Formally prove convolution with Dirac Delta returns the same function? Let $\delta$ be the Dirac Delta distribution and $f$ be Riemann-integrable. Then it is said $(f*\delta)(t) = f(t).$
But, I'm wondering if this can be proven. The dirac delta distribution converges to an object that's $0$ everywhere except at $y = 0.$
So this leave us with a question, what is the value of the integral at 0? We evaluate $\lim_{R \rightarrow 0} \int_{-R}^{R}f(t,y)\delta(y)dy$ which is
$$ \lim_{R \rightarrow 0} \lim_{n \rightarrow \infty} \sum_{i=1}^{n}f(t,y_i)\delta(y_i)\Delta y,$$ but I'm unsure what to do from here or where to find a reference.
 A: 
But, I'm wondering if this can be proven.

It's the very definition of the Delta Dirac distribution that integrating a product with it yields the original function evaluated at the integration variable at 0.
Plug that into the definition of convolution, and your statement literally is written there.
Your limiting approach to integration will not take you further – since  that is Riemann integration, and only bounded functions can be Riemann integrated. The Delta Dirac distribution is neither a function nor bounded.
Luckily, that's not a problem, since this is really just a matter of applying the definition of the Delta Dirac distribution with the integral.
A: If $f \in C_{c}^{\infty}$, then by definition of distributional convolution,
$$(\delta * f)(x) = \langle \delta(y), f(x - y) \rangle_y = f(x).$$
The above doesn't work for an arbitrary Riemann integrable $f$, as the definition of $\delta * f$ is different in this case. For such $f$ you can use the density of $C_c^{\infty}$ in the space of distributions to get a sequence of $C_c^{\infty}$ functions converging in the topology of the space of distributions to $f$ to get the result. Alternatively, for such $f$ you can use the definition of convolution to easily show that $\delta * f = f$ in the sense of distributions (which here means that $\delta * f = f$ a.e.): For all $\phi \in C_c^{\infty}$,
$$\langle \delta * f, \phi \rangle = \langle f(y), \langle \delta(x), \phi(x + y) \rangle_{x} \rangle_{y} = \langle f(y), \phi(y) \rangle_{y} = \langle f, \phi \rangle.$$
