Where 1 <= x <= 4:

f(x) = (x-1)/9 + 1, maps bijectively from [1,4] to [1,4/3]

g(x) = (x-1)/9 + 4/3, maps bijectively from [1,4] to [4/3,5/3]

h(x) = (x-1)/9 + 5/3 maps bijectively from [1,4] to [5/3,2]

Therefore, [1,2] must contain three times as many numbers as [1,4], right?

It doesn't work like that.