Trying to find two functions $\varphi$ measurable and $f$ continuous such that $\varphi\circ f$ isn't measurable

Question

A set $E \subseteq \mathbb{R}$ is measurable if for any subset $A\subset\mathbb{R}$, $$\mu(A)=\mu(A\cap E)+\mu(A\cap E^c).$$ where $\mu$ is the Lebesgue outer measure.

A function $\varphi : \mathbb{R} \to \mathbb{R}$ is measurable if for any $\alpha\in\mathbb{R}$ the set $E_\alpha:=\{x\in\mathbb{R} : \varphi(x)>\alpha\}$ is measurable.

Let $\varphi : \mathbb{R} \to \mathbb{R}$ be a measurable function, and let $\ \ f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function.

For two such functions we know that the composite fonction $f\circ\varphi$ is measurable because of a result that say $\varphi^{-1}(A)$ is measurable for any borelian set $A$ (our borelian here is $f^{-1}(]\alpha,\infty[)$).

Now I would like to know why for two functions $\ f$ and $\varphi$ like above $\varphi\circ f$ is not necessarily a measurable function.

Edit: The first answer of this question gives an example of such a pair of functions.

Answer 1

I believe my answer was incorrect. I'm leaving it here instead of deleting it as a warning: the reason it's wrong points out a peculiar aspect of the terminology. See the last paragraph.

(For a correct answer see the previous question that Carl Mummert found: While this queston is not really a duplicate of that question, it happens that the top-rated answer to that question also answers this one.)

Seems to me that there is no such example.

Fix $\alpha$, and let $E=\{x:\phi(x)>\alpha\}$. Then $$\{\phi\circ f>\alpha\}=f^{-1}(E),$$which is measurable since $f$ and $E$ are measurable.

No, that's not right. In the context of pure measure theory a function is measurable if the inverse image of any measurable set is measurable. But here "Lebesgue measurable" means just that the inverse image of any Borel-measurable set is Lebesgue measurable, not the inverse image of any Lebesgue measurable set.

This is not the first time this has tripped me: In the pure measure theory context it's trivial that the composition of two measurable functions is measurable, as above. Regardless the composition of two Lebesgue measurable functions need not be Lebesgue measurable... (doesn't contradict that theorem because "Lebesgue measurable" does not mean "Lebesgue-to-Lebesgue measurable".)

Answer 2

You may need to clarify the notations.

1. Does $\mu$ denote the Lebesgue outer-measure? If that is the case, your measurable really means Lebesgue measurable.

2. Let $\mathcal{B}$ be the $\sigma$-algebra of all Borel sets, i.e., the smallests $\sigma$-algebra on $\mathbb{R}$ that contains all open subsets of $\mathbb{R}$. Let $\mathcal{L}$ be collection of all Lebesgue measurable sets. It turns out that $\mathcal{L}$ is a $\sigma$-algebra and it is well-known that $\mathcal{B}\subsetneq\mathcal{L}$. Moreover, $\mathcal{L}$ is the completion of $\mathcal{B}$ with respect to the Lebesgue measure.

3. Under the general setting, if $(X,\mathcal{\mathcal{M}}_{X})$ and $(Y,\mathcal{M}_{Y})$ are two measurable spaces and $f:X\rightarrow Y$ is a mapping, we say that $f$ is $\mathcal{M}_{X}/\mathcal{M}_{Y}$-measurable if $f^{-1}(A)\in\mathcal{M}_{X}$ for each $A\in\mathcal{M}_{Y}$. (This definition is analogous to the definition of continuous mapping in general topology.) To be precise, your measurable function is actually $\mathcal{L}/\mathcal{B}$-measurable function.

4. If $\varphi:\mathbb{R}\rightarrow\mathbb{R}$ is $\mathcal{B}/\mathcal{B}$-measurable and $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous, then $\varphi\circ f$ is always $\mathcal{B}/\mathcal{B}$-measurable (and hence $\mathcal{L}/\mathcal{B}$-measurable).

5. However, if $\varphi:\mathbb{R}\rightarrow\mathbb{R}$ is $\mathcal{L}/\mathcal{B}$-measurable, then it is possible to find a continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $\varphi\circ f$ is not $\mathcal{L}/\mathcal{B}$-measurable. Although I forget the details, I remember that a counter-example can be constructed by choosing $f$ to be something related to Cantor set. Moreover, in that case, $f$ is also strictly increasing, a.e. differentiable with $f'(x)=0$ for $x$-a.e. You may consult the book ``Real Analysis ...'' written by Folland and the counter-example is included in the book as an exercise. (Sorry... I forget which chapter as I read and drilled it 13 years ago).