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Hey all, I got stuck on the proof of equation $(12)$. My consideration is as follows:
Assume that $A\in {\scr E}$ and $B\in\mathscr{E}$. Then $A=\bigcup_{s=1}^n I_s$, $B=\bigcup_{k=1}^m J_k$, where $I_s$ and $J_k$ are intervals in $R^p$ for $s=1,2,\dots,n$ and $k=1,2,\dots,m$.

Since $A\cup B=I_1\cup\dots\cup I_n\cup J_1\cup\dots\cup J_m$, $A\cup B\in \mathscr{E}$. It is left to show that $A-B\in\mathscr{E}$.

Note that $$ \begin{aligned} A-B&=A\cap B^c\\ &=A\cap\left(\bigcup_{k=1}^m J_k\right)^c\\ &=A\cap\left(\bigcap_{k=1}^m J_k^c\right)&& \text{(by De Morgan's Law})\\ &=\bigcap_{k=1}^m\left(A\cap J_k^c\right)\\ &=\bigcap_{k=1}^m \left[ \bigcup_{s=1}^n (I_s\cap J_k^c)\right] \end{aligned} $$ It can be shown that $\bigcap_{k=1}^m \left[ \bigcup_{s=1}^n (I_s\cap J_k^c)\right]=\bigcup_{s=1}^n \left[ \bigcap_{k=1}^m (I_s\cap J_k^c )\right]$, so we only need to show $\bigcap_{k=1}^m (I_s\cap J_k^c)$ is an interval.

Now we prove the intersection of any two intervals (namely, $A$ and $B$) is also an interval:
If $A\cap B=\emptyset$, then it is an interval by definition.
If $A\cap B\neq\emptyset$, then the points $\mathbf{x}=(x_1,x_2,\dots,x_p)$ in $A\cap B$ satisfy $$ c_i\leq x_i \leq d_i\quad (i=1,2,\dots,p), $$ so that $A\cap B$ is an interval.

If we can show that the complement of an interval is also an interval, then the theorem completes. This is where I got stuck. Furthermore, I have no idea about how to show $\mathscr{E}$ is not a $\sigma$-ring.

I would appreciate if you could explain in details.


1 Answer 1


You can't prove that the complement of an interval is also an interval: take $p=1$ and $A=[0,1]$. Then its complement is not an interval and is not an elementary set either, because by definition elementary sets are bounded.

After the big alignment, you say “we only need to show that $\bigcap_{k=1}^m(I_s\cap J_k^c)$ is an interval”, but this is false: take $I_s=[-2,2]$ and $J_k=[-1,1]$. Then $I_s\cap J_k$ is not an interval, but it is an elementary set nonetheless, namely $[-2,-1)\cup(1,2]$.

You need to fix this. Note that you cannot prove that $\mathscr{E}$ is closed under complements: you want to prove it is a ring (of sets) not an algebra (of sets). If it were closed under complements, then $R^p$ would belong to $\mathscr{E}$, which is certainly not the case.

Why isn't it a $\sigma$-ring? Again, consider $p=1$ and the intervals $[n,n]$ (a singleton is an interval according to the definition), for $n$ nonnegative integer. Then the union of these intervals is not a finite union of intervals. Generalizing to $R^p$ is easy.

  • $\begingroup$ In your second paragraph, do you mean $I_s\bigcap J_k^c$ is not an interval? Could you please give me some hints about how to prove ${\scr E}$ is a ring? Also, do you assume the unbounded set $[a,+\infty)$ is not an interval in $R^1$? $\endgroup$ Aug 26, 2020 at 18:22
  • $\begingroup$ @XiangdongMeng The example I give shows exactly this: if $A$ and $B$ are intervals, then $A-B=A\cap B^c$ need not be an interval. But it is an elementary set and this suffices to end the proof you want. $\endgroup$
    – egreg
    Aug 26, 2020 at 21:09

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