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Let $L^{1}(\mathbb{R})$ be defined as usual, with the equivalence relation : $f \approx g$ if and only if $f(x) = g(x)$ almost everywhere.

Is there a class in $L^{1}(\mathbb{R})$ such that every element is discontinuous almost everywhere ? Is there another class such that the every element is discontinuous everywhere ?

Thank you very much :)

EDIT : Now I know about the existence of a set $A$ that is dense, has measure greater than zero and its dense complement also has measure greater than zero. It solves this problem. (thanks to David Mitra)

EDIT 2 : Can you help me now with another question ? Is there a class in $L^1(\mathbb{R})$ such that every element is unbounded in every non-empty open interval in $\mathbb{R}$ ?

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    $\begingroup$ What is a representant? $\endgroup$ – Chris Janjigian May 6 '13 at 16:44
  • $\begingroup$ A member that defines a partition in the equivalence relation $\endgroup$ – thetruth May 6 '13 at 16:46
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    $\begingroup$ Shouldn't that be "a class in $L^1(\mathbb{R}) such that *every* element is discontinuous almost everywhere"?. It seems to me that in every class of $L^1$ there's a a.e. discontinuous representative, namely $f + \alpha\mathbf{1}_\mathbb{Q}$. $\endgroup$ – fgp May 6 '13 at 16:49
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    $\begingroup$ Isn't the characteristic function of a Cantor Set equivalent to the function identically zero ? (which is continuous) I mean, it differs from this last one in a set of measure zero, right ? Should we use Fat Cantor Set ? $\endgroup$ – thetruth May 6 '13 at 17:09
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    $\begingroup$ One can construct a measurable set $A\subset[0,1]$ such that for any interval $I$, both $A\cap I$ and $A\cap I^c$ have positive measure. Try the characteristic function of this set. $\endgroup$ – David Mitra May 6 '13 at 17:18
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If anyone is interested, here is a solution to the first question and what i think solves the second one :) Following David Mitra's cool suggestion : Let $A$ be a dense subset of $\mathbb{R}$ such that is dense and has measure greater than zero and its dense complement also has measure greater than zero. We can construct such $A$ applying the iterative process of Fat Cantor Set, also to the "every middle interval that is excluded on each step". Then, defining $f(x) = \frac{1}{x^{2}}$ for each $x \in A$ and $f(x) = 0$, otherwise, we have that every equivalent function to $f$ in $L^{1}(\mathbb{R})$, is discontinuous every where.
Regarding my second question, I THINK (not sure, though), that the following example works : Let $f(x)=x^{\frac{-1}{3}}$ and let $\mathbb{Q}=\lbrace q_{n} \rbrace$. Then define : $f(x) = \sum_{n=1}^{\infty} \frac{1}{n^{2}}f(x-q_{n})$. It's clear that is in $L^{1}(\mathbb{R})$ and it's unbounded in each open interval. Also, every other equivalent function is unbounded in each open interval, due to density of rationals.

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