Discontinuous Almost-Everywhere/ Unbounded in $L^{1}(\mathbb{R})$ 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 $\mathbb{R}$ ?
 A: 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.
