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.


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


1 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.

  • $\begingroup$ Right now it seems we are using claim 2 and contradiction to conclude that $f(V)$ is not Borel measurable... Is there a more direct way of seeing this? $\endgroup$
    – recmath
    Jun 14, 2015 at 3:25
  • 2
    $\begingroup$ @illysial: I don't think so. The problem is that the Borel sigma algebra is defined very "indirectly", i.e. as the smallest sigma algebra containing all open sets. This is somewhat similar to showing that a certain topology is not induced by a metric (say): You find a property which every topology knifed by a metric satisfies (e.g. The first axiom of countability), and prove that the given topology does not satisfy this, rather than taking an arbitrary metric and showing that it does not do the job. I consider the above proof eater "direct" given the indirect definition of Borel sets. $\endgroup$
    – PhoemueX
    Jun 14, 2015 at 17:54

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