Necessity of hypotheses (a theorem fails without them)

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Give an example to show that the Monotone Convergence Theorem (3.11) can fail if the hypothesis that f1,f2,…f_{1}, f_{2}, \ldots are nonnegative functions is dropped.

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Give an example to show that the Monotone Convergence Theorem can fail if the hypothesis of an increasing sequence of functions is replaced by a hypothesis of a decreasing sequence of functions.

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Remark.
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This exercise shows that the Monotone Convergence Theorem should be called the Increasing Convergence Theorem. However, see Exercise 20.

Let f:Rn∖{0}→Rf : \mathbb {R}^{n} \setminus \left\{ 0\right\} \rightarrow \mathbb {R} be a function which is continuously differentiable and whose partial derivatives are uniformly bounded:

∣∂f∂xi(x1,…,xn)∣≤M \left|\frac{\partial f}{\partial x_i}(x_1, \ldots , x_n)\right| \leq M

for all (x1,…,xn)≠(0,…,0)(x_1, \ldots , x_n) \neq (0, \ldots , 0). Show that if n≥2n \geq 2, then ff can be extended to a continuous function defined on all of Rn\mathbb {R}^{n}. Show that this is false if n=1n = 1 by giving a counterexample.

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Let p>1p>1 and an↓0a_{n} \downarrow 0. Prove that the convergence of the series ∑anp−1n−1/p\sum a_{n}^{p-1} n^{-1 / p} implies the convergence of the series ∑anp\sum a_{n}^{p}. Is monotonicity essential?

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Prove that Proposition 6.3.8 is not true if μ(X)=∞\mu (X)=\infty.

Proposition 6.3.8: Let (X,M,μ)(X, \mathfrak {M}, \mu ) be a finite measure space and let {fn}n=1∞\left\{ f_{n}\right\}_{n=1}^{\infty }, ff be measurable functions on XX with values in [−∞,+∞][-\infty , +\infty ] that are finite a.e. If {fn}\left\{ f_{n}\right\} converges to a real-valued function ff a.e. on XX, then {fn}\left\{ f_{n}\right\} converges to ff in measure on XX.

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Prove that Egorov's Theorem need not hold if μ(X)=∞\mu (X)=\infty.

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Show that the Monotone Convergence Theorem may not hold for decreasing sequences of functions.

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Give an example that the Monotone Convergence Theorem does not hold without the assumption that each fnf_{n} is nonnegative.

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The purpose of this exercise is to show that in the absence of σ\sigma finiteness the product measure need not be unique. Let Δ={(x,y)∈[0,1]2:x=y}\Delta =\left\{ (x, y) \in [0,1]^{2}: x=y\right\}, let mm be the Lebesgue measure on B([0,1])\mathcal{B}([0,1]), let ν\nu be the counting measure on P([0,1])\mathbb {P}([0,1]), and let m×νm \times \nu be the Carathéodory extension of the measure ω\omega defined on measurable rectangles by ω(E×F)=m(E)ν(F)\omega (E \times F)=m(E) \nu (F). Define λ\lambda on B([0,1])⊗P([0,1])\mathcal{B}([0,1]) \otimes \mathbb {P}([0,1]) by λ(E)=sup⁡{∑y∈Amy(([0,1]×{y})∩E):A is a finite subset of [0,1]}\lambda (E)=\sup \left\{ \sum_{y \in A} m_{y}(([0,1] \times \left\{ y\right\} ) \cap E): A \text{ is a finite subset of } [0,1]\right\}, where mym_{y} is the Lebesgue measure on [0,1]×{y}[0,1] \times \left\{ y\right\}. Prove that:

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(a)

λ\lambda is a measure.

(b)

m×νm \times \nu and λ\lambda agree on measurable rectangles.

(c)

(m×ν)(Δ)≠λ(Δ)(m \times \nu )(\Delta ) \neq \lambda (\Delta ).

Let ff and gg be nonnegative and continuous at cc.

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(a)

Prove that D+(fg)(c)≤g(c)D+f(c)+f(c)D+g(c)D^{+}(f g)(c) \leq g(c) D^{+} f(c)+f(c) D^{+} g(c).

(b)

Give an example to show that the inequality in (a) may be strict.

(c)

Give an example to show that the inequality in (a) is not true without the continuity and nonnegativity assumptions.

Give an example to show that the conclusions of Exercise 9.1.21 need not hold if the continuity assumption on ff is dropped.

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Show that the Vitali Covering Lemma does not extend to the case in which the covering collection contains some intervals that are not necessarily nondegenerate.

