# How to calculate $\int_{\Delta_n}\frac{\partial f}{\partial x_j}\,dx_1\cdots dx_n$?

Let $$f:\mathbb R^n\to\mathbb R$$ be smooth, $$1\leq j\leq n$$, and let $$\Delta_n=\{x\mid\sum_{i=1}^nx_i\leq1;x_i\geq0\}\subset\mathbb R^n$$ denote the standard $$n$$-simplex in $$\mathbb R^n$$. I'm trying to prove that $$\int_{\Delta_n}\frac{\partial f}{\partial x_j}\,dx_1\cdots dx_n\\=\int_{\Delta_{n-1}}f(x_1,\dots,1-\sum_{i\neq j}x_i,\dots,x_n)\,dx_1\cdots\widehat{dx_j}\cdots dx_n-\int_{\Delta_{n-1}}f(x_1,\dots,0,\dots,x_n)\,dx_1\cdots\widehat{dx_j}\cdots dx_n,$$ where the integrals are Lebesgue integrals (not that it makes much of a difference, I suppose). This makes sense to me intuitively, however I'm having difficulty showing it rigorously. My thoughts went something like this

Write $$\int_{\Delta_n}\frac{\partial f}{\partial x_j}$$ as $$\int_{\mathbb R^n}\frac{\partial f}{\partial x_j}\cdot\chi$$, where $$\chi$$ is the characteristic function of $$\Delta_n$$. Then by Fubini $$\int_{\mathbb R^n}\frac{\partial f}{\partial x_j}\cdot\chi=\int_{\mathbb R^{n-1}}\Big(\int_{\mathbb R}\frac{\partial f}{\partial x_j}(x_1,\dots,\widehat{x_j},\dots,x_n)\cdot\chi(x_1,\dots,\widehat{x_j},\dots,x_n)\,dx_j\Big)\,dx_1\cdots\widehat{dx_j}\cdots dx_n.$$ The outer integral "should" become $$\int_{\Delta_{n-1}}$$ because that's where the coordinates $$x_1,\dots,\widehat{x_j},\dots,x_n$$ "may vary" with $$x_j$$ fixed, while the inner integral "should" become $$\int_{0}^{1-\sum_{i\neq j}x_i}$$ because that's where the coordinate $$x_j$$ "may vary" with the rest fixed. The rest follows by the fundamental theorem of calculus.

How do I make this precise?

Write the integral as $$\int_{\Delta_n} \frac{\partial f}{\partial x_j}dx^1 \cdots dx^n = \int_{\Delta_{n-1}}\left(\int_0^{1 - \sum_{i \neq j}x_i} \frac{\partial f}{\partial x_i} dx^r\right)dx^1\cdots\widehat{dx^j}\cdots dx^n$$ Where $$\Delta_{n-1}$$ is the subset of $$\Delta_n$$ with $$x_j = 0$$. Then from the fundamental theorem of calculus, we have that $$\int_{\Delta_n} \frac{\partial f}{\partial x_j}dx^1 \cdots dx^n = \int_{\Delta_{n-1}}f(x_1, \dots, 1 - \sum_{i \neq j}x_i, \dots, x^n) - f(x_1, \dots, 0, \dots, x_n) dx^1 \cdots \widehat{dx^j} \cdots dx^n.$$ To completely justify this, it is sufficient to justify the first step, as the rest is just integration with respect to one variable. Let's assume that $$\int_{\Delta_n} \left|\frac{\partial f_i}{\partial x_i}\right| = M < \infty$$ Then as you said, from Fubini's theorem we have that $$\int_{\Delta^n}\frac{\partial f}{\partial x_i} = \int_{\mathbb R^{n-1}} \left(\int_\mathbb R \frac{\partial f}{\partial x_j} \chi_{\Delta_n}dx^j\right)dx^1 \dots \widehat{dx^j} \dots dx^n$$ where $$\chi_X$$ is the characteristic function on $$X \subseteq \mathbb R^n$$. So now we just need to manipulate the characteristic function inside the first integral. The main thing to note is that $$\chi_{\Delta_n} = \chi_{\Delta_{n-1}}\chi_{x_j \in [0, 1 - \sum_{j \neq j} x_i]},$$ where $$\chi_{\Delta_{n-1}}$$ only depends on $$\{x_i, i \neq j\}$$. So we have $$\int_{\Delta^n}\frac{\partial f}{\partial x_i} = \int_{\mathbb R^{n-1}} \left(\int_\mathbb R \frac{\partial f}{\partial x_j} \chi_{\Delta_{n-1}}\chi_{[0, 1 - \sum_{j \neq j} x_i]}dx^j\right)dx^1 \dots \widehat{dx^j} \dots dx^n$$ And we then note that $$\chi_{\Delta_{n-1}}$$ has no $$x_j$$ dependence and so can be pulled out of the integral. Hence $$\int_{\Delta^n}\frac{\partial f}{\partial x_i} = \int_{\mathbb R^{n-1}}\chi_{\Delta_{n-1}} \left(\int_\mathbb R \chi_{[0, 1 - \sum_{j \neq j} x_i]}\frac{\partial f}{\partial x_j} dx^j\right)dx^1 \dots \widehat{dx^j} \dots dx^n.$$
• How do you write $\chi_{\Delta_n} = \chi_{\Delta_{n-1}}\chi_{x_j \in [0, 1 - \sum_{j \neq j} x_i]}$, the last function doesn't make a lot of sense to me. May 5, 2022 at 0:41
• If $(x_1, \dots, \hat x_j \dots, x_n)$ are fixed then $g = \chi_{x_j \in [0, 1 - \sum_{i \neq j} x_i]}$ is just a standard characteristic function on the variable $x_j$. If we allow the $x_i$ to vary, and the $(x_i)$ are in $\Delta^{n-1}$ then $g = 1$ if and only if $(x_1, \dots, x_n) \in \Delta^n$, this follows from the definition of $\Delta^n$ May 5, 2022 at 10:47