Convolution: Integral vs. Discrete sum

In fact, the Convolution Integral between two functions $f(t)$ and $g(t)$ is defined as

$$(f*g)(t)=int_{-infty}^infty f(tau)g(t-tau),dtau$$

If both $f(t)$ and $g(t)$ are causal functions (i.e., $f(t)$ is causal if $f(t)=0$ for $t<0$), then

$$begin{align} (f*g)(t)&=int_{0}^t f(tau)g(t-tau),dtau\\ &=lim_{max_{iin[1,n]}(Delta tau_i)to 0}sum_{i=1}^n f(tau_i)g(t-tau_i)Delta tau_itag1 end{align}$$

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For the discrete case, we sample $f(t)$ and $g(t)$ at integer values of $t$. So, in $(1)$, we set $t=n$.

Moreover, we define $f(t)$ and $g(t)$ in terms of a train of Dirac Deltas $delta(t-k)$ with weights $f(k)$ and $g(k)$ so that $f(t)=sum_{m=1}^n f(m)delta(t-m)$, and $g(t)=sum_{ell=1}^n g(ell)delta(t-ell)$ in the integral in $(1)$. Proceeding, we find that

$$begin{align} (f*g)(n)&=int_{0}^t f(tau)g(t-tau),dtau\\ &=int_{0}^n sum_{m=1}^n f(m)delta(tau-m)sum_{ell=1}^n g(ell)delta(n-tau-ell),dtau\\ &=sum_{m=1}^n f(m)sum_{ell=1}^n g(ell)underbrace{int_0^n delta(tau-m)delta(n-tau-ell),dtau}_{=1,,text{for},,ell=n-m,,text{and otherwise},,=0}\\ &=sum_{m=1}^n f(m)g(n-m) end{align}$$