Fisher's example of a non-converging mean for N towards infinity In one of Fisher's classical paper [1] I stumbled over the following:

If the frequency with which the variate $x$ falls into the range $dx$, be given by $$df = \frac{1}{\pi}\frac{dx}{1+(x-m)^2}$$ where $m$ is the unknown parameter representing the centre of the symmetrical frequency curve of $x$, then it is not difficult to show that the arithmetic mean of any number of independent values of $x$, will be distributed in exactly the same distribution as a single value of $x$.

As Fisher called this 'not difficult', I would like to prove the statement ... but I fail :-).
I assume $\frac{df}{dx}$ represents the pdf $g(\cdot)$ of a random variable. If I put Fisher's statement in other words, it says: if there are two random variables $X$ and $Y$, both with pdf $g(\cdot)$, their mean has also $g(\cdot)$ as its pdf. So I guess that the convolution of $g(\cdot)$ with itself and divided by two - which represents the pdf of the mean - should be equal to $g(\cdot)$. However, when I try to calculate this convolution, I arrive at complex integrals which are not at all 'not difficult'.
Does anyone know how to actually prove this in a simple way?
[1] Fisher, R. A. (1925, July). Theory of statistical estimation. In Mathematical Proceedings of the Cambridge Philosophical Society (Vol. 22, No. 05, pp. 700-725). Cambridge University Press. 
 A: Well, the pdf in question is a form of the Cauchy distribution, which is a known stable distribution. One property of stable distributions is that linear combinations of stable distributions remain in the same distribution family. So, sums of cauchy are cauchy, and so is the sample mean, this is why the mean does not exist for a cauchy distribution, since the sample mean does not converge to a number ala SLLN. 
As for direct proofs, an easy was would be to use the characteristic function of the Cauchy. A useful property of characteristic functions is the the characteristic function of the sum of two independent random variables is the product of their characteristic functions. Now, since you want the mean, you also need to know that the characteristic function of aX, with a being a constant, is equal to the characteristic function of X, $\psi(x)$ but replacing the argument by $ax$ to get $\psi_{aX}=\psi(ax)$.
Now, you know enough to calculate the requried characteristic functions, multiply them, and see that you will get the characteristic function for the original cauchy distribution. Since there is a one-to-one relationship between density functions and characteristic functions, if X and Y have the same characteristic function, then they have the same density function.
BTW: Fisher was a GENIUS, so his version of "easy" is not helpful for us common folk...
