Sample and hold - LTI filter

I've found a material explaining the sample and hold operation pretty well, but there's one thing I can't understand.

Next, we would like to take this ideally sampled signal and hold the values between samples. As we have a train of impulses with the correct areas, we need a “block” that takes an impulse with area A, transforms it into a rectangular pulse of height A that starts at the time at which the delta function is input to the block, and persists for exactly Ts seconds. A little bit of thought shows that what we need is a linear, time-invariant (LTI) filter whose impulse response, h(t), is 1 between t = 0 and t = Ts and is zero elsewhere.

How do we know this is the impulse response that will give us a pulse of HEIGHT A, not area A? Could anyone explain this "a little bit of thought" part? Thanks in advance.

• It's sad to see, as late as 2008, that some author is still confused about the role and functions of the Sample-and-Hold (a device that sometimes precedes an A/D converter in the signal path) and the Zero-order Hold which is a hypothetical device (which is LTI) that is needed to model the hold state of a conventional D/A converter between sample instances. The author, S. Engelberg, is mistaken and repeats a mistake that was in early textbooks on DSP. Sample-and-Hold (S/H) is not the same as Zero-order Hold (ZOH). Mar 20 '14 at 3:07

An impulse of "area" $A$ means $A\delta(t)$. The nomenclature arises because $\displaystyle \int_{-\infty}^\infty A\delta(t) \,\mathrm dt = A,$ that is, the "area under the curve" is $A$.
Since the system is linear and time-invariant, and the response to input $\delta(t)$ is a pulse of height $1$ and duration from $0$ to $T_s$, the response to $A\delta(t)$ is a pulse of height $A$ and duration from $0$ to $T_s$. Its "area" is $AT_s$, that is, $A$ multiplied by the area $T_s$ of the response to the unit impulse.