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i. If $1 < p < \infty$ and $E = \{f_a, a \in A\}$ set of measurable functions of $\mathbb{R}$ and $\sup_{a \in A} ||f_a||_p < \infty$, I want to show that for $ 0 < q < p$, $\lim \limits_{x \rightarrow \infty} \sup \limits_{a \in A} \int_{|f_a| > x} |f_a|^q = 0$.

ii. For bounded Borel $f$ on $\mathbb{R}$ where $\int fg = 0$ for all $g \in C_c(\mathbb{R})$, I want to show $ f=0$ a.e.

iii. I want to show that $\forall \epsilon\gt0, \exists \delta\gt0$ s.t. $m(E \subseteq \mathbb{R}) < \delta \Rightarrow \int_E |f_a|^q < \epsilon$

My thoughts were for one to use Hölder's inequality to use Borel-Cantelli, but that didn't get me anywhere. For two, it seems obviously if $g \in C_c(\mathbb{R})$ then set of discontinuities are countable so then $f$ must be zero a.e.

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For i., can you bound the measure of $\{|f_a|\gt x\}$ uniformly on $a$ in $A$? –  Did Mar 16 '13 at 9:39
    
Could you try and make your title a little more informative? Measure theory sure is a broad subject. –  Pedro Tamaroff Mar 17 '13 at 1:09
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2 Answers 2

  1. Note that $$ \begin{align} \int_{|f_a|\gt x}|f_a(t)|^q\,\mathrm{d}t &\le\frac1{x^{p-q}}\int_{|f_a|\gt x}|f_a(t)|^p\,\mathrm{d}t\\ &\le\frac1{x^{p-q}}\int|f_a(t)|^p\,\mathrm{d}t\\ &\le\frac1{x^{p-q}}\|f_a\|_p^p\\ \sup_{a\in A}\int_{|f_a|\gt x}|f_a(t)|^q\,\mathrm{d}t&\le\frac1{x^{p-q}}\left(\sup_{a\in A}\|f_a\|_p\right)^p \end{align} $$ Now take the limit as $x\to\infty$.

  2. Given $\epsilon>0$ and $a<b\in\mathbb{R}$, find a $g\in C_c(\mathbb{R})$ so that $\|f-g\|_{L^2[a,b]}<\epsilon$. Then $$ \begin{align} \|f\|_{L^2[a,b]}^2+\|g\|_{L^2[a,b]}^2 &=\int_a^b\left(f(x)^2+g(x)^2\right)\,\mathrm{d}x\\ &=\int_a^b\left(f(x)-g(x)\right)^2\,\mathrm{d}x\\ &\le\epsilon^2 \end{align} $$ Thus, $\|f\|_{L^2[a,b]}\le\epsilon$ and since $\epsilon$ was arbitrary, we have $$ \|f\|_{L^2[a,b]}=0 $$

  3. Let $$ E_\lambda=\{x:|f(x)|\gt\lambda\} $$ Then by Dominated Convergence $$ \lim_{\lambda\to\infty}\int_{E_\lambda}|f(x)|^q\,\mathrm{d}x=0 $$ There are two possibilities, either there is some $\Lambda$ so that $\lambda\gt\Lambda\Rightarrow m(E_\lambda)=0$, or $\forall\lambda>0,m(E_\lambda)\gt0$. If $\Lambda$ exists, then $\delta\lt\epsilon/\Lambda$ will do since $f$ is essentially bounded. Otherwise, choose a $\lambda_\epsilon$ large enough so that if $\lambda\ge\lambda_\epsilon$, $$ \int_{E_\lambda}|f(x)|^q\,\mathrm{d}x\lt\epsilon $$ Such a $\lambda_\epsilon$ is guaranteed by the previous limit. Choose $\delta=m\left(E_{\large\lambda_\epsilon}\right)\gt0$. Integration over any set of measure $\delta$ will be at most the integral over $E_{\large\lambda_\epsilon}$.

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Your solution in 2. relies on the assumption that $C_c(\Bbb R)$ is dense in $L^2[a,b]$ for all $a$ and $b$. Is it simpler to establish this than a direct approach? –  Davide Giraudo Mar 16 '13 at 10:37
    
@DavideGiraudo: If we don't assume density, then there is more to do all around. If density is not a given, then perhaps a proof paralleling the proof of density would be shorter than proving density and then using the argument above. –  robjohn Mar 16 '13 at 11:10
    
For 1, I am having trouble seeing why the first line is true: $\int_{|f_a|>x} |f_a(t)|^q dt \le \frac{1}{x^{p-q}} \int_{|f_a|>x} |f_a(t)|^p dt$ –  chinaschool Mar 16 '13 at 20:14
    
@chinaschool: Wherever $|f_a|\gt x$, we also have that $|f_a|^{p-q}\gt x^{p-q}$. Thus, $|f_a|^p\frac1{x^{p-q}}\gt|f_a|^q$. –  robjohn Mar 16 '13 at 21:53
    
As a follow up, how would I show that $m(E\subseteq \mathbb{R}) < \delta \Rightarrow \int_E |f_a|^q < \epsilon$ –  chinaschool Mar 16 '13 at 23:23
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The answer to the first question has been already given. For the second one, first approximate pointwise the characteristic function of a compact set by continuous functions with compact support.

As we can choose the sequence monotonic, we can actually show that for each compact $K$, $$\int f\chi_Kd\lambda=0.$$ By regularity of Lebesgue measure, we can deduce that $\int f\chi_Sd\lambda=0$ for each measurable set $S$ of finite measure. Conclude.

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