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Let \(g\) be bounded and continuous on \(\mathbb R\) . Evaluate \(\displaystyle\lim_{n\to\infty}\int_{-\infty}^\infty g(x)\,\frac{n}{\pi(1+n^2x^2)}\,dx\) and justify every step.
Solution
Substitute \(u=nx\) . By the dilation rule of Exercise 4.13, applied to the positive and negative parts of the integrand,
\[ \int_{-\infty}^\infty g(x)\,\frac{n}{\pi(1+n^2x^2)}\,dx=\int_{-\infty}^\infty g\bigl(\tfrac un\bigr)\,\frac{1}{\pi(1+u^2)}\,du . \]For each fixed \(u\) , \(g(u/n)\to g(0)\) as \(n\to\infty\) because \(g\) is continuous at \(0\) . With \(B=\sup\lvert g\rvert<\infty\) , the integrand is bounded in absolute value by \(\frac{B}{\pi(1+u^2)}\) , which does not depend on \(n\) and is integrable, with integral \(B\) . By dominated convergence the limit is
\[ g(0)\int_{-\infty}^\infty\frac{du}{\pi(1+u^2)}=g(0). \]Before the substitution no dominator exists: the kernels \(\frac{n}{\pi(1+n^2x^2)}\) are the sequence \(v_n\) of Example 5.6, a spike of total mass \(1\) at the origin. The change of variable transfers the dependence on \(n\) from the spike to the slowly varying factor \(g\) , where domination is easy. For \(g(x)=\cos x\) the integral equals \(e^{-1/n}\) , which tends to \(1=g(0)\) .
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Let \(f_n\) be measurable on \([0,1]\) with \(\lVert f_n\rVert_2\le1\) for all \(n\) and \(f_n\to f\) almost everywhere. Prove that \(f_n\to f\) in \(L^1([0,1])\) , and show by example that \(\lVert f_n\rVert_1\le1\) would not suffice.
Solution
The limit is in \(L^2\) . By Fatou’s lemma applied to \(f_n^2\) , \(\int f^2\,d\lambda\le\liminf_n\int f_n^2\,d\lambda\le1\) . So \(\lVert f\rVert_2\le1\) and \(\lVert f_n-f\rVert_2\le2\) by Minkowski’s inequality.
Splitting the domain. Let \(\varepsilon>0\) . By Egorov’s theorem there is a measurable \(E\subseteq[0,1]\) with \(\lambda([0,1]\setminus E)<\varepsilon^2\) on which \(f_n\to f\) uniformly. On \(E\) ,
\[ \int_E\lvert f_n-f\rvert\,d\lambda\le\sup_E\lvert f_n-f\rvert\longrightarrow0 . \]On the complement use the Cauchy–Schwarz inequality with the indicator of \(E^c\) :
\[ \int_{E^c}\lvert f_n-f\rvert\,d\lambda\le\lVert f_n-f\rVert_2\,\lambda(E^c)^{1/2}<2\varepsilon . \]Hence \(\limsup_n\lVert f_n-f\rVert_1\le2\varepsilon\) for every \(\varepsilon\) , and \(f_n\to f\) in \(L^1\) .
The example. The spike \(f_n=n\mathbf 1_{(0,1/n)}\) has \(\lVert f_n\rVert_1=1\) and \(f_n\to0\) a.e., but \(\lVert f_n-0\rVert_1=1\) . Its \(L^2\) norms are \(n^{1/2}\) , unbounded. A bound in \(L^2\) prevents concentration, because a function of bounded \(L^2\) norm can carry only mass of order \(\lambda(A)^{1/2}\) on a small set \(A\) ; a bound in \(L^1\) does not.
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Evaluate \(\displaystyle\int_0^\infty\Bigl(\int_x^\infty\frac{e^{-y}}{y}\,dy\Bigr)dx\) , stating what must be checked before the order of integration is changed.
