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I'm helping a friend who is learning measure theory for the first time. We are looking at Real Analysis by Yeh. We just started reading it and got to this part:
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We are at the part about showing $\frak{A}$ is an algebra. This is the very beginning of the book so we pretty much only have the definition of algebra to work off of. We need to show

  • $\mathbb{R}^2 \in \frak{A}$
  • $A \in \frak{A}$ implies $A^c \in \frak{A}$
  • $A,B \in \frak{A}$ implies $A \cup B \in \frak{A}$

We figured out items 1 and 3 easily, so the question is if there's a nice way to do item 2?
Here was my proposal. $A=\cup_{i=1}^n R_i$ where each $R_i \in \frak{R}$ is a (possibly unbounded) rectangle. Then $A^c = \cap_{i=1}^n R_i^c$. 1st show $R_i^c$ is a finite union of other rectangles (this would involve up to 4 other rectangles), so $R_i^c \in \frak{A}$. But then we'd have to show $\frak{A}$ is closed to intersections, something like $(\cup_{i=1}^{n_1} R_i^{(1)}) \cap (\cup_{j=1}^{n_2} R_j^{(2)}) \in \frak{A}$, which seems rather tedious. And this is just $\mathbb{R}^2$...trying to do something like this in $\mathbb{R}^n$ seems even more daunting.

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    Isn't it enough to show that for $R_i, R_j \in \frak{R}$, $R_i \cap R_j \in \frak{R}$? – 2017-02-26
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    @FabioSomenzi Sorry, can you elaborate: why would that be enough? Thanks in advance – 2017-02-26
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    Because then you rely on distributivity of set union and intersection to claim that your intersection of two finite unions is a finite union of two-way intersections. – 2017-02-26
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    @FabioSomenzi Thanks for your help. I'm still worried about the rectangle complement part. I only figured out what $(a_1,b_1] \times (a_2,b_2]$ complement was in $\mathbb{R}^2$ by drawing a picture. How do we know $R^c$ is still a finite union of rectangles in $\mathbb{R}^n$? – 2017-02-26

2 Answers 2

1

Let $R = (a_1, b_1] \times \cdots \times (a_n, b_n]$ be a rectangle in $\mathbf{R}^n$. Let $C_i^1 = (b_i, \infty]$, $C_i^0 = (a_i,b_i]$, $C_i^{-1} = (-\infty, a_i]$. Notice that for each $i$, $\mathbf{R}$ is the disjoint union of $C_i^1, C_i^0,$ and $C_i^{-1}$. Possibly $C_i^1$ or $C_i^{-1}$ is empty, depending on whether $b_i = \infty$ or $a_i = -\infty$.

If $(x_1, ... , x_n) \in \mathbf{R}^n$ is not in $R$, this means that for some $i$, either $x_i \in C_i^1$ or $C_i^{-1}$. It follows that the complement of $R$ in $\mathbf{R}^n$ is the union of the rectangles

$$C_1^{m_1} \times \cdots \times C_n^{m_n}$$

where each $m_i$ is an integer equal to $-1,0,$ or $1$, and not all the $m_i$ are zero. If all the numbers $a_i, b_i$ are real, then this gives you $3^n-1$ rectangles comprising the complement, as you can see by drawing a picture in the case $n = 2$.

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    Thank you for showing how to generalize to $\mathbb{R}^n$ – 2017-02-26
2

If $R_1 = (a_{11},b_{11}] \times (a_{12},b_{12}]$ and $R_2 = (a_{21},b_{21}] \times (a_{22}, b_{22}]$, then

$$R_1 \cap R_2 = (\max(a_{11},a_{21}), \min(b_{11},b_{21})] \times (\max(a_{12},a_{22}), \min(b_{12},b_{22})] \in \frak{R} \enspace. $$

Since

$$ \bigg(\bigcup_{1 \leq i \leq n_1} R_i^{(1)}\bigg) \cap \bigg(\bigcup_{1 \leq j \leq n_2} R_j^{(2)}\bigg) = \bigcup_{1 \leq i \leq n_1, 1 \leq j \leq n_2} R_i^{(1)} \cap R_j^{(2)} \enspace, $$

we conclude that the complement of $A \in \frak{A}$ if the finite union of finite intersections of rectangles in $\frak{R}$. This proof is slightly abridged, but should give you the general idea.

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    Thank you for elaborating on your comments – 2017-02-26