Impulse Invariant mapping
consider an linear time-invariant system (LTI) in continuous time.  the input-output relationship is:

$$ y(t) = \int\limits_{-\infty}^{\infty} h(u) x(t-u) \, \mathrm{d}u $$

and, in the frequency domain

$$ Y(f) = H(f) X(f) $$

where

$$ X(f) \triangleq \mathscr{F} \big\{x(t)\big\} = \int\limits_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \, \mathrm{d}t  $$

and similarly for $H(f)$ and $Y(f)$.

now, suppose dimensionally $x(t)$ and $y(t)$ are the same species of animal, like they are both voltage or both position or both pure numbers.  then $X(f)$ and $Y(f)$ are also the same dimension of physical quantity.  then, dimensionally, what must $H(f)$ be?

$H(f)$, which is

$$ H(f) \triangleq \mathscr{F} \big\{h(t)\big\} = \int\limits_{-\infty}^{\infty} h(t) \, e^{-j 2 \pi f t} \, \mathrm{d}t  $$

must be dimensionless.  $H(f)$ must be a pure number with no units attached.  that is the only way it can multiply $X(f)$ and result in $Y(f)$ and the latter two have the same units or dimension.

now, in the integral above, what dimension must $h(t)$ have for the resulting integral to be the dimensionless $H(f)$?  consider the relationship between velocity and position:

$$ v(t) = \frac{\mathrm{d}r}{\mathrm{d}t} $$

or

$$ r(t) = r(t_0) + \int\limits_{t_0}^{t} v(u) \, \mathrm{d}u $$

$u$ is a dummy variable of integration and is time, just like $t$ is. 
dimensionally, what does the little differential element $\mathrm{d}u$ have?  $v(u)$ is velocity, length divided by time, and it is multiplied by tiny time increments, $\mathrm{d}u$, to become tiny length increments.  those tiny length increments are added up to become the length between positions $r(t_0)$ and $r(t)$.

so in the integral $\int_{-\infty}^{\infty} h(t) \, e^{-j 2 \pi f t} \, \mathrm{d}t$ , what must $h(t)$ be, dimensionally, so that the result of the integral is the dimensionless $H(f)$?

and what is the dimension of $h[n]$?

so this is the first step, understanding what must happen dimensionally to $h(t)$ to scale it to $h[n]$.

perhaps the best way to see why the dimensional scaler should be the sampling period, $T$ is by viewing the convolution integral as a Riemann summation:

$$\begin{align}
 y(t) &amp;= \int\limits_{-\infty}^{\infty} h(u) \, x(t-u) \, \mathrm{d}u \\
\\
  &amp;\approx \sum\limits_{i=-\infty}^{\infty} h(i \Delta u) \, x(t-i \Delta u) \, \Delta u \\
\end{align}$$

$\Delta u$ is a vanishingly small period of time.  If the sample rate $\frac1T$ is sufficiently high or $T$ is sufficiently small that it can take the place of $\Delta u$, then the Riemann summation is exactly the same as the discrete-convolution summation:

$$\begin{align}
 y(t) &amp;= \int\limits_{-\infty}^{\infty} h(u) \, x(t-u) \, \mathrm{d}u \\
\\
  &amp;\approx \sum\limits_{i=-\infty}^{\infty} h(i \Delta u) \, x(t-i \Delta u) \, \Delta u \\
\\
 \to y(t) &amp;= \sum\limits_{i=-\infty}^{\infty} h(i T) \, x(t - iT) \, T \\
\\
 \to y(nT) &amp;= \sum\limits_{i=-\infty}^{\infty} h(i T) \, x(nT - iT) \, T \\
\\
  &amp;= \sum\limits_{i=-\infty}^{\infty} T h(i T) \, x\big((n - i)T \big)  \\
\\
 \to y[n] &amp;= \sum\limits_{i=-\infty}^{\infty} \Big( T h(i T) \Big) \, x[n-i]  \\
\\
  &amp;= \sum\limits_{i=-\infty}^{\infty} h[i] \, x[n-i]  \\
\end{align}$$

there you can see (changing the dummy variable $i$ to $n$) that 

$$h[n] = T \, h(nT)\ .$$