Answer
The integrand \(\frac{e^{-y}}y\mathbf 1_{\{0<x<y\}}\) is nonnegative and Borel measurable, and Lebesgue measure is \(\sigma\) -finite, so Tonelli’s theorem applies with nothing else to check. Reversing, \(\int_0^\infty\frac{e^{-y}}y\bigl(\int_0^ydx\bigr)dy=\int_0^\infty e^{-y}\,dy=1\) .
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Let \(\mu\) be a finite measure and let \(A_n\) be measurable sets with \(\mu(A_n)\ge c>0\) for all \(n\) . Prove that the set of points lying in infinitely many \(A_n\) has measure at least \(c\) , and show that this fails for Lebesgue measure on \(\mathbb R\) .
Solution
The set of points lying in infinitely many \(A_n\) is \(\limsup_nA_n=\bigcap_n\bigcup_{k\ge n}A_k\) . Put \(C_n=\bigcup_{k\ge n}A_k\) . These sets decrease to \(\limsup_nA_n\) , and \(\mu(C_1)\le\mu(X)<\infty\) , so continuity from above applies:
\[ \mu\bigl(\limsup_nA_n\bigr)=\lim_n\mu(C_n). \]Since \(A_n\subseteq C_n\) , \(\mu(C_n)\ge\mu(A_n)\ge c\) for every \(n\) . Hence \(\mu(\limsup_nA_n)\ge c\) .
For Lebesgue measure on \(\mathbb R\) take \(A_n=[n,n+1]\) : each has measure \(1\) , and no point lies in more than two of them, so \(\limsup_nA_n=\emptyset\) . The step that fails is continuity from above, because \(\lambda(C_n)=\infty\) for every \(n\) . This is Exercise 1.8(b) again; it is the set version of the fact that mass cannot escape from a space of finite measure.
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Determine all \(p\in[1,\infty]\) for which \(f(x)=\dfrac1{x(\ln x)^2}\) belongs to \(L^p(0,\tfrac12)\) .
Answer
Only \(p=1\) . With \(u=\ln x\) , \(\int_0^{1/2}\frac{dx}{x(\ln x)^2}=\int_{-\infty}^{-\ln2}\frac{du}{u^2}=\frac1{\ln2}\) . For \(p>1\) , \(f^p=x^{-p}\lvert\ln x\rvert^{-2p}\) is not integrable at \(0\) , since a power \(x^{-p}\) with \(p>1\) outweighs any power of the logarithm; and \(f\) is unbounded, so \(f\notin L^\infty\) .
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Let \(\nu(E)=\int_E\bigl(\mathbf 1_{[0,1]}-2\cdot\mathbf 1_{(1,2]}\bigr)\,d\lambda\) on the Borel sets of \([0,2]\) . Find \(\nu^+\) , \(\nu^-\) , \(\lvert\nu\rvert([0,2])\) , and the Radon–Nikodym derivative of \(\nu^+\) with respect to \(\lvert\nu\rvert\) .
Answer
\(P=[0,1]\) and \(N=(1,2]\) form a Hahn decomposition. \(\nu^+\) is Lebesgue measure restricted to \([0,1]\) , \(\nu^-\) is twice Lebesgue measure restricted to \((1,2]\) , and \(\lvert\nu\rvert([0,2])=1+2=3\) . Since \(\lvert\nu\rvert\) has density \(\mathbf 1_{[0,1]}+2\cdot\mathbf 1_{(1,2]}\) and \(\nu^+\) has density \(\mathbf 1_{[0,1]}\) , the chain rule gives \(\dfrac{d\nu^+}{d\lvert\nu\rvert}=\mathbf 1_{[0,1]}\) .
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On \((0,1)\) let \(f_n(x)=x^{n-1}-2x^{2n-1}\) . Show that \(\sum_n\int_0^1f_n\,dx=0\) while \(\int_0^1\sum_nf_n\,dx=\ln2\) , and explain which hypothesis of which theorem fails.
Solution
Term by term. \(\int_0^1x^{n-1}\,dx=\tfrac1n\) and \(\int_0^12x^{2n-1}\,dx=\tfrac1n\) , so \(\int_0^1f_n\,dx=0\) for every \(n\) and \(\sum_n\int_0^1f_n\,dx=0\) .
