How does sampling rate affect discrete filters?
IF we take the sampling period as T.First I will convert the filter to time domain impulse response as 
$$
h(s)=\frac{1}{\tau s+1}\\h(s)=\frac{1/\tau}{s+\frac{1}{\tau}}
$$
this in time domain is
$$
h(t)=\frac{e^{-t/\tau}}{\tau}
$$
IF the above signal is sampled at time T the it can be written as
$$
h(Tn)=\frac{e^{\frac{-Tn}{\tau}}}{\tau}\\h(n)=\frac{e^{\frac{-Tn}{\tau}}}{\tau}
$$
if Z transform taken for the impulse response.Then the equation is
$$
H(z)=\frac{1/\tau}{1-e^{\frac{T}{\tau}}z^{-1}}
$$
Here we can see the sampling time coming in denominator which could put the pole inside or outside the unit circle