Howcome there is this simple counterexample to a result about measurable functions?

Question

There is a result that states:

If $\varphi$ is a measurable (in the sense that $\{x\ :\ \varphi(x)<\alpha\}\in\mathcal M\ \forall\alpha$) function, $f$ is a continuous function $\implies f\circ\varphi$ is measurable (but $\varphi\circ f$ is not necessarily measurable).

But if we take $\varphi(x)=x$ and $f$ a continuous non measurable function (Cantor's staircase function, or Cantor_function$(x)+x$ is we want it to be strictly increasing for whatever reason) then $f\circ\varphi=f$ is non measurable. What am I doing wrong?

Answer 1

There is no such thing as a continuous non-measurable function.

Answer 2

When talking about measurability, we have to be very careful how we construe the domain and codomain.

There are two reasonable notions of measurable function from $\mathbb{R}$ to $\mathbb{R}$:

• The preimage of every measurable set is measurable.

• The preimage of every Borel set is measurable. (EDIT: and as GEdgar comments below, if one doesn't state otherwise the general assumption is that "measurable function" refers to this sense - which is why the fact you've quoted is simply stated for "measurable functions.")

That is, in each case we think of the domain as equipped with the $\sigma$-algebra of measurable sets, but the codomain might be equipped with either of the $\sigma$-algebras of measurable sets or Borel sets. In one sense there are no continuous non-measurable functions, while in the other there are. See this old MSE question for details.

Your confusion comes because you're mixing two notions of measurability; the fact you quote in the question is true when we look at the codomain with the Borel algebra, while the existence of a continuous nonmeasurable function is true when we look at the codomain with the full algebra of measurable sets.

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To wrap up a possible loose end: the definition you refer to doesn't mention a $\sigma$-algebra at all, it just talks about the preimages of open sets. But it's easy to show that if the preimage of every open set is measurable, then the preimage of every Borel set is measurable since the Borel $\sigma$-algebra is generated by the open sets. By contrast, the $\sigma$-algebra of measurable sets is vastly bigger, and so merely having measurable preimages of open sets doesn't in any way guarantee measurable preimages of measurable sets.

Answer 3

A continuous function is always measurable, as you are talking about Lebesgue measure. This is because a function is continuous precisely when the sets $\{f <\alpha\} $ are open.

Your function is an example of a function where some preimages of measurable sets are not measurable. But all preimages of Borel sets are measurable.

This assymetry, that we consider measurable sets in the domain and Borel in the codomain, is due to integration. When you want to define the Lesbesgue integral, you notice that you need a notion of "near" in the codomain (to take limits), but in the domain you only need measurability.

Answer 4

We can only speak of a continuous function $X\to Y$ if $X$ and $Y$ are both topological spaces, hence are equipped with a topologies $\tau_X$ and $\tau_Y$.

Analogously we can only speak of a measurable function $X\to Y$ if $X$ and $Y$ are both measurable spaces, hence are equipped with a $\sigma$-algebra $\sigma_X$ and $\sigma_Y$.

If $f:X\to Y$ is continuous, then can we conclude that $f$ is measurable?

That depends on the question how these topologies and $\sigma$-algebras are related.

Commonly one works with $\sigma_X=\sigma(\tau_X)$ and $\sigma_X=\sigma(\tau_X)$ where $\sigma(\mathcal V)$ denotes the smallest $\sigma$-algebra that contains $\mathcal V$.

In that case $f$ can be shown to be measurable. This because it can be deduced in general that:$$f^{-1}(\sigma(\mathcal V))=\sigma(f^{-1}(\mathcal V))\tag1$$ For a proof of that see here.

Function $f$ is continuous if and only if $f^{-1}(\tau_Y)\subseteq\tau_X$ so that in that case: $$f^{-1}(\sigma_Y)=f^{-1}(\sigma(\tau_Y))=\sigma(f^{-1}(\tau_Y))\subseteq\sigma(\tau_X)=\sigma_X$$where the second equality is an application of $(1)$.

This states exactly that $f$ is measurable.