equality μΔ1{Δ1>Δ2}=μΔ2{Δ1>Δ2}\mu \Delta_{1}\left\{ \Delta_{1} > \Delta_{2}\right\} = \mu \Delta_{2}\left\{ \Delta_{1} > \Delta_{2}\right\}.

(ii)

Show that the assumption of finiteness of μΔ1\mu \Delta_{1} is not required for the conclusion Δ1=Δ2\Delta_{1} = \Delta_{2} a.e. [ μ\mu ] if μ\mu is a sigma-finite measure. Hint: For each positive rational number rr show that μ(Δ1−Δ2){Δ1>r≥Δ2}A=0\mu \left(\Delta_{1}-\Delta_{2}\right)\left\{ \Delta_{1} > r \geq \Delta_{2}\right\} A = 0 for each set with μA<∞\mu A < \infty.

Problem 3.4

Let PP and QQ be probability measures on (Ω,F\Omega , \mathcal{F}), and let E\mathcal{E} be a generating class for F\mathcal{F} that is stable under finite intersections and contains Ω\Omega. Suppose there exists a pp in L1(Q)\mathcal{L}^{1}(Q) such that PE=Q(pE)P E = Q(p E) for every EE in E\mathcal{E}. Show that PP has density pp with respect to QQ.

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Problem 3.5

Does Example ⟨3⟩\langle 3 \rangle have an analog for Lp\mathcal{L}^{p} convergence, for some p>1p > 1 : if vv is a finite measure dominated by a measure λ\lambda, and if {fn}\left\{ f_{n}\right\} is a sequence of measurable functions for which λ∣fn∣p→0\lambda \left|f_{n}\right|^{p} \rightarrow 0, does it follow that ν∣fn∣p→0\nu \left|f_{n}\right|^{p} \rightarrow 0 ?

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Problem 3.6

Let ν\nu and μ\mu be finite measures on the sigma-field σ(E)\sigma (\mathcal{E}) generated by a countable field E\mathcal{E}. Suppose that for each ϵ>0\epsilon > 0 there exists a δ>0\delta > 0 such that νE<ϵ\nu E < \epsilon for each EE in E\mathcal{E} with μE<δ\mu E < \delta. Show that vv is absolutely continuous with respect to μ\mu, as measures on σ(E)\sigma (\mathcal{E}).

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Problem 3.7

Let vv and μ\mu be finite measures on (X,A)(X, \mathcal{A}). Suppose vv has Lebesgue decomposition (Δ,N)(\Delta , \mathcal{N}) with respect to μ\mu. Find the Lebesgue decomposition of μ\mu with respect to vv.

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Problem 3.8

Show that a measure μ\mu is sigma-finite if and only if there exists a strictly positive, measurable function Ψ\Psi for which μΨ<∞\mu \Psi < \infty.

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Problem 3.9

Suppose (N1,Δ1)\left(\mathcal{N}_{1}, \Delta_{1}\right) and (N2,Δ2)\left(\mathcal{N}_{2}, \Delta_{2}\right) are both Lebesgue decompositions for a finite measure vv with respect to a finite measure μ\mu. Prove N1=N2\mathcal{N}_{1} = \mathcal{N}_{2} a.e. [ν+μ][\nu +\mu ] and Δ1N1c=Δ2N2c\Delta_{1} \mathcal{N}_{1}^{c} = \Delta_{2} \mathcal{N}_{2}^{c} a.e. [ν+μ][\nu +\mu ] by the following steps.

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(i)

Show that N1=N2\mathcal{N}_{1} = \mathcal{N}_{2} a.e. [ ν\nu ] by means of the equality v(N2N1c)=v(N2N1cN1)+μ(N2N1cΔ1)=0v\left(\mathcal{N}_{2} \mathcal{N}_{1}^{c}\right) = v\left(\mathcal{N}_{2} \mathcal{N}_{1}^{c} \mathcal{N}_{1}\right)+ \mu \left(\mathcal{N}_{2} \mathcal{N}_{1}^{c} \Delta_{1}\right) = 0, and a companion equality obtained by interchanging the roles of the two decompositions. Using the trivial fact that N1=N2\mathcal{N}_{1} = \mathcal{N}_{2} a.e. [ μ\mu ] (because both sets are μ\mu-negligible), deduce that N1=N2\mathcal{N}_{1} = \mathcal{N}_{2} a.e. [v+μ][v+\mu ].

