In my advanced calculus class, we were talking about convolution, and we were asked to prove that there is no real valued, Riemann integrable $f\in X$, where $X = \{f | \int\limits_{-\infty}^{\infty}|f(t)|dt < \infty\}$ such that for every continuous function $g\in X$, $f*g =g$ ($f$ doesn't have to be continuous). The integrals are all Riemann.
So, my attempt was to define for every $\varepsilon$ the function $$ g_{\varepsilon}(x) = \begin{cases} -\frac{1}{\varepsilon}|x| + 1 & x\in[-\varepsilon, \varepsilon] \\ 0 & \text{otherwise}\end{cases} $$
and now, assume $f$ is such identity element. Then $$ (f*g_{\varepsilon})(x) = \int\limits_{-\infty}^{\infty}f(t)g_{\varepsilon}(x-t)dt = \int\limits_{x-\varepsilon}^{x+\varepsilon}f(t)g_{\varepsilon}(x-t)dt = g_{\varepsilon}(x) $$ For $|x| \leq \varepsilon$, $$ \int\limits_{x-\varepsilon}^{x+\varepsilon}f(t)g_{\varepsilon}(x-t)dt = 1 $$ for all epsilon. The logic behind this attempt is to show that the integral on the left is not continuous which is a contradiction since integral functions are always continuous. I was unable to proceed from here.
I must mention that this class has nothing to do with measure theory, so we are suppose to prove this without the notions of measure, 'almost everywhere' etc.