I have an exercice in which the result
$$
\int_{[0,1]^n}\mathbf{1}_{\{0<x_n<\ldots<x_2<x_1<1\}}\prod_{k=1}^{n}\mathrm{d}x_k=\frac{1}{n!}
$$
is given in a development like it's obvious, which to me is not. So I've tried to verify this for $n=2, 3$ and it does indeed seem to hold: (for instance with $n=2$)
\begin{align*}
\int_{0}^{1}\int_{0}^{1}\mathbf{1}_{\{0<z<y<1\}}\mathrm{d}z\mathrm{d}y
&=
\int_{0}^{1}\left[\int_{0}^{y}\mathbf{1}_{\{0<y<1\}}\mathrm{d}z\right]\mathrm{d}y\\
&=
\int_{0}^{1}\left[z\mathbf{1}_{\{0<y<1\}} \right]_{0}^{y}\mathrm{d}y\\
&=
\int_{0}^{1}y\mathbf{1}_{\{0<y<1\}} \mathrm{d}y\\
&=
\int_{0}^{1}y \mathrm{d}y=\left[\frac{y^2}{2}\right]_{0}^{1}\\
&=
\frac{1}{2}
\end{align*}
Given (one of) Fubini's theorem, the order of integration shouldn't matter, but when I flip the order, I don't get the same result. I suppose something is off with the way I manipulate the indicator functions (quite possibly the presence of the integration variable in it), but anyway, here's my attempt:
\begin{align*}
\int_{0}^{1}\int_{0}^{1}\mathbf{1}_{\{0<z<y<1\}}\mathrm{d}y\mathrm{d}z
&=
\int_{0}^{1}\left[\int_{z}^{1}\mathbf{1}_{\{0<z<y\}}\mathrm{d}y\right]\mathrm{d}z\\
&=
\int_{0}^{1}\left[y\mathbf{1}_{\{0<z<y\}} \right]_{z}^{1}\mathrm{d}z\\
&=
\int_{0}^{1} (1-z)\mathbf{1}_{\{0<z<y\}} \mathrm{d}z\\
&=
\int_{0}^{y} (1-z) \mathrm{d}z=\left[z-\frac{z^2}{2}\right]_{0}^{y}\\
&=
y-\frac{y^2}{2}
\end{align*}
What am I doing wrong?
Best Answer
Don't leave it in when you take it out.
You have the bound variable of the inner integral occurring in the bounds of the outer integral. It certainly should not be there.
Your issue is that during the process of "integrating out" the bound variable $y$ you left it inside the indicator domain.
In short: $0<z<y<1$ exactly when we have $z\in(0..1)$ and $y\in(z..1)$, so:
$$\begin{align}\iint_{\Bbb R^2}\mathbf 1_{0<z<y<1} \,\mathrm d\langle y, z\rangle &=\int_\Bbb R \mathbf 1_{0<z<1}\int_z^1\mathrm dy\,\mathrm d z\\& =\int_0^1\int_z^1 \mathrm dy\,\mathrm d z\\[1ex] &= \int_0^1 (1-z)\,\mathrm d z\\[1ex]&= (z-\tfrac 12z^2)\big\vert_{z=0}^{z=1}\\[1ex]&= (1-\tfrac 12)-(0-\tfrac 120)\\[1ex]&= \tfrac 12 \end{align}$$
Compare that to how it was initially done, where it is observed that we have $0<z<y<1$ exactly when we have $y\in(0..1)$ and $z\in(0..y)$, so:
$$\begin{align}\iint_{\Bbb R^2}\mathbf 1_{0<z<y<1} \,\mathrm d\langle y, z\rangle &=\int_\Bbb R \mathbf 1_{0<y<1}\int_0^y\mathrm dz\,\mathrm d y\\& =\int_0^1\int_0^y \mathrm dz\,\mathrm d y\\[1ex] &= \int_0^1 y\,\mathrm d y\\[1ex]&= (\tfrac 12y^2)\big\vert_{y=0}^{y=1}\\[1ex]&= \tfrac 12 \end{align}$$