I know that this question is math related, but I still don't get why we can say the following formula holds:
I've a voltage \$V:\left[0,\infty\right]\to\mathbb{R}\$ and \$V\$ is periodic with \$T\$. Why can we say that the average voltage obays the following formula:
$$\overline{V}=\lim_{n\to\infty}\frac{1}{n}\int_0^nV(t)dt=\frac{1}{T}\int_0^TV(t)dt$$
While the other answers are correct, personally I have difficulty 'seeing' a proof's correctness from formulae. So let's get a little more hand-wavy.
Let's assume the average of a single period of the waveform is known. The average of the next period will be the same, this is the definition of period after all, each period is the same.
Given this, if we average over exactly one period, or two periods, or any integer number of periods, then we will get the same average.
However, what happens if we average over half a period? We'll get a different answer. So averaging over 1.5, or 2.5, or 10.5 periods will all give different answers, as the contribution from the half period is different to the contribution from the full period.
But, there is a trend. If we average over n.5 periods, we will have n contributions from the consistent average, and only one contribution from the different one. As n becomes larger, the contribution from the half period has a smaller effect, by the factor of n. As we allow n to grow without bound, the half period contribution shrinks to become a negligible part of the average. In the limit, as n tends to infinity, we drop all this 'tending to' language, and just say that they are equal, that the average is undisturbed by any non-period contributions.
Not a strictly formal proof, but... Assume \$n\$ as a multiple of \$T\$ so you can split the time \$n\$ into \$\frac{n}{T}\$ periods, where the integral will be the same \$I=\int_0^TV(t)dt\$. Then the infinite sum will be $$\int_0^nV(t)dt\ = \frac{n}{T}\cdot I = \frac{n}{T}\int_0^TV(t)dt$$. Now you can divide both sides by \$n\$ and have
$$\frac{1}{n}\int_0^nV(t)dt\ = \frac{1}{T}\cdot I = \frac{1}{T}\int_0^TV(t)dt$$
Now think that if \$n\$ is not multiple of \$T\$ there will be some bounded portion of the integral \$I\$ to be added, let's call it \$C\$. So in general case it will look like:
$$ \frac{1}{n}\int_0^nV(t)dt\ = \frac{C}{n} + \frac{1}{T}\int_0^TV(t)dt$$
ot we rewrite:
$$ \frac{1}{n}\int_0^nV(t)dt\ - \frac{C}{n} = \frac{1}{T}\int_0^TV(t)dt$$
But once we take \$n\$ to \$\infty\$, the \$\frac{C}{n}\$ term is going away, so we have the equality between the two sides.
you can say $$\overline{V}=\lim_{n\to\infty}\frac{1}{n}\int_0^nV(t)dt=\lim_{n'\to\infty}\frac{1}{n'}\sum_{i=1}^{n'}\frac{1}{T}\int_0^TV(t)dt=\frac{1}{T}\int_0^TV(t)dt$$
where $$n=n'\;T$$ or$$n'=\frac{n}{T}$$ So you can split the Integral into the summation over every periode, because $$V(t)=V(t+T)$$
The formula you have stated works on the following way: We use integral to calculate the area under a given line, that is the voltage function over given time interval from 0 to T. Once the area is calculated if it is divided by one side, we will get the average value. If we divide the area of the voltage with the total time, we will get the average voltage. (From a curvy line, we get constant line, on the same time interval) I will add pictures. Example: Let us suppose that we want to calculate the average voltage on the signal V= cos t for a given time period from a to b
Therefore we have to integrate the function in the period from a to b, calculating the area under the function 
We then divide the calculated area, with the length b-a and get another length value which is the average value
Then area from the average value and the time interval must be the same as the area under the function line
Another example: If we want to calculate the average value of the numbers 1,2,3,4,5 We would have to sum all of them and then divide them with 5 We have (1+2+3+4+5)(1/5)=3 In your example you replace the sum 1+2+... with integral and the total number of values with (1/5)=(1/t), then you get the average value which is 3 or Vavg We can also see that the sum of the values 1+2+3+4+5 are always the the same irrelevant of the position 5+1+2+3+4=15 the same holds true for integration too, because it is the same operation. Given the periodic function 1,2,3,4,5,1,2,3,4,5... we will Always have the same sum over the period of 5 numbers irrelevant of the starting number. So for a given periodic function with a period T, and integrating over the same period T we must always get the same result. If we try to calculate the infinite sum from 0 to n and start calculating 1+2+3+4+5+1... we will always know the numbers used and divide them with the sum. We must get the average value, because we will have navgsum/n, which is actually only the average value.
This is along the lines of MITU's comment above.
Let's say you wanted to know how much power your house consumed over your lifetime on average. Let's say your lifetime is n years.
Let's say you can recall a function that completely describes how much instantaneous power you were consuming during those n years. You want to add all of these values, which are infinite in number since the function is continuous in time.
In order to add these numbers, you need the integral. You then divide the integral by the time period you are interested in, n, to get the average over that period. This is the definition of the average value of a function from calculus. The limit to infinity is an extension to consider when the time period n goes to infinity.
So finally you get your number of average power consumed, so you've effectively applied the left side of the equation.
But then you think to yourself that you've (somehow) setup your house to consume exactly the same power every year, so that when you look at your function again, you notice that every year the function repeats.
Since you know that your function now repeats every year, there was no need to use ALL of the years to get your average value. You can get the same number by considering only one year, which is the period T of your function. This is the right side of the equation.
