I have seen that if $u$ is a summable function (in fact, I saw that if $u \in W^{1,p}$, but I think that summable is sufficient) in $\mathbb{R}^n$ then \begin{equation} \int_{\{u>j\}} (u-j) dx = \int_j^\infty | \{u>j\}| dt \end{equation} where $j$ is some positive integer and the notation $|A|$ denotes the Lebesgue measure of a set $A$.
I can't see this now, but I imagine that this is simple. I thank you.
It's
$$\int_{\{u > j\}} (u-j)\,dx = \int_j^\infty \lvert \{ u > t\}\rvert\,dt$$
actually, $\lvert \{ u > j\}\rvert$ is a misremembering.
Let $A = \{ (x,t) \in \mathbb{R}^n\times \mathbb{R} : j \leqslant t < u(x) \}$. Then we have
$$\begin{align} \int_{\{u > j\}} (u-j)\,dx &= \int_{\mathbb{R}^n} \left(\int_j^{u(x)} \chi_{\{ u > j\}}(x)\,dt \right)\,dx\\ &= \int_{\mathbb{R}^n} \left(\int_{\mathbb{R}} \chi_A(x,t)\,dt\right)\,dx \tag{Fubini}\\ &= \int_{\mathbb{R}} \left(\int_{\mathbb{R}^n}\chi_A(x,t)\,dx\right)\,dt\\ &= \int_{\mathbb{R}} \chi_{[j,\infty)}(t)\cdot \lvert \{ u > t\}\rvert\,dt\\ &= \int_j^\infty \lvert \{ u > t\}\rvert\,dt. \end{align}$$
