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 $a>0$, $f_a(t)=(ta)^{3/4}$ for $t>a$ and $0$ for $t \leq a$. Then, for $a,b>0$, convolution
$(f_a*f_b)(x):=\int_0^x f_a(xy)f_b(y)dy=\beta (xab)^{\frac{1}{2}}$ 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 $[0, \infty)$?

2$\begingroup$ You say $\int_0^x f_a(xy)f_b(y)dy=f_{a+b}(x)$. My calculations disagree. But the convolution $f_a * f_b$ is, indeed, unbounded at $a+b$. $\endgroup$ – Gerald Edgar Mar 31 '12 at 17:12

$\begingroup$ Sorry, I did mistake. I just have edited. $\endgroup$ – arc Mar 31 '12 at 18:21
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.

1$\begingroup$ Certainly $f*g$ is locally $L^1$, so almost all points are also Lebesgue points for $f*g$. For every such point $x$ and every $\varepsilon>0$ there exists a $\delta>0$ with almost all points $y\in B(x,\delta)$ satisfying $f*g(x)f*g(y)<\varepsilon$. (Why? Just assume not...) So there exists a version of $f*g$ continuous at $x$. With a little thought you can now devise a scheme to modify $f*g$ so it really is continuous except at countably many points. $\endgroup$ – Ollie Apr 1 '12 at 0:59
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?

$\begingroup$ An $L^1$ function can be discontinuous everywhere. $\endgroup$ – Gerald Edgar Mar 31 '12 at 23:36