Convolution as a $L^1$ limit of translates. I would like what convolution is, as a $L^1$ limit. Namely let $f,g\in L^1(\mathbb{R})$ (with some further conditions). Then what conditions on $f$ and $g$ ensure that $f\ast g$ is the $L^1$ limit for $h\rightarrow 0$ of finite sums of the form $\sum_{k} f(\cdot-kh)g(kh)h$?. I suppose this should hold if e.g. $f$ is general and $g$ compactly supported and continuous?
Is this a consequence of a more general result, expressing the Lebesgue integral as a limit of "Riemann sums" of sorts?
 A: This is really more a comment than an answer. I need your answer to a question below in order to figure out the right answer to your question. This won't fit into the comment box; read the following, answer the question at the bottom and then I'll post an answer to your question.
First I feel compelled to clean up the notation. For $x\in\Bbb R$ define the translation operator $\tau_x$ by $$\tau_xf(t)=f(t-x).$$
Now here's a fact:
Theorem 0 Suppose $f,g\in L^1(\Bbb R)$. For $h>0$ define $f_h$ by $$f_h=\sum_{k\in\Bbb Z}a_k\tau_{kh}f,$$where $$a_k=\int_{kh}^{(k+1)h}g.$$Then $||f_h-f*g||_1\to0$ as $h\to0$.
And now my question for you is
Q: Do you know Theorem 0, and how to prove it?
The reason I ask: If you're not aware of Theorem 0 then it seems quite possible to me that, depending on why you care about this, you really asked the wrong question; it seems quite likely that Theorem 0 is what you really want.
On the other hand if you're aware of Theorem 0 and you're wondering about precisely the question you asked, the appropriate answer is different. (It's not hard to show, using Theorem 0, that (for example) assuming $g$ is continuous with compact support is enough to do what you asked.)
