Let $f\left(x\right)$ and $g\left(x\right)$ two piecewise continuous function. I would like to prove (but I'm not sure it is true) that $$\left(f*g\right)\left(y\right)=\int_{0}^{y}f\left(x\right)g\left(y-x\right)dx$$ is continuous. I tried to search in some books but I find nothing.
My attempt: I fix $y>0$. Since $f,\,g$ are piecewise continuous then exists a partition $\left(a_{m},\,a_{m+1}\right),\,m=1,\dots,M $ of the interval $\left(0,y\right)$, with $a_{1}=0$ and $a_{M}=y$ such that $f\left(x\right)$ and $g\left(y-x\right)$ are continuous and since $$\int_{0}^{y}f\left(x\right)g\left(y-x\right)dx=\sum_{m=1}^{M-1}\int_{a_{m}}^{a_{m+1}}f\left(x\right)g\left(y-x\right)dx$$ then $\int_{0}^{y}f\left(x\right)g\left(y-x\right)dx$ is continuous in $y$ because is a finite sum of continuous function.
Is this proof correct?
If I remember correctly, the matching conditions at the discontinuities imposed on piecewise continuous functions (i.e. the limit of a piecewise continuous function at a discontinuity must exist from both sides) imply that a piecewise continuous function is bounded. Since the product of two piecewise continuous (hence bounded) functions is bounded we have $|fg| <M $ for all $x$, therefore $$\left| \int_0^{y-\delta} fg \ \text{d}x -\int_0^{y} fg \ \text{d}x \right|=\left| \int_{y-\delta} ^yfg \ \text{d}x \right|\le \int_{y-\delta} ^y|fg| \ \text{d}x< \int_{y-\delta} ^y M \ \text{d}x<M\delta$$ and the same can be done approaching $y$ from the right.