(ii)

Use the result from part (i) to show that

μ(fΔ1N1c)=ν(fN1c)=ν(fN2c)=μ(fΔ2N2c) for all f∈M+ \mu \left(f \Delta _{1} \mathcal{N}_{1}^{c}\right) = \nu \left(f \mathcal{N}_{1}^{c}\right) = \nu \left(f \mathcal{N}_{2}^{c}\right) = \mu \left(f \Delta _{2} \mathcal{N}_{2}^{c}\right) \quad \text{ for all } f \in \mathcal{M}^{+}

Deduce that Δ1N1c=Δ2N2c\Delta_{1} \mathcal{N}_{1}^{c} = \Delta_{2} \mathcal{N}_{2}^{c} a.e. [ μ\mu ] by arguing as in Problem [3].

(iii)

Use part (ii) to show that

ν{Δ1N1c>Δ2N2c}=ν{Δ1N1c>Δ2N2c}N1+μ{Δ1N1c>Δ2N2c}Δ1N1c=0 \nu \left\{ \Delta _{1} \mathcal{N}_{1}^{c} > \Delta _{2} \mathcal{N}_{2}^{c}\right\} = \nu \left\{ \Delta _{1} \mathcal{N}_{1}^{c} > \Delta _{2} \mathcal{N}_{2}^{c}\right\} \mathcal{N}_{1}+\mu \left\{ \Delta _{1} \mathcal{N}_{1}^{c} > \Delta _{2} \mathcal{N}_{2}^{c}\right\} \Delta _{1} \mathcal{N}_{1}^{c} = 0

Hint: What value does Δ1N1c\Delta_{1} \mathcal{N}_{1}^{c} take on N1\mathcal{N}_{1} ? Argue similarly for the companion equality. Deduce that Δ1N1c=Δ2N2c\Delta_{1} \mathcal{N}_{1}^{c} = \Delta_{2} \mathcal{N}_{2}^{c} a.e. [ν+μ][\nu +\mu ].

(iv)

Extend the argument to prove the analogous uniqueness assertion for the Lebesgue decomposition when vv and μ\mu are sigma-finite measures. Hint: Decompose XX into countably many sets {Xi:i∈N}\left\{ X_{i}: i \in \mathbb {N}\right\} with (ν+μ)Xi<∞(\nu +\mu ) X_{i} < \infty for each ii.

Problem 3.10

Suppose ν\nu and μ\mu are both sigma-finite measures on (X,A)(X, \mathcal{A}).

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(i)

Show that v+μv+\mu is also sigma-finite. Deduce via Problem [8] that there exists a strictly positive, measurable function Φ0\Phi_{0} for which νΦ0<∞\nu \Phi_{0} < \infty and μΦ0<∞\mu \Phi_{0} < \infty.

(ii)

Define ν0f=ν(fΦ0)\nu_{0} f = \nu \left(f \Phi_{0}\right) and μ0f=μ(fΦ0)\mu_{0} f = \mu \left(f \Phi_{0}\right) for each ff in M+\mathcal{M}^{+}. Show that ν0\nu_{0} and μ0\mu_{0} are both finite measures.

(iii)

From the Lebesgue decomposition ν0f=ν0(fN)+μ0(fΔ)\nu_{0} f = \nu_{0}(f \mathcal{N})+\mu_{0}(f \Delta ) for all f∈M+f \in \mathcal{M}^{+}, where μ0N=0\mu_{0} \mathcal{N} = 0 and Δ∈M+\Delta \in \mathcal{M}^{+}, derive the corresponding Lebesgue decomposition for λ\lambda with respect to μ\mu.

(iv)

From the uniqueness result of Problem [9], applied to λ0\lambda_{0} and μ0\mu_{0}, deduce that the Lebesgue decomposition for λ\lambda is unique up to the appropriate almost sure equivalences.

Problem 3.11

Let ν\nu be a finite measure for which ν=ν1+λ1=ν2+λ2\nu = \nu_{1}+\lambda_{1} = \nu_{2}+\lambda_{2}, where each νi\nu_{i} is dominated by a fixed sigma-finite measure μ\mu and each λi\lambda_{i} is singular with respect to μ\mu. Show that ν1=ν2\nu_{1} = \nu_{2} and λ1=λ2\lambda_{1} = \lambda_{2}.

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Problem 3.12

Let μ1\mu_{1} and μ2\mu_{2} be finite measures with densities m1m_{1} and m2m_{2} with respect to a sigma-finite measure λ\lambda. Define (μ1−μ2)+\left(\mu_{1}-\mu_{2}\right)^{+} as the measure with density (m1−m2)+\left(m_{1}-m_{2}\right)^{+} with respect to λ\lambda.

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(i)

Show that (m1−m2)+B=sup⁡π∑A∈π(m1−m2)AB\left(m_{1}-m_{2}\right)^{+} B = \sup_{\pi } \sum_{A \in \pi }\left(m_{1}-m_{2}\right) A B, the supremum running over all finite partitions of XX.

