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In some situations, you need to show Lebesgue-measurability of some function on $\mathbb{R}^n$ and the verification is kind of lengthy and annoying, and even more so because measurability is "obvious" because "why would it not be".

In such a situation, I have heard the argument: The function is clearly measurable, because the axiom of choice was not used to define it.

This argument makes some sense, because (as far as I know, I am not an expert) there are models of ZF (no C), where every function on $\mathbb{R}^n$ is Borel measurable. So, suppose that we have a function $f$ in our model of ZFC constructed without the axiom of choice, then it is also a function in the model of ZF constructed above. Hence it is measurable there, and "clearly" all functions that are measurable in the model above are also measurable in our model.

But the question is: The bold statement above is very "meta". So how rigorous is this argument? Can it be made rigorous?

/Edit: I changed "Borel"- to "Lebesgue"-Measurable.

In some situations, you need to show Lebesgue-measurability of some function on $\mathbb{R}^n$ and the verification is kind of lengthy and annoying, and even more so because measurability is "obvious" because "why would it not be".

In such a situation, I have heard the argument: The function is clearly measurable, because the axiom of choice was not used to define it.

This argument makes some sense, because (as far as I know, I am not an expert) there are models of ZF (no C), where every function on $\mathbb{R}^n$ is Lebesgue measurable. So, suppose that we have a function $f$ in our model of ZFC constructed without the axiom of choice, then it is also a function in the model of ZF constructed above. Hence it is measurable there, and "clearly" all functions that are measurable in the model above are also measurable in our model.

But the question is: The bold statement above is very "meta". So how rigorous is this argument? Can it be made rigorous?

/Edit: I changed "Borel"- to "Lebesgue"-Measurable.

In some situations, you need to show Borel-measurability of some function on $\mathbb{R}^n$ and the verification is kind of lengthy and annoying, and even more so because measurability is "obvious" because "why would it not be".

In such a situation, I have heard the argument: The function is clearly measurable, because the axiom of choice was not used to define it.

This argument makes some sense, because (as far as I know, I am not an expert) there are models of ZF (no C), where every function on $\mathbb{R}^n$ is Borel measurable. So, suppose that we have a function $f$ in our model of ZFC constructed without the axiom of choice, then it is also a function in the model of ZF constructed above. Hence it is measurable there, and "clearly" all functions that are measurable in the model above are also measurable in our model.

But the question is: The bold statement above is very "meta". So how rigorous is this argument? Can it be made rigorous?

In some situations, you need to show Lebesgue-measurability of some function on $\mathbb{R}^n$ and the verification is kind of lengthy and annoying, and even more so because measurability is "obvious" because "why would it not be".

In such a situation, I have heard the argument: The function is clearly measurable, because the axiom of choice was not used to define it.

This argument makes some sense, because (as far as I know, I am not an expert) there are models of ZF (no C), where every function on $\mathbb{R}^n$ is Borel measurable. So, suppose that we have a function $f$ in our model of ZFC constructed without the axiom of choice, then it is also a function in the model of ZF constructed above. Hence it is measurable there, and "clearly" all functions that are measurable in the model above are also measurable in our model.

But the question is: The bold statement above is very "meta". So how rigorous is this argument? Can it be made rigorous?

/Edit: I changed "Borel"- to "Lebesgue"-Measurable.