Why does the amplitude of a discrete convolution depend on the time step? I've just finished covering convolutions in my signals class, and I've been playing around with the conv function in MATLAB, but there's something I don't quite understand. Say I have two discrete signals, $a$ and $b$ sampled at $dt$. As $dt \rightarrow 0$, the amplitude of the result increases. Since the convolution integral is just
$$
y(t)=\int_{-\infty}^\infty f(\tau)h(t-\tau)d\tau,
$$

why does the resulting convolution seem to depend on what my value is for $dt$? By decreasing the sampling rate, shouldn't I just have a better approximation?
Your computer doesn't compute the continuous integral, it does discrete convolution, which is just a sum of products at each time step.  When you increase dt, you get more points in each signal vector, which increases the sum at each time step. You must normalize the result of conv() according to the length of the vectors involved.