Consider a density of the form
$$ p(x|\sigma)=\frac{1}{\sigma}f(\frac{x}{\sigma}). $$
where $\sigma>0$. Note that this will be a normalized density provided $f(x)$ is correctly normalized. The parameter $\sigma$ is known as a scale parameter, and the denstiy exhibits scale invariance because if we scale $x$ by a constant to give $\hat{x}=cx$, then $$ p(\hat{x}|\hat{\sigma})=\frac{1}{\hat{\sigma}}f(\frac{\hat{x}}{\hat{\sigma}}). $$
where we have defined $\hat{\sigma}=c\sigma$. This transformation corresponds to a change of scale, for example from meters to kilometers if $x$ is a length, and we would like to choose a prior distribution that reflects this scale invarance. If we consider an interval $A\leqslant \sigma\leqslant B$, and a scaled interval $A/c\leqslant \sigma\leqslant B/c$, then the prior should assign equal probability mass to these two intervals. Thus we have
$$ \int_A^Bp(\sigma)d\sigma=\int_{A/c}^{B/c}p(\sigma)d\sigma=\int_A^Bp(\frac{1}{c}\sigma)\frac{1}{c}d\sigma. $$
and because this must hold for choices of $A$ and $B$, wwe have
$$ p(\sigma)=p(\frac{1}{c}\sigma)\frac{1}{c}. $$
and hence $p(\sigma)\propto 1/\sigma$. Please prove that this is an improper prior because the integral of the distribution over $0\leqslant \sigma\leqslant \infty$ is divergent.
For $n \in \mathbb Z$, we have $$\int_{c^n}^{c^{n+1}} p(\sigma) \, d \sigma = \int_{c^n/c}^{c^{n+1}/c} p(\sigma) \, d \sigma = \int_{c^{n-1}}^{c^{n}} p(\sigma) \, d \sigma$$ and $$(0, \infty) = \cdots \cup [c^{-2}, c^{-1}] \cup [c^{-1}, 1] \cup [1, c^1] \cup [c^1, c^2] \cup \cdots$$ so $$\int_0^\infty p(\sigma) \, d\sigma = \sum_{n=-\infty}^\infty \int_{c^n}^{c^{n+1}} p(\sigma) \, d\sigma = \sum_{n=-\infty}^\infty \int_1^c p(\sigma) \, d\sigma = \infty$$ assuming $\int_1^c p(\sigma) \, d\sigma \neq 0$.