Borel algebra on $\mathbb R$: why not the discrete one? I'm only beginning with measure theory so I'm sorry if my question has an obvious and immediate answer. My teacher motivated the introduction of the  Borel algebra on $\mathbb{R}$ by claiming that the discrete $\sigma$-algebra on $\mathbb{R}$ is too big and that it contains sets that aren't measurable and thus we need the more appropriate Borel algebra; but he did not justify this claim. Can someone shed some light on his claim?
 A: A $\sigma$-algebra is merely a tool towards defining a measure - which is a function whose domain is the $\sigma$-algebra. The first non-trivial example of a measure you are likely to study is the Lebesgue measure, which generalizes the idea of the length of an interval
$$
l([a,b]) = b-a
$$
The length has a nice property called translation-invariance: ie. if you a shift an interval by a fixed distance $d$, then the length of the new interval $[a+d,b+d]$ is the same as that of $[a,b]$.
This is a desirable property for many reasons, and this restricts the kind of sets that can be "measured". Wait until you come across the standard construction of a non-measurable set. Although that uses the Axiom of Choice, it should indicate why "translation-invariance" does not allow all sets to be measurable.
A: Here's an example, which is a minor modification of two well-known constructions:

There exists a subset of $A$ of  $\Bbb R$ that is both a Vitali set and a Bernstein set.

Proof. Let $\mathfrak{c} = \aleph_\alpha$,thus we can define a subset of points of $\Bbb R$ ex post and index all closed sets of $\Bbb R$ ex ante by $\{x_\kappa: \kappa < \aleph_\alpha\}$ and $\{F_\lambda: \lambda < \aleph_\alpha\}$ respectively.
Suppose that, for $\xi < \mathfrak{c}$, $\{x_\lambda : \lambda < \xi\}$ has been defined, and so is $Z_\xi = \bigcup \{x_\lambda + \Bbb Q : \lambda < \xi\}$.
Of course, $|Z_\xi| \leq \max \{ \xi, \aleph_0\} < \mathfrak{c} $ and $|F_\xi| = \mathfrak{c}$(shown by that it contains an isomorphic copy of a Cantor set), so by an uncountable form of $\sf AC$, we're able to choose an element $x_\xi$ from $F_\xi \setminus Z_\xi $. Continuing this process of transfinite recursion lead us to the construction of $A = \{x_\kappa: \kappa < \aleph_\alpha\}$.
Remark. $A$ is a Vitali set in the sense that $|A \cap x+\Bbb Q| \leq 1$. $A$ is a Bernstein set in the sense that $A$ doesn't contain any perfect set.
