1设 f<script id="MathJax-Element-1" type="math/tex">f</script> 是实直线 R<script id="MathJax-Element-2" type="math/tex">\bbR</script> 上的实函数, 若有常数 M>0<script id="MathJax-Element-3" type="math/tex">M>0</script> 使得对任何有限个两两不同的实数 x1,?,xn<script id="MathJax-Element-4" type="math/tex">x_1,\cdots,x_n</script> 都有 ∣∣∑ni=1f(xi)∣∣≤M<script id="MathJax-Element-5" type="math/tex">\dps{\sev{\sum_{i=1}^nf(x_i)}\leq M}</script>. 证明: {x; f(x)≠0}<script id="MathJax-Element-6" type="math/tex">\sed{x;\ f(x)\neq 0}</script> 是至多可数的.

解答: 首先说明对 ? n∈N<script id="MathJax-Element-7" type="math/tex">\forall\ n\in \bbN</script>, An={x; f(x)>1/n}<script id="MathJax-Element-8" type="math/tex">A_n=\sed{x;\ f(x)>1/n}</script> 是有限集 (个数不超过 n([M]+1)<script id="MathJax-Element-9" type="math/tex">n([M]+1)</script>). 若不然,

∑x∈Anf(x)>1n∑n∈An1>1n?n([M]+1)>M,

<script id="MathJax-Element-10" type="math/tex; mode=display">\bex \sum_{x\in A_n} f(x) >\frac{1}{n}\sum_{n\in A_n}1 >\frac{1}{n}\cdot n([M]+1)>M, \eex</script> 这是一个矛盾. 其次, 同上论述, Bn={x; f(x)<?1/n}<script id="MathJax-Element-11" type="math/tex">B_n=\sed{x;\ f(x)<-1/n}</script> 也是有限集. 于是

{x; f(x)≠0}=∪∞n=1(An∪Bn)

<script id="MathJax-Element-12" type="math/tex; mode=display">\bex \sed{x;\ f(x)\neq 0}=\cup_{n=1}^\infty\sex{A_n\cup B_n} \eex</script> 是至多可数的.

 

 

2设E<script id="MathJax-Element-13" type="math/tex">E</script> 是实直线 R<script id="MathJax-Element-14" type="math/tex">\bbR</script> 上的 Lebesgue<script id="MathJax-Element-15" type="math/tex">Lebesgue</script> 可测集, 且 m(E)<∞<script id="MathJax-Element-16" type="math/tex">m(E)<\infty</script>. 证明:

<script id="MathJax-Element-17" type="math/tex; mode=display">\bex \lim_{n\to\infty}\int_E e^{inx}\rd x=0. \eex</script> 这里 m<script id="MathJax-Element-18" type="math/tex">m</script> 表示 Lebesgue<script id="MathJax-Element-19" type="math/tex">Lebesgue</script> 测度.

解答: 由 m(E)<∞<script id="MathJax-Element-20" type="math/tex">m(E)<\infty</script> 知 χE∈L1(R)<script id="MathJax-Element-21" type="math/tex">\chi_E\in L^1(\bbR)</script>, 而所证即为标准的 Riemann?Lebesgue<script id="MathJax-Element-22" type="math/tex">Riemann-Lebesgue</script> 引理.

 

 

3设 f<script id="MathJax-Element-23" type="math/tex">f</script> 是 [0,1]<script id="MathJax-Element-24" type="math/tex">[0,1]</script> 上实的 Lebesgue<script id="MathJax-Element-25" type="math/tex">Lebesgue</script> 可测函数, 并且 Z<script id="MathJax-Element-26" type="math/tex">\bbZ</script> 是整数集. 证明:

<script id="MathJax-Element-27" type="math/tex; mode=display">\bex \lim_{n\to\infty}\int_0^1 \sev{\cos f(x)}^n\rd x =m\sex{f^{-1}(\pi \bbZ)}. \eex</script>

证明: 注意到 |cosf(x)|n≤1<script id="MathJax-Element-28" type="math/tex">\sev{\cos f(x)}^n\leq 1</script> 及

<script id="MathJax-Element-29" type="math/tex; mode=display">\bex \sev{\cos f(x)}^n\stackrel{a.e.}{\to} \chi_{f^{-1}(\pi \bbZ)}, \eex</script> 我们由 Lebesgue<script id="MathJax-Element-30" type="math/tex">Lebesgue</script> 控制收敛定理得到结论.

 

 

4对 σ<script id="MathJax-Element-31" type="math/tex">\sigma</script>-有限的测度空间 (X,Σ,μ)<script id="MathJax-Element-32" type="math/tex">(X,\varSigma,\mu)</script>, 设 f<script id="MathJax-Element-33" type="math/tex">f</script> 是 X<script id="MathJax-Element-34" type="math/tex">X</script> 上的非负可测函数, 记

<script id="MathJax-Element-35" type="math/tex; mode=display">\bex \mu\sex{f>t}=\mu\sed{x;\ f(x)>t}. \eex</script> 证明: <script id="MathJax-Element-36" type="math/tex; mode=display">\bex \int_X f\rd\mu=\int_0^\infty \mu\sex{f>t}\rd t. \eex</script>

证明: 由 Fubini<script id="MathJax-Element-37" type="math/tex">Fubini</script> 定理,

<script id="MathJax-Element-38" type="math/tex; mode=display">\bex \int_X f\rd\mu =\int_X \int_0^{f}\rd t\rd\mu =\int_0^\infty \int_X \chi_{f>t}\rd\mu \rd t =\int_0^\infty \mu(f>t)\rd t. \eex</script>