This is actually a good example of the difference between pointwise and uniform convergence. Consider the characteristic function of the vase at time t, g_t : N -> {0, 1}. Then g_t(n) = 1 if and only if ball n is in the vase at time t. It will actually help to convert g_t to a function on the real numbers, by making f_t(x) = g_t(floor(x)).
Now, for each ball n, there is a time when it will be destroyed, and therefore will never be in the vase after that time. So the characteristic function f_t(x) converges pointwise to f(x) = 0. This is presumably what you mean by the limit of the vase being the empty set.
But the criterion of uniform convergence is that for any epsilon>0 there is a t such that f_t is within epsilon of the limit everywhere. Which is obviously not true, because at any time t there are some balls in the vase, and so the characteristic function is 1 somewhere. So f_t(x) does not uniformly converge to anything.
As it happens, without uniform convergence, the limit of the integrals of f_t(x) (which just so happens to be the number of balls in the vase, by my setup) is not generally equal to the integral of the limiting function f(x). So, in a way it is not really true that you can say
in the limit as T approaches infinity, there are infinitely many balls in the vase
This is actually a good example of the difference between pointwise and uniform convergence. Consider the characteristic function of the vase at time
t,g_t : N -> {0, 1}. Theng_t(n) = 1if and only if ballnis in the vase at timet. It will actually help to convertg_tto a function on the real numbers, by makingf_t(x) = g_t(floor(x)).Now, for each ball
n, there is a time when it will be destroyed, and therefore will never be in the vase after that time. So the characteristic functionf_t(x)converges pointwise tof(x) = 0. This is presumably what you mean by the limit of the vase being the empty set.But the criterion of uniform convergence is that for any
epsilon>0there is atsuch thatf_tis withinepsilonof the limit everywhere. Which is obviously not true, because at any timetthere are some balls in the vase, and so the characteristic function is 1 somewhere. Sof_t(x)does not uniformly converge to anything.As it happens, without uniform convergence, the limit of the integrals of
f_t(x)(which just so happens to be the number of balls in the vase, by my setup) is not generally equal to the integral of the limiting functionf(x). So, in a way it is not really true that you can sayas the integral does not transfer to the limit.