How inflation solves the horizon problem

Hi,
I'm trying to figure out how inflation (just deSitter) solves the horizon problem, but I am stuck. I understand the solution in terms of conformal coordinates, allowing for a negative conformal time lets the lightcones of CMB intersect. Fine. But how do I see "physically" what is going on?
In most reviews I studied they compare some (comoving) scale L to the comoving Hubble scale 1/(H a(t)) (a(t) being the scale factor, here a~exp(H t)), and since this Hubble radius shrinks down, the horizon problem is no more.
BUT: I don't get why we compare the scale L to the Hubble radius in the first place. None of my reviews provide a proper meaning of 1/Ha (well, besides some handwaving scaling arguments...), so this seems fishy to me. If I try to do it the way I thought it was right, comparing the scale to the integral over 1/a from the beginning of inflation to time t, it comes out wrong, since this integral still increases with time, i.e. the "horizon" does not shrink down.

Any help very much appreciated!

Regards,
torus

Chalnoth
The reason why the scale L is compared to the Hubble radius is that the Hubble radius sets (somewhat roughly) the possible interaction length: on longer scales, speed of light limitations prevent any interaction.

Inflation solves this particular issue because during inflation, the Hubble scale was nearly constant, such that if we take any length scale L today, and consider that L scales as a*L going into the past, at some point during inflation a*L < c/H(a) (since H is nearly a constant during inflation, and a*L decreases monotonically into the past).

Does that help?

Well, the question is exactly: Why does the Hubble radius set the interaction length?
Sure, it has the right dimension, but we could still multiply by a or something. Or it could be an integral.

Chalnoth
$$\int_{a_1}^{a_2} \frac{1}{aH} d\ln a = \frac{1}{H a_1} - \frac{1}{H a_2}$$