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Show that the condition imposed on the set AA in Exercise 9.4.37 (a) is not necessary, i.e., that there is a set that does not satisfy it, yet F(x):=m([0,x]∩A)F(x):= m([0, x] \cap A) is an absolutely continuous, strictly increasing function on [0,1][0,1] such that F′=0F^{\prime }=0 on a set of positive measures.

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Give an example to show that Proposition 9.5.3 fails if ν\nu is a σ\sigma-finite measure.

Proposition 9.5.3: Let (X,M,μ)(X, \mathfrak {M}, \mu ) be a measure space and ν\nu a finite measure on (X,M)(X, \mathfrak {M}). Then ν\nu is absolutely continuous with respect to μ\mu if and only if for every ε>0\varepsilon > 0 there is δ>0\delta > 0 such that for every E∈ME \in \mathfrak {M}, μ(E)<δ\mu \left(E\right) < \delta implies ν(E)<ε\nu \left(E\right) < \varepsilon.

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Let XX be an uncountable set, and let M\mathfrak {M} consist of those sets AA such that either AA or ACA^{C} is countable. Define two measures μ,ν\mu , \nu on M\mathfrak {M} by μ(A)=ν(A)=0\mu (A)= \nu (A)=0 if AA is countable, and μ(A)=1,ν(A)=∞\mu (A)=1, \nu (A)=\infty if ACA^{C} is countable.

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(a)

Find a measurable function f:X→[0,+∞]f: X \rightarrow [0,+\infty ] such that ν(E)=∫Efdμ\nu (E)=\int_{E} f d \mu, for any E∈ME \in \mathfrak {M}.

(b)

Prove that there does not exist a measurable function g:X→[0,+∞]g: X \rightarrow [0,+\infty ] such that μ(E)=∫Egdν\mu (E)=\int_{E} g d \nu, for any E∈ME \in \mathfrak {M}.

Give an example that the Radon-Nikodym theorem fails when ν\nu is not σ\sigma-finite.

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Prove or disprove: The Lebesgue Decomposition Theorem remains true if measures μ\mu and ν\nu are not assumed to be σ\sigma-finite.

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Show that Theorem 2.36 and its Corollary become false (in R1\mathbb {R}^{1}, for example) if the word "compact" is replaced by "closed" or by "bounded".

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Find continuous functions fn:[0,1]→[0,∞)f_{n}:[0,1] \rightarrow [0, \infty ) such that fn(x)→0f_{n}(x) \rightarrow 0 for all x∈[0,1]x \in [0,1] as n→∞n \rightarrow \infty, ∫01fn(x)dx→0\int_{0}^{1} f_{n}(x) d x \rightarrow 0, but sup⁡nfn\sup_{n} f_{n} is not in L1L^{1}. (This shows that the conclusion of the dominated convergence theorem may hold even when part of its hypothesis is violated).

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Let Ω={ω1,ω2,ω3,ω4}\Omega =\left\{ \omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}\right\}, let A\mathcal{A} consist of ∅,{ω1,ω2}\varnothing ,\left\{ \omega_{1}, \omega_{2}\right\}, {ω1,ω3},{ω2,ω4},{ω3,ω4}\left\{ \omega_{1}, \omega_{3}\right\} ,\left\{ \omega_{2}, \omega_{4}\right\} ,\left\{ \omega_{3}, \omega_{4}\right\}, and Ω\Omega, and let μ:A→R\mu : \mathcal{A} \rightarrow \mathbb {R} be defined as follows: μ(Ω)=6,μ(∅)=0\mu (\Omega )=6, \mu (\varnothing )=0, and μ({ω1,ω2})=μ({ω1,ω3})=μ({ω2,ω4})=μ({ω3,ω4})=\mu \left(\left\{ \omega_{1}, \omega_{2}\right\} \right)=\mu \left(\left\{ \omega_{1}, \omega_{3}\right\} \right)=\mu \left(\left\{ \omega_{2}, \omega_{4}\right\} \right)=\mu \left(\left\{ \omega_{3}, \omega_{4}\right\} \right)= 3.

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(a)

A\mathcal{A} is neither a π\pi-system nor a semiring, and σ(A)=2Ω\sigma (\mathcal{A})=2^{\Omega }.

(b)

μ\mu is a measure.

