thermodynamics - Why does the air pressure at the surface of the earth exactly equal the weight of the entire air column above it

Why does the air pressure at the surface of the earth (resulting from collisions of molecules on the surface of the earth which has to do with the velocity of the particles) exactly equal the weight of the entire air column above it (which just has to do with the number and mass of the molecules in the air column)?

Here are the beliefs that give rise to my perplexity:

1. 
   

   The weight of the atmosphere is the mass of the molecules (in a column with cross section 1 square inch, let's say) times the force of gravity.

   
   


2. 
   

   Thus if we cooled the air column to the point where it was a solid and set the solid on a scale, we would get the weight of that column of air which, I am told, should read 14.7 lbs.

   
   


3. 
   

   The pressure from the air on a patch of ground 1 square inch in area is due to the collisions of the air molecules with that patch of ground. This is a function of the number of particles hitting the ground per unit time (which is itself a function of the density of the air at the ground) and the average kinetic energy of those molecules (their temperature). The air pressure I am told is 14.7 lbs/square inch.

   
   



4. 
   

   If I enclose some the air at the ground in a rigid container (say I screw the cap on to a glass jar on the ground), thus isolating it from the effects of the column of air above, the pressure of the gas inside the jar is 14.7 psi and would remain that (assuming the container doesn't change shape/volume and I keep it at the same temperature) even if I took it too the top of a mountain or in to space.

   
   


5. 
   

   The molecules of air are sufficiently far apart so that inter-molecular forces are negligible.

   
   

So why does the air pressure at the surface of the earth (resulting from collisions of molecules on the surface of the earth which has to do with the density and velocity of the particles) exactly equal the weight of the entire air column above it (which just has to do with the number and mass of the molecules in the air column)?

Answer

Suppose the pressure at the Earth's surface is $P$. Consider an air column of cross-sectional area $A$. The upward force on the column is $F_{\text{up}}=PA$. Denote the weight of the column as $W$. By definition of "weight", the downward force on the column is $F_{\text{down}}=W$.

Suppose the pressure is too low, such that $F_{\text{up}}

In other words, the pressure is such as to balance the weight of the column because that's the only situation which won't immediately change.