Sample median of Cauchy distribution is consistent. How? When we use chebyshev's inequality to show whether an estimator is consistent or not, we require the mean square error of the estimator and I do not know sample median's probability distribution. So please advice how this can be shown.  
 A: Partial answer
For uneven samples $n=2k+1$ the distribution of the median of a distribution with probability density function $f(x)$ and cumulative density function F(x) can be expressed as:
$$f_{\text{median n= 2k+1}}(x) = \frac{(2k+1)!}{k! k!} F(x)^kf(x) (1-F(x))^k$$
which relates to the sampling of a number at $x$ (the middle term $f(x)$), having $k$ samples that are below $x$ (the first term $F(x)^k$), having $k$ samples above $x$ (the last term $F(x)^k$), and the number of ways that these numbers can be arranged (the factor at the beginning $(2k+1)!/(k!k!)$).
For the Cauchy distribution this becomes:
$$f_{\text{Cauchy median $n=2k+1$}}(x) =  \frac{(2k+1)!}{k! k!} \left(  \frac{1}{\pi \gamma \left[1 + \left(\frac{x-x_0}{\gamma} \right)^2 \right] } \right) \times \left(  0.25 - \frac{\text{tan}^{-1}\left(\frac{x-x_0}{\gamma} \right)}{\pi^2}   \vphantom{\frac{1}{\pi \gamma \left[1 + \left(\frac{x-x_0}{\gamma} \right)^2 \right] }}\right)^k $$
The variance does seem to be finite for $k \geq 2$ (at least checking in wolfram alpha). I can imagine it might be possible to obtain an explicit expression. It seems like the indefinite integral can be expressed in terms of inverse tangent, logarithm and polylogarithm functions. But it is not a very pretty looking expression.
