Suppose $n \ge 2$ and $S$ a standard $n$-simplex in $\Bbb R^n$ with base length $a$, where $a >0$. So $$S=\{(x_1, \ldots, x_n) \in \Bbb R^n\mid x_i \ge0, \sum_{i=1}^n x_i \le a\}$$

Find the Lebesgue integral $\int_{\Bbb R^n}1_S$, where $1_S$ is the characteristic function on $S$.

I'm thinking of using the Fubini Theorem and induction but I don't know how to start.

  • 1
    $\begingroup$ Well, if you want to use induction... have you proved a base case yet? Or maybe a couple? What do you think the answer will be? $\endgroup$ – Nick Peterson Mar 20 '18 at 23:29
  • $\begingroup$ The volume of the unit simplex ($a=1$) is $\frac{1}{n!}$. So, for any $a$ the volume is $\frac{a^n}{n!}.$ Interesting that the total of all these volumes (including $0$-simplex) is $e^a$. $\endgroup$ – szw1710 Mar 20 '18 at 23:43

Fubini's theorem says that if the integral of the absolute value is finite then the multiple integral is equal to every one of the iterated integrals $$ \int_\mathbb R \left( \int_\mathbb R \cdots\cdots\left( \int_\mathbb R \cdots \cdots \right) \cdots \right) $$ In this case we can write the multiple integral and one of these iterated integrals as follows: \begin{align} & \int\limits_{\left\{ \begin{array}{c} (x_1,\,\ldots,\,x_n)\,:\, \forall i\, x_i\,\ge\,0\ \&\ \sum\limits_i x_i \, \ge\, 0 \end{array} \right\}} 1_s(x_1,\ldots,x_n)\, d(x_1,\ldots ,x_n) \\[15pt] = {} & \int_0^a \left( \int_0^{a-x_1} \left( \int_0^{a-x_1-x_2} \left( \int_0^{a-x_1-x_2-x_3} \cdots\cdots \, dx_4 \right) \, dx_3 \right) \, dx_2 \right) \, dx_1 \end{align}

If you try this when $n=3$ you get $$ \int_0^a \int_0^{a-x_1} \int_0^{a-x_1-x_2} 1 \, dx_2\,dx_2\,dx_1 = \frac{a^3} 6. $$ It you try it with $n=0,1,2,3,4$ you will guess that in general it is $\dfrac{a^n}{n!}.$ So that is what you would need to prove by induction. If it's true in case $n-1$ then you have $$ \int_0^{a-x_1} \left( \int_0^{a-x_1-x_2} \left( \int_0^{a-x_1-x_2-x_3} \cdots\cdots \, dx_4 \right) \, dx_3 \right) \, dx_2 = \frac{(a-x_1)^{n-1}}{(n-1)^!}. $$ The induction step then consists of computing $$ \int_0^a \frac{(a-x_1)^{n-1}}{(n-1)!} \, dx_1. $$

  • $\begingroup$ So usually for the integral of two measures, one types \,dy\,dx? Thanks for editing that for me, it looks nicer. $\endgroup$ – user284331 Mar 21 '18 at 0:13

For $n=2$, the integral is $\displaystyle\int_0^a\int_0^{a-x} 1 \,dy\,dx$. Actually one can verify that, for example, $n=3$, $S=\{(x,y,z): 0\leq x\leq a, 0\leq y\leq a-x, 0\leq z\leq a-x-y\}$.


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