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Can we find a function such that $f\in L^1([0,1])$ and for any $0\leq a<b\leq 1$ we have that $||f||_{L^{\infty}([a,b])}=\infty$?

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Try constructing appropriate nonnegative functions $f_n$ so that for each $n$, the support of $f_n$ is contained in the $n$'th dyadic interval $D_n$ and $f_n\in B(L_1)\setminus L_\infty$. Then consider $f=\sum 2^{-n} f_n$. – David Mitra Jan 24 '13 at 22:59
    
@DavidMitra: I am slightly confused - do you mean $f_{n}\in L_{\infty}-B(L_{1})$? For example in Davide's answer all $f_{n}$ is in $L^{\infty}$. – Bombyx mori Jan 24 '13 at 23:13
    
@user32240 I think what I wrote is right. We want each $f_n$ to not be in $L_\infty$. This is the case below, with $f_j(x)={1\over\sqrt{|x-r_j|}}$. – David Mitra Jan 24 '13 at 23:21
    
@user32240, they are not in $L^\infty$, they are in $L^1$ – Tomás Jan 24 '13 at 23:21
    
@DavidMitra: I think the essential superum for $f_{n}$ is $\infty$ right? They are in $L^{1}$ but not in $L^{\infty}$..I see. – Bombyx mori Jan 24 '13 at 23:39

Yes, we can. Consider $\{r_j,j\in\Bbb N\}$ an enumeration of rational numbers of $[0,1]$ and $$f(x):=\sum_{j=1}^{+\infty}\frac{2^{—j}}{\sqrt{|x-r_j|}}.$$

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+1 Very nice Answer... – Tomás Jan 24 '13 at 23:08
    
And $f^2$ is an example of a function that is finite a.e., but nowhere locally integrable, as was pointed out here. – Martin Jan 25 '13 at 8:54

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