Can convolution on $R_+$  be discontinuous everywhere ? Let $L^+$ be a set of all real valued functions  defined on a real line which are Lebesgue integrable on each $[0,c]$, where $c>0$,
and are zero for $x<0$. 
Let for $a>0$,  $f_a(t)=(t-a)^{-3/4}$ for $t>a$ and $0$ for $t \leq a$. Then, for $a,b>0$, convolution 
$(f_a*f_b)(x):=\int_0^x f_a(x-y)f_b(y)dy=\beta (x-a-b)^{-\frac{1}{2}}$ for $x> a+b$ and $0$ for $x \leq a+b$, where $\beta=B(\frac{1}{4}, \frac{1}{4})$.
It shows that convolution of two functions from $L^+$ need not be continuous. Is it possible, maybe by condensation
of singularities and above example to show existence of two functions from $L^+$ which convolution is discontinuous everywhere
on $[0, \infty)$?
 A: How about this.  If $g \in L^+$, then almost every point is a Lebesgue point for $g$.  And if $h \in L^+$, then almost every point is a Lebesgue point for $h$.  We should then be able to show that the convolution $g * h$ is differentiable almost everywhere.  Or something like that.
A: You can't have a convolution of two functions in $L_1(0,c)$ that is not in $L_1(0,c)$, else you violate Young's inequality, or the fact that $L_1(0,c)$ (i.e. the space you refer to as $L^{+}$) is a Banach algebra under convolution, so the answer is  no. I assume you are taking $a < c$ and $b < c$ (otherwise you can't work in $L_1(\mathbb{R})$ because $f_a$ and $f_b$ don't belong there).
EDIT: Excuse me, Arc. I do apologize. I did not make the relevant distinction between an everywhere discontinuous function and a function defined only on a set of measure zero. I now understand the definition, so my answer is largely unhelpful.
However, if you are trying to construct a function that is discontinuous on a dense set, I do get the feeling that your construction wont do because the output of your convolution is (as you would expect) better behaved than the inputs, it is locally in $L_{2-\epsilon}$ and still has only one point of discontinuity. Perhaps I am missing the point here, though. Can you explain a bit more about what you want to achieve and your approach? 
