Change of Variable Trick The thing I need to prove is that
$$\int_{0}^{\infty}\int_{s}^{2s} \psi_\sigma(s)g(t)\frac{dt}{t}\frac{ds}{s}=\int_{\sigma}^{2\sigma}\int_{0}^{\infty} \psi_u(s)g(s)\frac{ds}{s}\frac{du}{u}$$
Suppose that
$$\psi_\sigma(s):=\begin{cases} (\frac{s}{\sigma})^N  &\text{ if } 0<s\leq\sigma ,\\ 0 &\text{ if } \sigma<s<\infty. \end{cases}$$
where $N$ is any fixed natural number, and  we are given a property that $\psi_\sigma(a^{-1}s)=\psi_{a\sigma}(s)$ $\forall a$.
Now suppose that $g\geq 0$ is an integrable function on $[0,\infty)$ with respect to the measure $\frac{dx}{x}$.
Show that
$$\int_{0}^{\infty}\int_{s}^{2s} \psi_\sigma(s)g(t)\frac{dt}{t}\frac{ds}{s}=\int_{\sigma}^{2\sigma}\int_{0}^{\infty} \psi_u(s)g(s)\frac{ds}{s}\frac{du}{u}$$
What I have tried:
I am not sure if those tries are helpful for solving this... but here is what I have tried,,,
Apparently, I need to change the domain from
$$\left\{ (s,t): 0<s<\infty, s<t<2s \right\} \implies \{ (s,u): 0<u<\infty, \sigma<s<2\sigma  \}$$.
Thus, I strongly think that I need to change the coordinate to polar coordinate... But seems difficult since there is something like $st$ and we don't know the exact structure of function $g$ helping me to change the variable.
Secondly, by Toneli, since $\psi_\sigma(s)=0$ for $\sigma < s<\infty$,
$$\int_{0}^{\infty}\int_{s}^{2s} \psi_\sigma(s)g(t)\frac{dt}{t}\frac{ds}{s}= \int_{0}^{\sigma} \int_{s}^{2s} \psi_\sigma(s)g(t)\frac{dt}{t}\frac{ds}{s}\\ =  \int_{0}^{\sigma} \int_{\frac{t}{2}}^{t} \psi_\sigma(s)g(t) \frac{ds}{s}\frac{dt}{t}+\int_{\sigma}^{2\sigma}\int_{\frac{t}{2}}^{\sigma} \psi_\sigma(s)g(t)\frac{ds}{s}\frac{dt}{t}. $$
Seems something needless tries....
Thus, thirdly, I am forced to use polar coordinate. However, it also failed...
$$\int_{0}^{\infty} \int_{s}^{2s} \psi_\sigma(s)g(t) \frac{dt}{t}\frac{ds}{s} = \int_{\frac{\pi}{4}}^{\frac{3\pi}{8}}\int_{0}^{\infty}\psi_\sigma(s)g(t) \frac{r dr d\theta}{ts}.$$
Forthly, if I just try Toneli at the first place,
$$ \int_{0}^{\infty} \int_{s}^{2s} \psi_\sigma(s)g(t)\frac{dt}{t}\frac{ds}{s}=\int_{0}^{\infty}\int_{\frac{t}{2}}^{t}\psi_\sigma(s)g(t)\frac{ds}{s}\frac{dt}{t}$$
In all the trial, I got stuck....
Hoping I can get a help from here!