(ii)

Show that ∣μ1−μ2∣=(μ1−μ2)++(μ2−μ1)+\left|\mu_{1}-\mu_{2}\right| = \left(\mu_{1}-\mu_{2}\right)^{+}+\left(\mu_{2}-\mu_{1}\right)^{+}.

Problem 3.13

Let PP and QQ be probability measures with densities pp and qq with respect to a sigma-finite measure λ\lambda. For fixed α≥1\alpha \geq 1, show that Δα(P,Q):=λ∣p1/α−q1/α∣α\Delta_{\alpha }(P, Q): = \lambda \left|p^{1 / \alpha }-q^{1 / \alpha }\right|^{\alpha } does not depend on the choice of dominating measure.

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Problem 3.14

Adapt the argument from the previous Problem to show that the relative entropy D(P∥Q)D(P \| Q) does not depend on the choice of dominating measure.

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Problem 3.15

Let PP be the standard Cauchy distribution on the real line, and let QQ be the standard normal distribution. Show that D(P∥Q)=∞D(P \| Q) = \infty, even though PP and QQ are mutually absolutely continuous.

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Problem 3.16

Let PP and QQ be probability measures defined on the same sigma-field F\mathcal{F}. A randomized test is a measurable function ff with 0≤f≤10 \leq f \leq 1. (For observation ω\omega, the value f(ω)f(\omega ) gives the probability of rejecting the "hypothesis" PP.) Find the test that minimizes Pf+Q(1−f)P f+Q(1-f).

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Problem 3.17

Let PP and QQ be finite measures defined on the same sigma-field F\mathcal{F}, with densities pp and qq with respect to a measure μ\mu. Suppose X0X_{0} is a measurable set with the property that there exists a nonnegative constant KK such that q≥Kpq \geq K p on X0X_{0} and q≤Kpq \leq K p on X0cX_{0}^{c}. For each F\mathcal{F}-measurable function with 0≤f≤10 \leq f \leq 1 and Pf≤PX0P f \leq P X_{0}, prove that Qf≤QX0Q f \leq Q X_{0}.

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Problem 3.18

Let μ=μ1−μ2\mu = \mu_{1}-\mu_{2} and μ′=μ1′−μ2′\mu^{\prime } = \mu_{1}^{\prime }-\mu_{2}^{\prime } be signed measures, in the sense of Section 3. Show that v:=(μ1+μ2′)∧(μ1′+μ2)−(μ2+μ2′)v: = \left(\mu_{1}+\mu_{2}^{\prime }\right) \wedge \left(\mu_{1}^{\prime }+\mu_{2}\right)-\left(\mu_{2}+\mu_{2}^{\prime }\right) is the largest signed measure for which v≤μv \leq \mu and v≤μ′v \leq \mu^{\prime }.

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Problem 3.19

Let f(x)=(1+x)log⁡(1+x)−xf(x) = (1+x) \log (1+x)-x, for x≥−1x \geq -1. Use the representations

f(x)=x∫{0≤s≤1}f′(xs)ds=x2∬{0≤t≤s≤1}f′′(xt)dtds f(x) = x \int \left\{ 0 \leq s \leq 1\right\} f^{\prime }(x s) d s = x^{2} \iint \left\{ 0 \leq t \leq s \leq 1\right\} f^{\prime \prime }(x t) d t d s

to show that f(x)≥1/2x2(1+x/3)−1f(x) \geq 1 / 2 x^{2}(1+x / 3)^{-1}.

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Problem 3.20

Show that the function x−1cos⁡(x−2)x^{-1} \cos \left(x^{-2}\right) is not integrable on [0,1].

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Problem 3.21

Prove Theorem ⟨19⟩\langle 19 \rangle by establishing the following assertions.

Theorem 19: Let HH be a continuous function defined on an interval [a,b][a, b], with a (real-valued) right-hand derivative h(x):=lim⁡δ↓0(H(x+δ)−H(x))/δh(x): = \lim_{\delta \downarrow 0}\left(H(x+\delta )-H(x)\right) / \delta existing at each point of [a,b)[a, b). If hh is Lebesgue integrable then H(x)=H(a)+∫axh(t)dtH(x) = H(a)+\int_{a}^{x} h(t) d t for each xx in [a,b][a, b], and hence HH is an absolutely continuous function on [a,b][a, b].

