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Give an example of a measurable space $(X, S)$ and a function $f:X \to \mathbb{R}$ such that $|f|$ is $S$-measurable but $f$ is not $S$-measurable.

I have considered a function on $(\mathbb{R},\mathcal B)$ where $\mathcal B$ is the collection of Borel sets. If you define a new set $A\subseteq X$ such that $A$ is not measurable, then I think you can arbitrarily define a function that maps

  • $f(x) = 1$ for $x\in A$
  • $f(x) = -1$ for $x \notin A$

Then $|f|$ maps everything to $1$, and $f$ inverse$(a, \infty)$ maps back to the measurable space, right? Or, does this then become NOT a function, because the elemtens of $A$ and not $A$ map to $1$?

drhab
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DJCham
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    A constant function is always measurable. Your idea is okay. The only thing needed is a non-measurable set. – drhab Jan 24 '20 at 13:08
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    You're making it more complicated than it needs to be. Just let $A$ be any nonmeasurable subset of $\mathbb{R}$ and define $f=\textbf{1}A-\textbf{1}{\mathbb{R}\setminus A}$, where the $\textbf{1}$'s are just indicator functions. – Ben W Jan 24 '20 at 13:08

2 Answers2

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"Or, does this then become NOT a function, because the elemtens of $A$ and not $A$ map to $1$?..."

Your idea is fine but what you state in your last line makes me think that you have some confusion with the concepts of inverse of a function $f$ and the preimage of a set wrt function $f$.

For being measurable in this context it is enough if for every $a\in\mathbb R$ the set $f^{-1}((a,\infty)):=\{x\in X\mid f(x)\in(a,\infty)\}$ is an element of $\sigma$-algebra $S$.

Now if $f$ is a constant function (as in your case) taking some value $c$ then thereare only two possibilities for $f^{-1}((a,\infty))$:

  • $f^{-1}((a,\infty))=X$ if $c>a$
  • $f^{-1}((a,\infty))=\varnothing$ if $c\leq a$

By definition a $\sigma$-algebra $S$ on $X$ contains both sets so the function is measurable.

Do not confuse with something like "inverse function" $f^{-1}$ which is not a relevant issue here.

drhab
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I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
This exercise is Exercise 9 on p.38 in Exercises 2B in this book.

I used the following example in this book to solve Exercise 9.

2.36 Example measurable functions

  • If $\mathcal{S}=\{\emptyset,X\}$,$\dots$
  • If $\mathcal{S}$ is $\dots$
  • If $\mathcal{S}=\{\emptyset,(-\infty,0),[0,\infty),\mathbb{R}\}$ (which is a $\sigma$-algebra on $\mathbb{R}$), then a function $f:\mathbb{R}\to\mathbb{R}$ is $\mathcal{S}$-measurable if and only if $f$ is constant on $(-\infty,0)$ and $f$ is constant on $[0,\infty)$.

My solution:

Let $(X,\mathcal{S}):=(\mathbb{R},\{\emptyset,(-\infty,0),[0,\infty),\mathbb{R}\})$.
Then $(X,\mathcal{S})$ is a measurable space.
Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that $f(x)=1$ if $1\leq x$ and $f(x) = -1$ if $x<1$.
Then, $|f|(x)=1$ for any $x\in\mathbb{R}$.
Since $|f|$ is constant on $(-\infty,0)$ and $|f|$ is constant on $[0,\infty)$, $|f|$ is $\mathcal{S}$-measurable by 2.36.
Since $f(0)=-1$ and $f(1)=1$, $f$ is not constant on $[0,\infty)$.
So, $f$ is not $\mathcal{S}$-measurable by 2.36.

tchappy ha
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