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Let $f\in L^p$. Show that $f=f\chi_E+g$ where $m(E)<\infty$ and $|g|\le1$. Assume that $m$ is the Lebesgue measure on $\mathbb R$.

Using Chebychev's inequality, I can find a set of finite measure $E$ such that $m(E)=m(x:|f(x)|>1)\le\|f\|_p^p<\infty$. I can't go further.

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You are practically done! Just choose the right $g$... hint: it "complements" $f \chi_E$. –  lazyhaze Apr 4 '12 at 14:29
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@lazyhaze I see, thanks. It suffices to take $g=f\chi_{\mathbb{R} \setminus E}$. –  Nicolas Essis-Breton Apr 4 '12 at 14:38
    
Yes, precisely. –  lazyhaze Apr 4 '12 at 15:39
    
@NicolasEssis-Breton, you can answer your own question, that can helps to do your post more useful. –  leo Apr 11 '12 at 5:07
    
@leo Good idea. It will enhance the post. I did it. Hope it's correct. –  Nicolas Essis-Breton Apr 11 '12 at 19:51
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up vote 1 down vote accepted

Per leo comment. Here is my attempt at this question.
Let $E=\{x: |f(x)|>1\}$. Since $|f|>1$ if and only if $|f|^p>1$, by Chebychev's inequality, $m(E)=m(\{x: |f(x)|^p>1\} \le \int |f|^p$. Since $f \in L^p$, the previous integral is finite and hence $E$ as finite measure. Since $\mathbb R \setminus E=\{x: |f(x)|\le 1\}$, define $g=f \chi_{\mathbb R \setminus E}$, then $|g| \le 1$. As $E$ and $\mathbb R \setminus E$ are complementary set, $f=f\chi_E+g$. The result follows.

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Good :-) ....... –  leo Apr 11 '12 at 21:35
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