An approximation to the cdf of the normal from a pdf? In this paper (p. 36), authors wrote

$$p(n,T) = \Phi \Big(\frac{n}{T},\mu,\sigma \Big) - \Phi \Big (\frac{n-1}{T},\mu,\sigma \Big)\; (3) $$
Bellow we will use the approximation
$$p(n,T) = \frac{1}{T}N \Big(\frac{n}{T},\mu,\sigma \Big)\; (4)$$

On which $p(n,T)$ is their notation for $p(n = T)$, the probability that $n$ is $T$; $\Phi(\cdot)$ is the cdf and $N(\cdot)$ the pdf. I don't get why they use (4) to approximate (3), nor how they arrived to that equation.
Is this kind of approximation to the cdf standard? Sorry if the notation is a bit confusing. Let me know if I can improve the question to make it clearer.
 A: This is a simple approximation where the integral over the density is approximated by taking the integrand to be equal to its upper value.  Using a slight variation of their notation (to make the conditioning on parameters clearer), you have:
$$\begin{equation} \begin{aligned}
p(n,T) 
&= \Phi \Big( \frac{n}{T} \Big| \mu, \sigma \Big) - \Phi \Big( \frac{n-1}{T} \Big| \mu, \sigma \Big) \\[6pt]
&= \int \limits_{(n-1)/T}^{n/T} \text{N}(r | \mu, \sigma) \ dr \\[6pt]
&\approx \int \limits_{(n-1)/T}^{n/T} \text{N} \Big( \frac{n}{T} \Big| \mu, \sigma \Big) \ dr \\[6pt]
&= \text{N} \Big( \frac{n}{T} \Big| \mu, \sigma \Big) \int \limits_{(n-1)/T}^{n/T} \ dr \\[6pt]
&= \Bigg[ \frac{n}{T} - \frac{n-1}{T} \Bigg] \cdot \text{N} \Big( \frac{n}{T} \Big| \mu, \sigma \Big). \\[6pt]
&= \frac{1}{T} \cdot \text{N} \Big( \frac{n}{T} \Big| \mu, \sigma \Big). \\[6pt]
\end{aligned} \end{equation}$$
Notice that the approximation step involves replacing the integrand density function with its value at the upper bound of the range of integration.  So long as the range of integration is small (i.e., $T$ is large), the integrand will not vary much over the range of integration, so the approximation will be reasonably accurate.
