Is the set A measurable?

Question

I am having problem with the following question:

If $A=\{\text{sin }x^2 | x \in [0,1]\cap\mathbb{Q}\}$, then is $A$ measurable?

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My work so far:

Let $f:[0,1]\to\mathbb{R}$ be defined as $f(x) =\text{sin }x^2\quad$ $\forall x\in[0,1]\cap\mathbb{Q}$ Then, $f$ is absolutely continuous, and $m([0,1]\cap\mathbb{Q})=0.$ As absolutely continuous functions on closed intervals take measurable sets to measurable sets, and $[0,1]\cap\mathbb{Q}\subset[0,1]$ then, $m(f([0,1]\cap\mathbb{Q}))=0$ $\Rightarrow m(A)=0$ Thus, $A$ is measurable.

Is this work valid? Also I would love to have some inputs on how to make this proof better, by improving some subtle mistakes.

Answer 1

Your overall argument strategy is sound. However, some thoughts:

1. Along the way to proving that $A$ is measurable, you've concluded that it has measure 0, which is a stronger statement. While it's great to prove something stronger than your original goal, it should prompt you to go back and check that your argument actually supports it. In this case, it really doesn't work -- you justify why $[0,1] \cap \mathbb{Q}$ being measurable implies that $A = f([0, 1] \cap \mathbb{Q})$ is measurable (ie by the lemma you reference), but not anything about how $[0,1] \cap \mathbb{Q}$ having measure 0 implies anything about the measure of $A$. If you really want, there's probably a way to justify that, but I would just drop that from your argument.
2. Be careful with your definitions. Your argument assumes that $f$ is absolutely continuous on a closed domain, but your definition of it refers to $[0,1] \cap \mathbb{Q}$. Also, it is important to specify that $f$ is absolutely continuous on $[0,1]$. While just saying ''$f$ is absolutely continuous'' is valid, in that it by definition refers to $f$ being absolutely continuous on its domain $[0,1]$, stating that explicitly is good practice.
3. Lastly, be careful about your assumptions. You explicitly refer to two, that $f$ is absolutely continuous and that absolutely continuous functions on closed intervals map measurable sets to measurable sets. These are fine, since they are hard to miss and thus more likely to be caught if they are incorrect. Next, you state that $m([0, 1] \cap \mathbb{Q}) = 0$. While technically this is an explicit assumption, to me it reads as a reflection of the implicit assumption that the measure $m$ is a complete measure, using that $[0,1] \cap \mathbb{Q}$ is a subset of $\mathbb{Q}$. I assume you are thinking with respect to the Lesbegue measure, so it is not incorrect, but personally I would write out why $[0,1] \cap \mathbb{Q}$ is measurable and has measure zero. Lastly, in concluding that $m(A) = 0$, you implicitly assume that absolutely continuous functions map measure-zero sets to measure-zero sets. This is probably true, but personally I see it as a big red flag -- I can't tell you how many times I've run into an innocuous statement like that and found a substantial logic issue hiding underneath (in my own work).