Sum first. For \(0<x<1\) both geometric series converge:
\[ \sum_{n=1}^\infty f_n(x)=\frac1{1-x}-\frac{2x}{1-x^2}=\frac{(1+x)-2x}{1-x^2}=\frac1{1+x}, \]and \(\int_0^1\frac{dx}{1+x}=\ln2\) .
The failed hypothesis. The two answers differ, so no theorem on term-by-term integration can apply. Theorem 4.5 requires nonnegative terms, and \(f_n\) changes sign at \(x=2^{-1/n}\) . Theorem 5.5 requires \(\sum_n\int\lvert f_n\rvert<\infty\) . Here \(\int_0^1\lvert f_n\rvert\,dx\ge\int_0^{2^{-1/n}}f_n\,dx=\frac1{2n}-\frac1{4n}=\frac1{4n}\) , and \(\sum\frac1{4n}=\infty\) . Each \(f_n\) has a positive part and a negative part of mass about \(\tfrac1{4n}\) each, which cancel exactly in the integral but not in the pointwise sum, where the positive parts of different terms sit at different places. It is the function analogue of the array in Example 6.3.
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Let \(f\) be locally integrable on \(\mathbb R\) with \(\int_If\,d\lambda=0\) for every bounded interval \(I\) . Prove that \(f=0\) almost everywhere, by two different methods.
Hint
By differentiation. The averages \(A_rf(x)\) are all \(0\) , and they tend to \(f(x)\) at almost every \(x\) by Theorem 10.4. By uniqueness of measures. On \([-N,N]\) the finite measures with densities \(f^+\) and \(f^-\) agree on the \(\pi\) -system of intervals and have equal total mass, so they agree on all Borel sets by Theorem 1.17; test them on \(\{f>0\}\) .
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Let \(E\subseteq[0,1]^2\) be a Borel set every vertical section of which has length at least \(\tfrac12\) . Prove that the set of heights \(y\) whose horizontal section has length at least \(\tfrac14\) has measure at least \(\tfrac13\) .
Solution
By Theorem 6.4 the two slicings of \(E\) give the same number:
\[ \int_0^1\lambda(E^y)\,dy=\int_0^1\lambda(E_x)\,dx\ge\frac12 . \]Let \(S=\{y\in[0,1]:\lambda(E^y)\ge\tfrac14\}\) , a measurable set since \(y\mapsto\lambda(E^y)\) is measurable. On \(S\) the integrand is at most \(1\) , and off \(S\) it is less than \(\tfrac14\) . Hence
\[ \frac12\le\int_0^1\lambda(E^y)\,dy\le1\cdot\lambda(S)+\frac14\bigl(1-\lambda(S)\bigr)=\frac14+\frac34\lambda(S), \]which gives \(\lambda(S)\ge\tfrac13\) . The constant \(\tfrac13\) cannot be improved. Fill the strip of heights \([0,s]\) completely, and above it let the horizontal section at height \(y\) be an interval of length \(a<\tfrac14\) that slides once around \([0,1]\) , modulo one, as \(y\) runs from \(s\) to \(1\) ; then each \(x\) is covered for a proportion \(a\) of those heights. Every vertical section has length \(s+(1-s)a\) , which equals \(\tfrac12\) for \(s=\frac{1/2-a}{1-a}\) , and \(S=[0,s]\) has measure \(s\) , which tends to \(\tfrac13\) as \(a\to\tfrac14\) . The argument is Markov’s inequality run backwards: a lower bound on an average of a bounded function forces the function to be large on a set of definite size.
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Let \(E\subseteq\mathbb R\) be a set, not assumed measurable, with \(\lambda^*(E\cap I)\le\tfrac12\lambda(I)\) for every bounded interval \(I\) . Prove that \(\lambda^*(E)=0\) .
