Non-informative prior for regression model I'm looking at p. 355 of Gelman's Bayesian Data Analysis (3rd ed.), for which there is no errata, and I see this:

In the normal regression model, a convenient non-informative prior
  distribution is uniform on $(\beta,\, \log\sigma)$ or, equivalently,
  $$p(\beta,\, \sigma^2|X) \propto \sigma^{-2}$$

Shouldn't it say "uniform on $(\beta, \log\sigma^2)$"?
If $W = g(X)$ with $g$ monotonic increasing, $h \equiv g^{-1}$, don't we have $$f_{W}(w) = f_{X}(h(w))\;h^\prime(w)$$
Let $X=\log\sigma^2$, $g=\exp$, $h=\log$; the density of $W=g(X)=\sigma^2$ is $f(w) = w^{-1}$ or, if I understand Gelman's notation correctly, $p(\sigma^2)=\sigma^{-2}$.
Am I right or am I confused about notation... or just completely wrong?
 A: It could say uniform on $(\beta, \log \sigma^2)$, but what is written is also correct. These are equivalent, because
\begin{equation}
\log \sigma^2 = 2\log\sigma
\end{equation}
and reparametrization by multiplying by a constant keeps a uniform pdf uniform (in your notation, $h'(w)$ is constant).
A: The original is correct.
If you try to take the (improper) uniform prior on the half line (say by considering it as a limit of a sequence of proper priors), then at any point you choose, consider the ratio of the probability to its right over the probability to its left.
No matter how high you choose that point ($10^{10^{10}}$? ), you're effectively saying you're almost certain that $\sigma^2$ is larger than that and that it has almost no chance of being smaller. That's not really in accord with the concept of 'uninformative'.
It's better (for a number of reasons) to work on the log-scale, which covers the whole real line (and so uniformity doesn't 'push all the probability to the end'); uniformity on the log-scale results in the $1/\sigma^2$ prior you have there. I believe this is also the Jeffreys prior for $\sigma^2$ in that model. 
