problem of proving one member of union is sigma-algebra Suppose $\mathscr{A}$ is $\sigma$-algebra. Let $A\in\mathscr{A}$. If $B\cup A\in \mathscr{A}$, is $B\in \mathscr{A}$? Can someone give hint?
 A: My hint was a little bit more advanced for beginning measure theory, so here's a trivial example. Let $\mathcal{A}=\{\emptyset, X\}$ for some $X$ nonempty. You can show that this defines a sigma algebra. Then $X$ nonempty means there is $x\in X$, so $\{x\}\subset X$. Then $\{x\}\cup X=X$ but $\{x\}\not\in \mathcal{A}$
Edit: Yes, you need to assume $|X|\geq 2$, or else $X=\{x\}$ and the result does hold.
Now, for my comment. This is going to be based on Chapter 1 of Folland's Real Analysis, but you can also read about it here, it's called the Vitali Set.
Here's the basic idea: We have a $\sigma-$algebra on $\mathbb{R}$ which is generated by open intervals (unions, intersections, complements) called the Borel $\sigma-$algebra. Now, to every interval, we associate a natural size $m((a,b))=b-a$, and we want to extend this so that points $m(\{x\})=0$ and $m(\bigsqcup_n U_n)=\sum_nm(U_n)$, where this union is disjoint (this is an intuitive description of Lebesgue measure). From topology, every open set is a countable disjoint union of open intervals, and a little more work can show then that every element of our Borel $\sigma-$algebra has a well defined size (possibly infinite). We're also going to want translational invariance $m(r+U)=m(U)$ for $r$ real and if $U\subset V$, then $m(U)\leq m(V)$
If we agree on this, we proceed. We're going to construct a terrible subset of $[-1,2]$. Define an equivalence relation on $(0,1)$ by $x\sim y\iff x-y\in\mathbb{Q}$. Now consider $(0,1)/\sim$, the equivalence classes, and for each one pick an element from each class, and put them all into a set $N\subset (0,1)$. Now, enumerate the rationals $r_n=\mathbb{Q}\cap [-1,1]$ and consider the sets $N_n=N+r_n$ and take the union $\mathcal{N}=\bigsqcup N_n$. I leave it to you to check the union is disjoint. You also need to check that we have $[0,1]\subset \mathcal{N}\subset [-1,2]$.
Now here's where things start to break down. $m([0,1])=1$, $m([-1,2])=3$. Then $$1\leq m(\mathcal{N})\leq 3\Rightarrow 1\leq \sum_n m(N_n)\leq 3\Rightarrow 1\leq m(N)\sum_n\leq 3$$Now we're in trouble. The sum doesn't converge so we need $m(N)=0$, but this implies $1\leq 0$. If $m(N)>0$, then $\infty\leq 3$. Both are impossible, so $m(N)$ must not exist. But we agreed above that every element of our $\sigma-$algebra has a size. So $N$ cannot be a member of this. But $N\subset (0,1)$, so $N\cup (0,1)=(0,1)$ is a member of the set. 
Edit 2: I did a fairly standard construction of the Vitali set, but it relies heavily on the idea of Lebesgue measure and accepting certain things as fact. Here's an example of a non-Borel set due to Lusin using continued fractions.
A: Hint: Suppose it were true. Then it would be true whenever $B \subset A$. Is every subset of a set in a $\sigma$-algebra also contained in the $\sigma$-algebra. You should be able to choose appropriate $A$ and $\sigma$-algebra to figure this out.
