So I am currently enrolled in a measure theory class and as such, a number of homework questions I've come across so far have been along the lines of 'show that some collection is a $\sigma$-algebra on some set' or 'determine the $\sigma$-algebra generated by some collection.' I seem to have no trouble showing this when the collection is something simple, like $\{ \emptyset,\{a\},\{b\}, \{a,b\} \}$ for $a,b\in X=\{a,b \}\subset\mathbb{R}$. However, things apparently are more difficult for me to grasp when the collection is very large.
For instance, on my previous homework, I was asked to show that $\left\{\bigcup_{n\in K} [n,n+1)\Bigg{|}K\subseteq\textbf{Z} \right\}$is a $\sigma$-algebra on $\mathbb{R}$. My approach was as follows (and this is exactly what I handed in):
First, note that\begin{align*} \textbf{R}\setminus\bigcup_{n\in K}[n,n+1) &= \textbf{R}_{<\inf K}\cup \textbf{R}_{\geq(\sup K)+1} \\ &= \cdots\cup[\inf K -2,\inf K-1)\cup[\inf K-1,\inf K)\cup[\sup K+1,\sup K+2)\cup[\sup K+2,\sup K+3)\cup\cdots\end{align*}This shows closure under complementation. Now, let $K_1,K_2,\cdots$ be a sequence of subsets of \textbf{Z}. Then\begin{equation*} \bigcup_{i=1}^\infty \bigcup_{n\in K_i}[n,n+1) = \bigcup_{n\in\bigcup_{i=1}^\infty K_i}[n,n+1)\end{equation*}But $\bigcup_{i=1}^\infty K_i\subset\textbf{Z}$, so $\bigcup_{n\in\bigcup_{i=1}^\infty K_i}[n,n+1)\in \left\{\bigcup_{n\in K} [n,n+1)|K\subseteq\textbf{Z} \right\}$. This shows closure under countable unions. Finally, since $\emptyset\subset\textbf{Z}$, $\bigcup_{n\in\emptyset}[n,n+1)=\emptyset$. So the collection is indeed a $\sigma$-algebra.
I lost a number of points on this last homework, and I presume it's due to a question like this.
Similarly, another question I had was to determine the $\sigma$-algebra on R generated by(a) $A:=\{(r,r+1)|r\in\textbf{Q} \}$ and (b) $B:=\{[r,\infty)|r\in\textbf{Q} \}$. I answered as follows:
Any $\sigma$-algebra containing $A$ must contain $\emptyset, \bigcup_{k=1}^\infty (r_k,r_k+1)$, and $(r,r+1)^\complement=\textbf{R}\setminus(r,r+1)$. For $r_1<r_1+1<r_2\in\textbf{Q}$, we have that $(r_1,r_1+1)\cup(r_2,r_2+1)=(r_1,r_2+1)$, which is just another open rational interval. We could do this union for countably many open rational intervals and still get another open rational interval. Now,\begin{equation*} (r,r+1)^\complement = \textbf{R}\setminus(r,r+1)=\left\{\bigcup_{i=1}^\infty (r_i,r_i+1)\Bigg{|}r_i\neq r, r_i+1\neq r+1 \right\} = \text{ another open rational interval.}\end{equation*}Finally, $\bigcap_{i=1}^\infty (r_i,r_i+1)=\emptyset$. Since in this $\sigma$-algebra rational open intervals are closed under the $\sigma$-algebra operations and $\overline{\textbf{Q}}=\textbf{R}$, the smallest $\sigma$-algebra containing $A$ is the open sets of $\mathbb{R}$, that is the Borel $\sigma$-algebra.
For part (b), note that\begin{equation*} \bigcup_{k=1}^\infty [r_k,\infty) = \bigcup_{k=1}^\infty\bigcap_{j=1}^\infty \left(r_k+\frac{1}{j},\infty \right) = \bigcap_{j=1}^\infty\left(\bigcup_{k=1}^\infty \left(r_k+\frac{1}{j},\infty \right) \right)\end{equation*}Similarly,\begin{equation*} [r,\infty)^\complement = \bigcup_{j=1}^\infty \textbf{R}\setminus\left(r_k+\frac{1}{j},\infty \right)=\lim_{N\to\infty}\bigcup_{j=1}^N \textbf{R}\setminus\left(r+\frac{1}{j},\infty \right)=\textbf{R}\setminus(r,\infty)=(-\infty,r)\end{equation*}which can be expressed as a countable union of rational open intervals which is another rational open interval. Finally, $(-\infty,r)\cap[r,\infty)=\emptyset$. Since the intervals in $B$ can be expressed as countable intersections of open rational intervals which we have just demonstrated are closed under countable unions and complementation, we conclude that the smallest $\sigma$-algebra containing $B$ is again the Borel $\sigma$-algebra.
I feel like my approaches are all wrong. If anyone could give some tips on how best to approach problems like this, I'd appreciate it.