(c)

Define two new distinct measures vv and ξ\xi on the σ\sigma-field 2Ω2^{\Omega } by the following: v({ω1})=v({ω4})=1,v({ω2})=v({ω3})=2,ξ({ω2})=ξ({ω3})=1v\left(\left\{ \omega_{1}\right\} \right)=v\left(\left\{ \omega_{4}\right\} \right)=1, v\left(\left\{ \omega_{2}\right\} \right)=v\left(\left\{ \omega_{3}\right\} \right)=2, \xi \left(\left\{ \omega_{2}\right\} \right)=\xi \left(\left\{ \omega_{3}\right\} \right)=1, and ξ({ω1})=ξ({ω4})=2\xi \left(\left\{ \omega_{1}\right\} \right)=\xi \left(\left\{ \omega_{4}\right\} \right)=2. Then vv and ξ\xi are distinct measure extensions of μ\mu from A\mathcal{A} to σ(A)=2Ω\sigma (\mathcal{A})=2^{\Omega }.

(d)

Let μ∗:2Ω→R\mu^{*}: 2^{\Omega } \rightarrow \mathbb {R} denote the outer measure induced by μ\mu. Then μ∗=μ\mu^{*}=\mu on A\mathcal{A} and μ∗\mu^{*} is a measure.

(e)

μ∗,v\mu^{*}, v, and ξ\xi are distinct measures.

Show that the result in Exercise 12 can fail without the hypothesis that VV is finite-dimensional.

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Remark.
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Exercise 12: suppose VV is finite-dimensional and S,T,U∈L(V)S, T, U \in \mathcal{L}(V) and STU=ISTU = I. Then TT is invertible and T−1=UST^{-1} = US.

Prove that the condition of monotonic nondecrease of the function ff in Problem 524 is essential. If it is dropped, the statement may fail. ⊛\circledast

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Find counterexamples to show that Theorem 1.25 is incorrect if we remove the requirements that uu must be a nonfinal vertex and vv must be a noninitial vertex.

Theorem 1.25: Let rr be a regular expression. Then, an ε\varepsilon-edge (u,v)(u, v) in G(r)G(r) which is a unique out-edge from a nonfinal vertex uu or a unique in-edge to a noninitial vertex vv can be shrunk into a single vertex, still preserving the property of Theorem 1.23. (If one of the endpoints of the ε\varepsilon-edge is the initial vertex or the final vertex, then so is the resulting vertex.)

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Prove that the new initial state ss and the final state ff in the construction of Example 2.22 are necessary. That is, find NFA's M1M_{1} and M2M_{2} such that the NFA's M1′M_{1}^{\prime } and M2′M_{2}^{\prime } of Figure 2.28 have the property L(M1′)≠L(M1)∗L\left(M_{1}^{\prime }\right) \neq L\left(M_{1}\right)^{*} and L(M2′)≠L(M2)∗L\left(M_{2}^{\prime }\right) \neq L\left(M_{2}\right)^{*}.

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Give a counterexample to show that the following construction fails to prove Theorem 1.49, the closure of the class of regular languages under the star operation. 1 Let N1=(Q1,Σ,δ1,q1,F1)N_{1}=\left(Q_{1}, \Sigma , \delta_{1}, q_{1}, F_{1}\right) recognize A1A_{1}. Construct N=(Q1,Σ,δ,q1,F)N=\left(Q_{1}, \Sigma , \delta , q_{1}, F\right) as follows. NN is supposed to recognize A1∗A_{1}^{*}.

Footnotes

  1. In other words, you must present a finite automaton, N1N_{1}, for which the constructed automaton NN does not recognize the star of N1N_{1} ’s language. ↩

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(a)

The states of NN are the states of N1N_{1}.

(b)

The start state of NN is the same as the start state of N1N_{1}.

(c)

F={q1}∪F1F=\left\{ q_{1}\right\} \cup F_{1}. The accept states FF are the old accept states plus its start state.

(d)

Define δ\delta so that for any q∈Q1q \in Q_{1} and any a∈Σεa \in \Sigma_{\varepsilon },

δ(q,a)={δ1(q,a)q∉F1 or a≠εδ1(q,a)∪{q1}q∈F1 and a=ε. \delta (q, a)= \begin{cases} \delta _{1}(q, a) & q \notin F_{1} \text{ or } a \neq \varepsilon \\ \delta _{1}(q, a) \cup \left\{ q_{1}\right\} & q \in F_{1} \text{ and } a=\varepsilon . \end{cases}

(Suggestion: Show this construction graphically, as in Figure 1.50.)

Give a counterexample to show that the following construction fails to prove that the class of context-free languages is closed under star. Let AA be a CFL that is generated by the CFG G=(V,Σ,R,S)G=(V, \Sigma , R, S). Add the new rule S→SSS \rightarrow S S and call the resulting grammar G′G^{\prime }. This grammar is supposed to generate A∗A^{*}.

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