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Here is the definition of measurable function from Stein and Shakarchi's Real Analysis: Measure Theory, Integration, and Hilbert Spaces (p.28):

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  1. Talking about [-∞,a] for all a seems like [-∞,∞]. So one would think why not simply say $f^{-1}([-∞,∞])$ (which is actually E) is measurable (and this would make a useless definition). Could you please provide examples showing how these two definitions differ?

  2. Does the range of f need to be measurable? If not, please give me an example.

  3. What is a neat example of a non-measurable function (whose domain is measurable)?

Edit: I think this would make a better definition. However it's apparently limited to finite valued functions.

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    $\begingroup$ Since for a function $f: E \to \mathbb{R}$ the set $f^{-1}[-\infty, +\infty] = E$, requiring that this set is measurable is really no extra condition at all. $\endgroup$
    – Joppy
    Oct 16, 2018 at 11:50
  • $\begingroup$ @Joppy, Exactly, I added that to my post. $\endgroup$
    – asmani
    Oct 16, 2018 at 11:51
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    $\begingroup$ Right, so can you now see why requiring that all the preimages of $[-\infty, a)$ for various $a$ is a different condition? They should be subsets of $E$, and requiring that each of these subsets are measurable is somehow saying that $f$ is cutting up $E$ into measurable bits. $\endgroup$
    – Joppy
    Oct 16, 2018 at 11:53

2 Answers 2

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"Measurable" is the minimal requirement of a function to even think about its Lebesgue integral. You might have seen Wikipedia's page, and in particular the following picture;

Lebesgue integration

The lowest picture describes Lebesgue's integration. There, the base of every rectangle is a set of the form $$\tag{1} \{x\ :\ \lambda_1\le f(x)\le \lambda_2\},$$ and the integral is defined as the "sum" of all such rectangles, with a suitable limiting process involved.

In particular, to even think about integrals it is necessary that the area of the sets (1) is well-defined, and so, it is necessary that these sets are all measurable. It turns out that, to ensure such measurability, it is necessary and sufficient that the sets $$ \tag{2} \{x\ :\ f(x)< a\} $$ are all measurable. Therefore, we say that $f$ is measurable if all the sets (2) are.

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1.

Saying that some expression $p(a)$ that depends on $a$ is true "for all $a$" mean, in mathematics, the following:

Whatever value of $a$ you choose to put into the expression, the statement will always be true.

Therefore, the statement

For all $a\in\mathbb R: f^{-1}([-\infty, a))$ is measurable

is the statement:

Whatever value you choose $a$ to be, the set $f^{-1}([-\infty,a))$ is measurable.

This is not the same as saying that $f^{-1}([\infty, \infty])$ is measurable. The statement says that $f$ is measurable if $f^{-1}([-\infty, 0))$ is measurable and $f^{-1}([-\infty, 2))$ is measurable and $f^{-1}([-\infty, 3242))$ is measurable and $f^{-1}([-\infty, \sqrt2)$ is measurable and $f^{-1}([-\infty, \pi))$ is measurable and $f^{-1}([-\infty, e))$ is measurable and $f^{-1}([-\infty, 2348+\frac{\sqrt[e]{\pi})}{\ln(398453)})$ is measurable and so on and so on.

2.

The definition does not demand that the range of $f$ is measurable.

3.

An example of a non-measurable function would be any indicator function of a non-measurable set.

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  • $\begingroup$ Is the domain of the function in part 3 is not a measurable set, right? So that's not what I asked. $\endgroup$
    – asmani
    Oct 16, 2018 at 11:57
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    $\begingroup$ @Asmani The domain is $\mathbb R$, a measurable set. $\endgroup$
    – 5xum
    Oct 16, 2018 at 11:58
  • $\begingroup$ ahhh, I see, thanks! $\endgroup$
    – asmani
    Oct 16, 2018 at 11:59
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    $\begingroup$ @Asmani I am talking about the function $\chi_A$, defined as $$\chi_A(x)=\begin{cases}1&\text{ if} x\in A\\ 0&\text{ if} x\notin A\end{cases}$$ where $A$ is a non-measurable set. Clearly, the domain of $\chi_A$ is the whole of $\mathbb R$, since $\chi_A(x)$ is defined for every real number. $\endgroup$
    – 5xum
    Oct 16, 2018 at 12:01

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