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I would like to prove or give a counterexample to the following. Thanks for any help in advance.

Let $f$ : R → R be a function.

a) Suppose $f^{2}$ is Lebesgue measurable. Does it follow that $f$ is Lebesgue measurable?

b) Suppose $f^{3}$ is Lebesgue measurable. Does it follow that $f$ is Lebesgue measurable?

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  • $\begingroup$ Does $f^2$ here mean $f\circ f,$ as your title seems to imply? Or does it mean the function $x \mapsto (f(x))^2,$ the way $f^2$ would usually be interpreted? (And the same question for $f^3,$ of course.) $\endgroup$
    – Mitchell Spector
    Sep 28, 2016 at 16:37
  • $\begingroup$ Oops. I have made a mistake with the title. I meant the second meaning. $\endgroup$
    – shmiggens
    Sep 28, 2016 at 17:30
  • $\begingroup$ Thanks -- I noticed people were answering assuming the second meaning. $\endgroup$
    – Mitchell Spector
    Sep 28, 2016 at 17:59

3 Answers 3

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Consider the function

$ f(x) =\begin{cases} &\frac{1}{|x|} & x>1 \\ &1 & |x| \le 1. \end{cases} $

It's easy to see that this is not integrable where as its square and its cube is.

Edit: Sorry I miss-read integrable instead of measurable. To answer the actual question...consider

$ f(x) =\begin{cases} &-1 & x\in \{\text{Some non-measurable set}\} \\ &1 &\text{otherwise.} \end{cases} $

Clearly $f^2$ is measurable where as $f$ is not.

For the second part: Note that $x^\frac{1}{3}$ is monotonic increasing hence measurable and since the composition of measurable functions is measurable we can deduce $f^3$ measurable implies $f$ measurable.

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  • $\begingroup$ Whoops sorry! I'll change it. $\endgroup$
    – Zestylemonzi
    Sep 28, 2016 at 16:11
  • $\begingroup$ I am unsure about the second part. You have picked a specific $f$ and showed that $f^{3}$ is measurable. I don't see how that is a proof. $\endgroup$
    – shmiggens
    Sep 28, 2016 at 16:32
  • $\begingroup$ I am not choosing $f(x) = x^\frac{1}{3}$. I am pointing out that $g(x) = x^\frac{1}{3}$ is measurable. Since the composition of measurable functions is measurable, if $f^3$ is measurable then $g \circ f^3 = f$ is measurable. $\endgroup$
    – Zestylemonzi
    Sep 28, 2016 at 16:36
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Hints: a) think about $$f = \chi_E - \chi_{\mathbb R \setminus E}$$ b) if $g$ is Lebesgue measurable, then so is $h\circ g$ for every continuous $h$ on $\mathbb R.$

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By a composition being Lebesgue measurable I assume you mean $f \circ g$ is $(\mathcal{L}, \mathcal{B})$-measurable where $\mathcal{L}$ is the Lebesgue sigma algebra and $\mathcal{B}$ is the Borel sigma algebra. We want to know whether $f$ is $(\mathcal{L}, \mathcal{B})$-measurable.

Let $g(y) = y^2$ and $f(x): \mathbb{R} \to \mathbb{R}$. Then for any $B \in \mathcal{B}$ we necessarily have $f^{-1}(g^{-1}(B)) \in \mathcal{L}$ by assumption. To show $f$ is $(\mathcal{L}, \mathcal{B})$-measurable we need that $f^{-1}(A) \in \mathcal{L}$ for any $A \in \mathcal{B}$. But, if we let $B = g(A)$ for some $A \in \mathcal{B}$ then we have only that $A \subset g^{-1}(B)$ and know only that $f^{-1}(g^{-1}(B)) \in \mathcal{L}$; there is no guarentee that $f^{-1}(A) \in \mathcal{L}$. In other words, there exist Borel sets $A$ for which we cannot be sure $f^{-1}(A) \in \mathcal{L}$, so $f$ is not necessarily $(\mathcal{L}, \mathcal{B})$-measurable.

If $g$ were injective on $\mathbb{R}$ like $g(y) = y^3$ then we would have $A = g^{-1}(g(A))$ and then $f$ would be $(\mathcal{L}, \mathcal{B})$-measurable.

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