"Dirac Comb" vs "Ones Comb" While learning sampling theory - I noticed that examples of continuous signal sampling always achieved the goal via multiplying the signal with a "Dirac Comb".
I was intrigued by the requirement to use a Dirac Comb instead of a "Ones Comb" - I.E: multiply the signal by the value '1' at the sampling times. 

The explanation I got: "That's necessary because multiplying by a Dirac function is equivalent to Convolution in the frequency domain which maintains the spectrum of the original sampled signal".     

This made a lot of sense to me...But than, when I proceeded to learn about discrete domain sampling - I was surprised to find out that examples used a "Ones Comb" to sample a signal (not the "Dirac Comb").

Why the difference ? How does discrete time sampling get away with multiplying by a train of discrete ones ?
the integral

$$ \int\limits_{-a}^{a} \delta(x) \, \mathrm{d}x = 1  \qquad \forall a&gt;0$$

but this integral

$$ \int\limits_{-a}^{a} g(x) \, \mathrm{d}x = 0  \qquad \forall a&gt;0$$

where 

$$ g(x) = \begin{cases}
1 \qquad &amp; \text{for } x=0 \\
0 \qquad &amp; \text{for } x\ne0 \\
\end{cases} $$

this is why we need this dirac delta thing for the sampling or sifting property:

the integral

$$ \int\limits_{-\infty}^{\infty} f(x) \, \delta(x-x_0) \, \mathrm{d}x = f(x_0) $$

but this integral

$$ \int\limits_{-\infty}^{\infty} f(x) \, g(x-x_0) \, \mathrm{d}x = 0 $$

can you see what's wrong with using $g(x)$ for sampling?