Is every subset of $\mathbb{R}$ Lebesgue measurable?

Question

Take any subset $X$ of $\,\mathbb{R}$.

Define a constant function $f$ for some $c \in \mathbb{R}$ $f\colon X \to \mathbb{R}\quad \forall x \in X\ f(x) = c.$

$f$ is continuous, thus preimage of any closed set in $\mathbb{R}$ is closed. Take $f^{-1}(\{c\}) = X$

Since $\{c\}$ was closed in $\mathbb{R}$ $X$ is closed in $\mathbb{R}$ too, but every closed set in $\mathbb{R}$ is Lebesgue measurable thus $X$ is measurable.

Answer 1

No. One example of a non-measurable set is a so called Vitali set. The construction is complicated, and I think that it is fair to direct you to the wikipedia page, for example.

The reason that your reasoning doesn't work is quite simple: Consider the open unit interval $(0,1)$, and take $f:(0,1) \rightarrow \mathbb{R}, x \mapsto 0$.

This clearly does not give us that $(0,1)$ is closed in $\mathbb{R}$, because it isn't. It tells us that $(0,1)$ is closed in $(0,1)$.

Answer 2

The answer is no. Consider the interval $X=\left[0, 1\right)$. For $x, y \in X$ define $x\approx y$ to mean $x-y$ is rational. It is easy to see that this is an equivalence relation.

For any set $S\subseteq X$ and $a\in X$ define the translation of $S$ by $a$ to be:

$$t(S,a) = \{s+a: s\in S\ \&\ s+a<1\} \\ \cup \{s+a-1: s\in S\ \&\ s+a\ge 1\}$$

Writing $\left\| Y\right\|$ for the Lebesgue measure of $Y$, it is easy to see:

$$\left\| t(S,a)\right\| = \left\| S\right\|$$

If $E$ is an equivalence class under $\approx$ it is easy to see that:

$$E=t(E, q)\ \mbox{for all}\ q\in X\cap\mathbb{Q}$$

Moreover if $e\in E$ we easily see:

$$E=\bigcup_{q\in X\cap\mathbb{Q}}t(\{e\},q)$$

Using the Axiom of Choice, construct the set $U$ by choosing one element $e$ from each equivalence class $E$.

From our previous observations it follows that:

$$\left[0,1\right) = \bigsqcup_{q\in \left[0,1\right)\cap\mathbb{Q}} t(U,q)$$

where $\sqcup$ denotes disjoint union. But Lebesgue measure is countably additive so if $U$ is measurable:

$$\left\| \left[0,1\right) \right\| =\sum_{q \in \left[0,1\right)\cap\mathbb{Q}} \left\| t(U,q)\right\| $$

But the sets on the rhs all have the same measure as $U$. If this measure is zero the last equation tells us the unit interval has measure $0$, while if this measure is non zero it tells us that the unit interval has infinite measure. Either case contradicts it having measure $1$.

Hence $U$ is unmeasurable.