# Definiton of the measurable function and singleton.

INTRO PART

Definition: We say that function $$f:\mathbb{R}^d\rightarrow \mathbb{R}$$ is measurable if the set $$\{x:f(x) is measurable.

Next we can get many equivalent definitions using set-theoretic operations.

For example $$\{x: f(x) \geq a\}=\mathbb{R}^{d}\setminus \{x: f(x) and this is also the definition of the measurable function.

We can go further and get equivalent definitions like: $$f$$ is measurable if the set $$\{x:a is measurable.

Okey. I think it is all correct and I have not made any mistake.

QUESTION PART

But now let's combine this ideas and get the next "definition": $$f$$ is measurable if the set $$\{x:f(x)=a\}$$ is measurable. And we can prove this. Indeed $$\{x:f(x)=a\}=\{f(x)

But we know that this is the necessary but not sufficient condition for measurability of the function.

Question: How is this possible? How we can get so many equivalent definitions which are correct and in the same time we get wrong "definition" although we use the same method to get it? Where am I wrong?

• Take the goal in showing $f$ is measurable using the first definition. In the equivalent definitions you list, you can show that $\{x : f(x) > a\}$ is measurable for all $a$ with set operations. However, if $\{x : f(x) = a\}$ is measurable for all $a$, it is not always the case that $\{x : f(x) > a\}$ is measurable. Sep 9, 2021 at 6:17
• No. You dont answer my question. Sep 9, 2021 at 6:59
• Your idea of 'getting' one definition from another makes no sense. Sep 9, 2021 at 7:18

The problem, intuitively, with your suggestion, that $$f$$ is measurable if $$\{f=a\}$$ is measurable for all $$a$$, is that the singletons don't generate the $$\sigma$$-algebra. What this means is that you cannot construct the other measurable sets from them with the operations that the $$\sigma$$-algebra provides in order to prove that $$f^{-1}(A)$$ is measurable for every measurable $$A$$.
Now let's actually prove that this cannot work from a rigorous perspective. By Vitalis theorem there is a non-measurable subset $$V\subseteq [0,1]$$. Define $$f:[0,1]\to[-1,1], x\mapsto \begin{cases} x,&x\in V\\ -x,&x\notin V\end{cases}$$ Now note that $$f$$ is injective and thus $$f^{-1}(\{a\})$$ is a singleton or empty for every $$a$$ and thus measurable. However $$f^{-1}([0,1])=V$$ and since $$[0,1]$$ is measurable, this means that $$f$$ is not.