# Is the composition of two Lebesgue Measurable functions necessarily Lebesgue measurable?

In class, my professor said that this was not the case, but I'm not understanding why not? What is the intuition for this, and can you give an example of two Lebesgue measurable functions whose composition is non-Lebesgue measurable? I'm seeing some answers on here that involve Borel Measure which we have not gone over yet, so I'm wondering how this works without talking about Borel measurability.

For reference, we've defined a function $$f$$ on measurable set $$\Omega \subset \mathbb{R}$$ to be Lebesgue measurable if $$f^{-1}((a, \infty))$$ is measurable for any $$a \in \mathbb{R}$$.

• The problem is that a Lebesgue measurable set (i.e. what your pre-image set is) is not going to be in the "Borel Hierarchy" (i.e. a nice set that can be built out of countably many intersections, unions, and complements of open sets). Thus you don't know if the preimage of a Lebesgue measurable set is going to be Lebesgue measurable. There are Lebesgue measurable sets which are not in the Borel Hierarchy. – Robert Wolfe May 9 at 16:48
• A more succinct way to say it might be that the definition of a Lebesgue measurable function is not "symmetric" in domain and codomain (i.e. the preimage of a Borel measurable set must be a LEBESGUE measurable set---which is a wider class of sets). So these functions might not be closed under composition. – Robert Wolfe May 9 at 16:52

The basic reason is that a Lebesgue measurable function is one with the property that preimages of Borel sets are Lebesgue measurable. The main reason why we define things that way is because not all continuous functions are Lebesgue-Lebesgue measurable, so if we defined measurability through Lebesgue-Lebesgue measurability then the Lebesgue integral would not be an extension of the Riemann integral.

Anyway, the structural reason why this has anything to do with composition is because:

$$(f \circ g)^{-1}(A)=g^{-1}(f^{-1}(A)).$$

Thus if $$A$$ is Borel and $$f,g$$ are Lebesgue-Borel measurable, then $$f^{-1}(A)$$ might be Lebesgue measurable but not Borel measurable, and so $$g^{-1}(f^{-1}(A))$$ could fail to even be Lebesgue measurable. To do this construction, it suffices to take $$g$$ to be a Lebesgue-Borel measurable function which is not Lebesgue-Lebesgue measurable, find a set $$M$$ which is Lebesgue measurable such that $$g^{-1}(M)$$ is not Lebesgue measurable, and then set $$f=1_M,A=\{ 1 \}$$.

As you may be aware, making up Lebesgue nonmeasurable sets is never really "constructive", but there is still a way to build such a $$g$$ and show it has the necessary properties, provided we can conjure up a suitable Lebesgue nonmeasurable set in the course of the proof. To do that, we look at

$$h : [0,1] \to [0,2], h(x)=c(x)+x$$

where $$c$$ is the standard Cantor function. It is easy to see that $$h$$ is a strictly increasing continuous function. It's perhaps less easy to see that $$\mu(h(C))=1$$ where $$C$$ is the standard Cantor set; you might call this lemma 1. With that in hand, you can select a Lebesgue nonmeasurable set $$N \subset h(C)$$ (you might call this ability to construct a nonmeasurable subset of a positive measure set lemma 2).

Now $$h^{-1}(N)$$ is a Lebesgue measurable set (because it is a subset of $$C$$). But its preimage under $$h^{-1}$$ (i.e. $$N$$) isn't Lebesgue measurable. It turns out this is only possible because it is not Borel, because $$h(B)$$ is Borel for every Borel $$B$$ since $$h$$ is strictly increasing.

In the end we can take $$g=h^{-1},M=h^{-1}(N)$$ in the notation of the third paragraph.

I wrote this answer essentially by following along with https://www.math3ma.com/blog/lebesgue-but-not-borel but the answer is complete even if this link dies.