$\int_{-\infty}^{\infty}f(\xi)d\xi = \lim_{\delta \to 0 }\sum_{n=-\infty}^{\infty}\delta f(\delta n)$? Assume that $f$ is continuous and moderate decrease, show that
$$\int_{-\infty}^{\infty}f(\xi)d\xi = \lim_{\delta \to 0,~  \delta>0}\sum_{n=-\infty}^{\infty}\delta f(\delta n)$$
From the condition, we can get the $f$ is Riemann integral, so 
$$\int_{-\infty}^{\infty}f(\xi)d\xi =\lim_{||T|| \to 0}\sum_{n=-\infty}^{\infty}f(x_n ^*)\Delta x_n$$
Of course, the Riemann sum seems to be familiar with $ \lim_{\delta \to 0,~  \delta>0}\sum_{n=-\infty}^{\infty}\delta f(\delta n)$.
But how to argue this?
 A: (There is a missing '$\delta$' in the formula above.)
This is straightforward to demonstrate with the Lebesgue integral.
Let $f_\delta(x) = f(\delta \lfloor \frac{x}{\delta} \rfloor )$. Since $f$ is continuous, you have $\lim_{\delta \downarrow 0} f_\delta(x) = f(x)$. Since $f$ satisfies a 'moderate decrease' condition, we have $|f_\delta(x)| \le \frac{A}{1+(\delta \lfloor \frac{x}{\delta} \rfloor)^2} \le g(x) = \max(\frac{A}{1+x^2}, \frac{A}{1+(x-\delta)^2})$. Since $g$ is integrable, we can use the dominated convergence theorem to conclude that $\lim_{\delta \downarrow 0} \int f_\delta = \int f$.
To finish, we need to compute $\int f_\delta$:
\begin{eqnarray}
\int f_\delta (x) dx  &=& \sum_{n=-\infty}^\infty \int_{[n \delta, (n+1)\delta]} f_\delta (x) dx\\
&=&  \sum_{n=-\infty}^\infty \int_{[n \delta, (n+1)\delta]} f(n \delta) dx \\
&=&  \delta \sum_{n=-\infty}^\infty f(n \delta)
\end{eqnarray}
Aside:
Without the 'moderate decrease' condition. the result is not necessarily true. For example,
Let $\tau(x) = \max(1-|x|,0)$, and define $f(x) = \sum_{k \in \mathbb{Z}} \tau(k^2(x-k))$. It is easy to check that $\tau(x) \ge 0, f(x) \ge 0$, $\int \tau = 1$, and $\int f = \sum_{k \in \mathbb{Z}} \frac{1}{k^2} < \infty$. Hence $f$ is Lebesgue or improper-Riemann integrable. However, $f(k) = 1$ for all $k \in \mathbb{Z}$, hence if $\delta = \frac{1}{n}$, we have $\sum_{k \in \mathbb{Z}} f(\delta k) \ge \sum_{k \in \mathbb{Z}} f( k)  = \infty$, so the formula above doesn't hold (even if multiplied by $\delta$, of course).