 A: Begin by writing
$$\int_{0}^{\infty}\int_{s}^{2s} \psi_\sigma(s)g(t)\frac{dt}{t}\frac{ds}{s} = \int_0^\infty \int_0^\infty \left( \frac s \sigma \right)^N g(t) \chi_{\{s \le \sigma\}} \chi_{\{s \le t \le 2s\}} \frac{dt}{t}\frac{ds}{s}$$
Tonelli's theorem gives you
$$ \int_0^\infty \int_0^\infty \left( \frac s \sigma \right)^N g(t) \chi_{\{s \le \sigma\}} \chi_{\{s \le t \le 2s\}} \frac{dt}{t}\frac{ds}{s} = \int_0^\infty \left(\int_0^\infty \left( \frac s \sigma \right)^N g(t) \chi_{\{s \le \sigma\}} \chi_{\{s \le t \le 2s\}}\frac{ds}{s} \right) \frac{dt}{t}$$
Make the substitution $v = \dfrac s\sigma$ in the inner integral. Then $s = \sigma v$ and $\dfrac{ds}{s} = \dfrac{dv}{v}$ so that
$$\int_0^\infty \left(\int_0^\infty \left( \frac s \sigma \right)^N g(t) \chi_{\{s \le \sigma\}} \chi_{\{s \le t \le 2s\}}\frac{ds}{s} \right) \frac{dt}{t} = \int_0^\infty \left(\int_0^\infty v^N g(t) \chi_{\{\sigma v \le \sigma\}} \chi_{\{\sigma v \le t \le 2\sigma v\}}\frac{dv}{v} \right) \frac{dt}{t}$$
Apply Tonelli again:
$$\int_0^\infty \left(\int_0^\infty v^N g(t) \chi_{\{\sigma v \le \sigma\}} \chi_{\{\sigma v \le t \le 2\sigma v\}}\frac{dv}{v} \right) \frac{dt}{t} = \int_0^\infty \left(\int_0^\infty v^N g(t) \chi_{\{\sigma v \le \sigma\}} \chi_{\{\sigma v \le t \le 2\sigma v\}} \frac{dt}{t} \right) \frac{dv}{v}$$
and make the substition $u = \dfrac tv$ (treating $v$ as fixed) in the inner integral. As before $t = uv$ and $\dfrac {dt} t = \dfrac{du}u$ so that 
$$\int_0^\infty \left(\int_0^\infty v^N g(t) \chi_{\{\sigma v \le \sigma\}} \chi_{\{\sigma v \le t \le 2\sigma v\}} \frac{dt}{t} \right) \frac{dv}{v} = \int_0^\infty \left(\int_0^\infty v^N g(uv) \chi_{\{\sigma v \le \sigma\}} \chi_{\{\sigma v \le uv \le 2\sigma v\}} \frac{du}{u} \right) \frac{dv}{v}.$$
The integrand can be cleaned up a bit. Note
$$\chi_{\{\sigma v \le \sigma\}} \chi_{\{\sigma v \le uv \le 2\sigma v\}} = \chi_{\{v \le 1\}} \chi_{\{\sigma \le u \le 2\sigma\}}$$
Tonelli one more time:
$$\int_0^\infty \left(\int_0^\infty v^N g(uv) \chi_{\{v \le 1\}} \chi_{\{\sigma \le u \le 2\sigma\}} \frac{du}{u} \right) \frac{dv}{v} = \int_0^\infty \left(\int_0^\infty v^N g(uv) \chi_{\{v \le 1\}} \chi_{\{\sigma \le u \le 2\sigma\}}  \frac{dv}{v}  \right) \frac{du}{u}$$
Now make the substitution $s = uv$ (treating $u$ as fixed) in the inner integral. Then $v = \dfrac su$ and $\dfrac{ds}s = \dfrac{dv}v$ giving you finally
$$\int_0^\infty \left(\int_0^\infty v^N g(uv)  \chi_{\{v \le 1\}} \chi_{\{\sigma \le u \le 2\sigma\}}   \frac{dv}{v}  \right) \frac{du}{u} = \int_0^\infty \left(\int_0^\infty \left( \frac su \right)^N g(s) \chi_{\{ s \le u \}} \chi_{\{\sigma \le u \le 2\sigma\}}  \frac{dv}{v}  \right) \frac{du}{u} $$
This last expression can be rewritten as the expression you are looking for.
A: The desired equality can be written
$$\tag 1\int_{0}^{\sigma}\int_{s}^{2s} (s/\sigma)^N g(t)\frac{dt}{t}\frac{ds}{s}=\int_{\sigma}^{2\sigma}\int_{0}^{u} (s/u)^Ng(s)\frac{ds}{s}\frac{du}{u}.$$
We need to verify $(1)$ for all $g\in L^1([0,2\sigma],dt/t).$
Suppose $g(t)= t^n$ on $[0,2\sigma],$ where $n$ is a positive integer. Then it is routine to verify $(1)$ holds for such a $g.$ It follows that $(1)$ holds for any polynomial $p$ with $p(0)=0.$ But polynomials that vanish at $0$ are dense in $L^1([0,2\sigma],dt/t),$ an exercise I'll leave to the reader for now. This gives $(1)$ for all $g\in L^1([0,2\sigma],dt/t).$ 