Solution
Bounded sets. Suppose first \(E\subseteq[-N,N]\) , so that \(\lambda^*(E)<\infty\) . Let \(\varepsilon>0\) and choose bounded open intervals \(I_k\) covering \(E\) with \(\sum_k\lambda(I_k)<\lambda^*(E)+\varepsilon\) . Since \(E=\bigcup_k(E\cap I_k)\) , subadditivity and the hypothesis give
\[ \lambda^*(E)\le\sum_k\lambda^*(E\cap I_k)\le\frac12\sum_k\lambda(I_k)<\frac12\bigl(\lambda^*(E)+\varepsilon\bigr). \]Because \(\lambda^*(E)\) is finite it may be subtracted: \(\tfrac12\lambda^*(E)<\tfrac12\varepsilon\) . As \(\varepsilon\) was arbitrary, \(\lambda^*(E)=0\) .
General sets. The set \(E\cap[-N,N]\) satisfies the same hypothesis, by monotonicity of \(\lambda^*\) , so it has outer measure zero for each \(N\) , and \(E\) is the countable union of these sets.
No measurability was needed, and no differentiation theorem: the hypothesis is self-improving under covering. For a measurable set the statement also follows from Corollary 10.5, since a set of positive measure has points of density \(1\) ; the direct argument is both more elementary and more general.
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A function satisfies \(\lVert f\rVert_1=8\) and \(\lVert f\rVert_3=1\) . Give the best bound for \(\lVert f\rVert_2\) that follows from these two numbers alone, and name the inequality behind it.
Answer
Interpolation (Proposition 7.9, a consequence of Hölder’s inequality) with \(\tfrac12=\tfrac\theta1+\tfrac{1-\theta}3\) , that is \(\theta=\tfrac14\) : \(\lVert f\rVert_2\le\lVert f\rVert_1^{1/4}\lVert f\rVert_3^{3/4}=8^{1/4}\approx1.682\) .
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Let \(g_n\in L^2(\mu)\) with \(\lVert g_n\rVert_2\le n^{-2}\) . Prove that \(\sum_ng_n\) converges almost everywhere and in \(L^2\) , and say whether the same conclusion holds under \(\lVert g_n\rVert_2\le n^{-1}\) .
Hint
\(\sum_nn^{-2}<\infty\) , so Lemma 8.1 applies with \(p=2\) . Under \(\lVert g_n\rVert_2\le n^{-1}\) the conclusion can fail: on \([0,1]\) take the constant functions \(g_n=\tfrac1n\) , whose sum diverges at every point.
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Let \(\nu\) be the measure on \([0,1]\) defined by \(\nu(B)=\lambda(\{x\in[0,1]:x^2\in B\})\) . Show that \(\nu\ll\lambda\) , find \(\dfrac{d\nu}{d\lambda}\) , and compute \(\int t\,d\nu(t)\) in two ways.
Answer
For \(0\le t\le1\) , \(\nu([0,t])=\lambda(\{x:x^2\le t\})=\sqrt t=\int_0^t\frac{ds}{2\sqrt s}\) . So \(\nu\) agrees on the intervals \([0,t]\) with the measure of density \(\frac1{2\sqrt t}\) , hence everywhere, and \(\dfrac{d\nu}{d\lambda}=\dfrac1{2\sqrt t}\) on \((0,1]\) . Then \(\int t\,d\nu=\int_0^1\frac{t}{2\sqrt t}\,dt=\tfrac13\) , and directly from the definition of \(\nu\) , \(\int t\,d\nu(t)=\int_0^1x^2\,dx=\tfrac13\) .
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Suppose \(f:\mathbb R\to\mathbb R\) has the property that \(\{f=c\}\) is Lebesgue measurable for every real \(c\) . Must \(f\) be Lebesgue measurable? Prove it or give a counterexample.
Answer
No. Let \(V\) be a non-measurable set and put \(f(x)=e^x\) for \(x\in V\) and \(f(x)=-e^x\) for \(x\notin V\) . The function is injective, so each \(\{f=c\}\) is empty or a single point, hence measurable; but \(\{f>0\}=V\) is not. Measurability must be tested on the uncountably large sets \(\{f>a\}\) , which are not countable unions of level sets.