With no loss of generality, assume a=0a = 0 and b=1b = 1 and H(0)=0H(0) = 0. Write mm for Lebesgue measure on B[0,1]\mathcal{B}[0,1]. Fix ϵ>0\epsilon > 0. Define An:={nϵ≤h<(n+1)ϵ}A_{n}: = \left\{ n \epsilon \leq h < (n+1) \epsilon \right\} for each integer nn, both positive and negative. Note that

h+(x)=∑n≥0h(x){x∈An} and h−(x)=−∑n<0h(x){x∈An} h^{+}(x) = \sum _{n \geq 0} h(x)\left\{ x \in A_{n}\right\} \quad \text{ and } \quad h^{-}(x) = -\sum _{n < 0} h(x)\left\{ x \in A_{n}\right\}
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(i)

There exist open sets GnG_{n} and compact sets KnK_{n} with Kn⊆An⊆GnK_{n} \subseteq A_{n} \subseteq G_{n} and m(Gn\Kn)<ϵn\mathfrak {m}\left(G_{n} \backslash K_{n}\right) < \epsilon_{n}, with ϵn\epsilon_{n} so small that the functions

f(x):=∑n≥0(n+1)ϵ{x∈Gn} and g(x):=∑n<0∣n+1∣ϵ{x∈Kn} f(x): = \sum _{n \geq 0}(n+1) \epsilon \left\{ x \in G_{n}\right\} \quad \text{ and } \quad g(x): = \sum _{n < 0}\left|n+1\right| \epsilon \left\{ x \in K_{n}\right\}

satisfy the inequalities f≥h+f \geq h^{+} and g≤h−g \leq h^{-} and m(f−h+)+m(h−−g)<ϵ\mathfrak {m}\left(f-h^{+}\right)+\mathfrak {m}\left(h^{-}-g\right) < \epsilon.

(ii)

The function hϵ:=ϵ+f−∑n=−N−1∣n+1∣ϵ{x∈Kn}h_{\epsilon }: = \epsilon +f-\sum_{n = -N}^{-1}\left|n+1\right| \epsilon \left\{ x \in K_{n}\right\}, for a large enough positive integer NN, has the properties hϵ≥h+ϵh_{\epsilon } \geq h+\epsilon and m(hϵ−h)<3ϵ\mathfrak {m}\left(h_{\epsilon }-h\right) < 3 \epsilon.

(iii)

The function hϵh_{\epsilon } is lower semicontinuous, that is, for each real number rr the set {hϵ>r}\left\{ h_{\epsilon } > r\right\} is open.

(iv)

For 0≤x≤10 \leq x \leq 1, define a continuous function G(x):=∫0xhϵ(t)dt−H(x)G(x): = \int_{0}^{x} h_{\epsilon }(t) d t-H(x). It achieves its maximum at some point yy in [0,1][0,1]. We cannot have y<1y < 1, for otherwise there would be a small δ>0\delta > 0 for which hϵ(t)>h(y)+ϵh_{\epsilon }(t) > h(y)+\epsilon for y≤t≤y+δ<1y \leq t \leq y+\delta < 1, and (H(y+δ)−H(y))/δ<h(y)+ϵ/2(H(y+\delta )-H(y)) / \delta < h(y)+\epsilon / 2, from which we could conclude that (G(y+δ)−G(y))/δ≥h(y)+ϵ−(h(y)+ϵ/2)=ϵ/2(G(y+\delta )-G(y)) / \delta \geq h(y)+\epsilon -(h(y)+\epsilon / 2) = \epsilon / 2, a contradiction. Thus G(1)>G(0)=0G(1) > G(0) = 0. That is, H(1)<∫01hϵ(t)dt≤∫01h(t)dt+4ϵH(1) < \int_{0}^{1} h_{\epsilon }(t) d t \leq \int_{0}^{1} h(t) d t+4 \epsilon, for each ϵ>0\epsilon > 0.

(v)

Similarly H(x)≤∫0xh(t)dtH(x) \leq \int_{0}^{x} h(t) d t for 0≤x≤10 \leq x \leq 1.

(vi)

An analogous argument applied to −H-H gives the reverse inequality, letting us conclude that H(x)=∫0xh(t)dtH(x) = \int_{0}^{x} h(t) d t for 0≤x≤10 \leq x \leq 1. The function HH is absolutely continuous on [0,1][0,1].

Problem 3.22

Let HH be a continuous real function defined on [0,1). Suppose the right-hand derivative hh exists and is finite everywhere. If hh is an increasing function, show that HH is convex.

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Problem 3.23

Let hh be integrable with respect to Lebesgue measure m\mathfrak {m} on Rd\mathbb {R}^{d}. Show that

lim⁡r→01mB(z,r)λx({∣x−z∣<r}h(x))=h(z) a.e. [m]. \lim _{r \rightarrow 0} \frac{1}{\mathfrak {m} B(z, r)} \lambda ^{x}(\left\{ \left|x-z\right| < r\right\} h(x)) = h(z) \quad \text{ a.e. }[\mathrm{m}] .

Generalize by replacing open balls by decreasing sequences of closed sets Fn↓{z}F_{n} \downarrow \left\{ z\right\} that are regular in the sense of Section 5.

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