Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I got back an assignment for a first course in analysis and I have made a very basic error, and I'm having a lot of trouble pinpointing exactly what piece of information I'm missing.

You have two measure spaces $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\mu)$. Assume that $X$ and $Y$ are disjoint. Let $Z=X\cup Y$

Show that $\mathcal{C}=\{A\cup B:A\in \mathcal{A} \,and\,B\in \mathcal{B}\} $ is $\sigma$-algebra on $Z$.

Showing that $\emptyset \in \mathcal{C}$ and that $\mathcal{C}$ is closed under countable unions was easy, but it's the

If $C\in \mathcal{C}$, then $C^c \in \mathcal{C}$

which I've gotten wrong.

My thinking is like this: We have $\mathcal{A} \subset X$ and $\mathcal{B} \subset Y$, where $A^c \in \mathcal{A}$ and $B^c \in \mathcal{B}$. Since $X$ and $Y$ are disjoint, the intersection between two elements that are respectively in $X$ and $Y$ is just the empty set.

$C=(A\cup B)$, so then $C^c=(A \cup B)^c=A^c \cap B^c=\emptyset$

I guess I'm making a very basic error because the feedback I got was just "Wrong!", so I'd really appreciate if someone pointed it out for me.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

For $C \subseteq Z$ you have $$C^c = Z \backslash C$$ i.e. the complement of the set is taken with respect to the set $Z$. Hence $$A^c = Z \backslash A$$ since you consider $A$ as a subset of $Z$ (and not $X$!). Therefore in general $A^c \cap B^c \not= \emptyset$.

Example: $X=[0,\frac{1}{2})$, $Y= \left[\frac{1}{2},1 \right]$, $A=\left(\frac{1}{4},\frac{1}{2}\right)$, $B=\left[\frac{1}{2},\frac{3}{4}\right]$, then $$A^c \cap B^c = \left[0,\frac{1}{4}\right] \cup \bigg(\frac{3}{4},1 \bigg]$$ where the complement is taken with respect to to $Z=[0,1]$.

share|improve this answer
    
I see, but doesn't that make $C^c=(A\cup B)^c=(A^c \cap B^c)=(Z\setminus A\cap Z\setminus B)=Z\setminus (A\cup B)$? Doesn't this imply that $C^c$ is not in $\mathcal{C}$? –  john.abraham May 12 '13 at 14:56
1  
@john.abraham No. Note that $$Z \backslash A = \underbrace{Y}_{\in \mathcal{B} \subseteq \mathcal{C}} \cup \underbrace{(X \backslash A)}_{\in \mathcal{A} \subseteq \mathcal{C}}$$ i.e. $Z \backslash A \in \mathcal{C}$ for $A \subseteq X$. And this implies $(Z \backslash A) \cap (Z \backslash B) \in \mathcal{C}$. –  saz May 12 '13 at 15:05
    
How do you then interpret $Z\setminus (A\cup B)$? –  john.abraham May 12 '13 at 15:44
    
@john.abraham There's nothing to interpret - it's simply the complement of $A \cup B$...? I don't see your point, unfortunately. –  saz May 12 '13 at 15:54
    
Well $(Z\setminus A) \cap (Z\setminus B) = Z\setminus (A\cup B)$ right? Which is the set $\{s\in Z|s\notin (A\cup B)\}$. Why does it follow that this set is an element of $\mathcal{C}$? (I'm really struggling to wrap my mind around this, it's embarassing, frankly, so thank you for your patience) –  john.abraham May 12 '13 at 16:48

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.