[Math] Composition of 2 Lebesgue measurable functions is not lebesgue measurable: Are these two functions Borel Measurable

Question

Background

Let $(X,\mathcal{A})$ be a measure space and $\mathcal{B}$ the Borel $\sigma-$algebra. A function $f: (X,\mathbb{\mathcal{A}}) \rightarrow (\mathbb{R},\mathcal{B})$ is $\mathcal{A}$ measurable if for all real $\alpha$, $\{x:f(x)>\alpha\}\in \mathcal{A}$. In particular, if $\mathcal{L}$ is the Lebesgue $\sigma-$ algebra, $f:(\mathbb{R},\mathcal{L}) \rightarrow (\mathbb{R},\mathcal{B})$ is Lebesgue measurable if it satisfies the above definition.

• Claim 1:

  If $f:\mathbb{R} \rightarrow \mathbb{R}$ is Lebesgue measurable and
  $g:\mathbb{R} \rightarrow \mathbb{R}$ is continuous, then $g \circ f$
  is Lebesgue measurable.

  Proof: Let $\alpha \in \mathbb{R}$. Since the inverse image under a continuous function of an open set is an open set,
  $G=\{x:g(x)>\alpha\}$ is open and therefore a Borel set. Since
  $\mathcal{B} \subset \mathcal{L}$, and $f$ is Lebesgue measurable, we
  see that $f^{-1}(G) \in \mathcal{L}$. Therefore $g \circ f$ is
  Lebesgue measurable.

• Claim 2:

  A slight modification of the above proof shows that the above result holds if $g$ is Borel measurable instead of continuous.

• Claim 3 If $f:\mathbb{R} \rightarrow \mathbb{R}$ and $g:\mathbb{R} \rightarrow \mathbb{R}$ are both Lebesgue measurable,
  then $g \circ f$ need not be Lebesgue measurable.

  Proof Let $C$ be the cantor set. Note that $C$ has Lebesgue measure $0$. Let $f: [0,1] \rightarrow C$ be the well known strictly
  increasing function that maps the interval $[0,1]$ onto $C$. Then $f$
  is Borel measurable and therefore Lebesgue measurable. Let $V \subset
  [0,1]$ be the vitali set. Since the Lebesgue measure is complete, we
  find that $f(V) \subset C$ is lebesgue measurable (with measure
  zero). Let $g:C \rightarrow \{0,1\}$ be the indicator function for
  $f(V)$, i.e, $g=\chi_{f(V)}$. Then $g$ is Lebesgue measurable by
  direct verification. However, $\{x:g\circ f>0\}=V$ which is not
  Lebesgue measurable.

Question

Doesn't the counterexample in claim 3 contradict claim 2, due to $f$ and $g$ both being Borel measurable?

Answer 1

The point is precisely that $g$ is Lebesgue measurable, but not Borel measurable.

If it was Borel measurable, we would get the contradiction you describe.

Also note the way the measurability is proved: Completeness of the Lebesgue measure (on the Lebesgue measurable sets) is invoked. When restricted to the Borel sets, this completeness fails.