Is this argument correct?
$\left \| f -g\right \|_{p}^{p}=$
$\int_{0}^{\infty}m(|f-g|>t^{\frac{1}{p}}) dt\stackrel{c.o.v.}{=}$
$p\int_{0}^{\infty}m(|f-g|>t) t^{p-1}dt\stackrel{t=|u-s|}{=}$
$p\int_{0}^{\infty}m(|f-g|>t-s) |t-s|^{p-1}dt\stackrel{h=\int\frac{dh}{ds}ds}{=}$
$p(p-1)\int_{0}^{\infty}\int_{0}^{t}[m(\{x:f>t\}-\{x:g>s\})+ m(\{x:g>t\}-\{x:f>s\})]|t-s|^{p-2}dsdt$
f,g are non-negative functions in $L^{p}(R^{n})$ where $1<p<\infty$.
thanks
$\left \| f-g \right \|_{p}^{p}=$
$\int_{\{f>g\}}(f-g)^{p}dx+\int_{\{g>f\}}(g-f)^{p}dx$. Let's look at $\int_{\{f>g\}}(f-g)^{p}dx$.
$=p\int_{\{f>g\}}\int_{g(x)}^{f(x)}(t-g(x))^{p-1}dtdx$
$\stackrel{h=\int \frac{dh}{ds}ds}{=}(-1)p(p-1)\int_{\{f>g\}}\int_{g(x)}^{f(x)}\int_{t}^{g(x)}(t-s)^{p-2}dsdtdx=$.
$(-1)p(p-1)\int_{\{f>g\}}\int_{0}^{\infty}\chi_{f(x)\geq t\geq g(x)}\int_{t}^{0} \chi_{g(x)\geq s} (t-s)^{p-2}dsdtdx=$
$(-1)p(p-1)\int_{0}^{\infty}\int_{t}^{0}\int_{\{f>g\}} \chi_{\{f(x)>t\}\cap \{g(x)\geq s\}} (t-s)^{p-2}dsdtdx=$
$(-1)p(p-1)\int_{0}^{\infty}\int_{t}^{0}m(\{f(x)>t\}\cap \{g(x)\geq s\}) (t-s)^{p-2}dsdtdx\stackrel{switch~bounds}{=}$
$p(p-1)\int_{0}^{\infty}\int_{0}^{t}m(\{f(x)>t\}\cap \{g(x)\geq s\}) (t-s)^{p-2}dsdtdx$.
We get similar result for other i.e. $p(p-1)\int_{0}^{\infty}\int_{0}^{t}m(\{g(x)>t\}\cap \{f(x)\geq s\}) (t-s)^{p-2}dsdtdx$. Thus,
$p(p-1)\int_{0}^{\infty}\int_{0}^{t}[m(\{f>t\}/\{g>s\})+ m(\{g>t\}/\{x:f>s\})]|t-s|^{p-2}dsdt$