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

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

For example $$\{x: f(x) \geq a\}=\mathbb{R}^{d}\setminus \{x: f(x) <a\}$$ 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 $\{x:a<f(x)<b\}$ is measurable.

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


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 $$\{x:f(x)=a\}=\{f(x)<a\}\setminus \{f(x)\leq a\}.$$

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?

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    $\begingroup$ 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. $\endgroup$
    – Daniel
    Sep 9, 2021 at 6:17
  • $\begingroup$ No. You dont answer my question. $\endgroup$ Sep 9, 2021 at 6:59
  • $\begingroup$ Your idea of 'getting' one definition from another makes no sense. $\endgroup$ Sep 9, 2021 at 7:18

1 Answer 1


I shall reformulate essentially what my (now deleted) comment said in this answer with more details.

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 $A$.

The "proof" in your answer of this post does not work because you did not finish your argument about how exactly your condition for measurability implies that every preimage of a measurable set is measurable.

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.

Finally, to answer your question: This is not possible, the definitions are not equivalent because they define different classes of functions.

  • $\begingroup$ I did not know that we must show that "condition for measurability implies that every preimage of a measurable set is measurable.". Thanks! $\endgroup$ Sep 9, 2021 at 8:51

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