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I've read this and many other questions in here, but while the question I've just posted is very similar, it is not exactly the same as mine.

I have (as in the linked question) found two definitions of a measurable function:

Definition 1 (see here): Let $(X, \Sigma)$ and $(Y, T)$ be measurable spaces. A function $$f:X\longrightarrow Y$$ is said to be measurable if $\,\,\forall E\in T$ $$f^{-1}(E) := \{x\in X \,\, | \,\, f(x)\in E\} \in \Sigma $$ That is, if the preimage of any $T$-measurable set is $\Sigma$-mesurable.

Definition 2 (see here): Let $(X, \Sigma)$ be a measurable space and $E\in\Sigma$ be a $\Sigma$-measurable set. A function $$f:E\longrightarrow \mathbb{R}$$ is said to be $\Sigma$-measurable on $E$ iff $\,\,\forall \alpha \in \mathbb{R}$ $$\{x\in E \,\, | \,\, f(x) \leq \alpha\}\in \Sigma$$ That is, if given any $\Sigma$-measurable set $E$, all the sets of elements of $E$ whose image under $f$ is less than or equal than some real number $\alpha$.


To me these definitions are miles away, super different. How can they be the same?


"Equivalence" proof

Here I try to show that they are somewhat similar, following the answer from Yanko.

Let $(X, \Sigma)$ and $(Y, T)$ be measurable spaces and let $f:X\longrightarrow Y$ be a measurable function. Now, we set $Y = \mathbb{R}$ and so we work with $(X, \Sigma)$ and $(\mathbb{R}, T \subseteq\mathcal{P}(\mathbb{R})$ where $\mathcal{P}(\mathbb{R})$ indicates the power set of $\mathbb{R}$. The our function $f$ becomes $$f: X\longrightarrow \mathbb{R}$$ and by definition of a measurable function it is such that $\forall E\in T \subseteq \mathcal{P}(\mathbb{R})$ $$f^{-1}(E) := \{x\in X \,\, | \,\, f(x)\in E\} \in \Sigma$$

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    $\begingroup$ The equivalence can only be proved if $\mathbb R$ is equipped with the Borel $\sigma$-algebra. In other cases you can prove only one side. $\endgroup$ – Vera Jul 6 '18 at 13:36
  • $\begingroup$ @Vera could you please show me a proof? or direct me to a proof? $\endgroup$ – Euler_Salter Jul 6 '18 at 13:38
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    $\begingroup$ If you write the second version as $f^{-1}((-\infty,\alpha)) \in \Sigma$ for all $\alpha$, it's not that different than the first version. $\endgroup$ – littleO Jul 6 '18 at 13:39
  • $\begingroup$ Cannot find it that easily (but probably it can be found somewhere here). This answer might help if you replace $\mathcal O$ by the collection of intervals of the form $(-\infty,x]$. $\endgroup$ – Vera Jul 6 '18 at 13:49
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They're not the same. The first definition is much more general than the second.

As you can see, the second definition is only defined for functions which take values in $\mathbb{R}$ while the first definition is defined for arbitrary measure spaces.

Indeed, if in definition 1 you take $Y=\mathbb{R}$ then you get (almost) the same definition as in definition 2. The difference is that instead of a general measurable $\sigma \subseteq\mathbb{R}$ you take sets of the form $(-\infty,\alpha)$. It is a good exercise to check why this suffices.

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  • $\begingroup$ Thanks for your answer, I am still quite new to this. So if the first definition is the general definition of a measurable function, what do we call functions satisfying the second definition? $\endgroup$ – Euler_Salter Jul 6 '18 at 13:08
  • $\begingroup$ @Euler_Salter The first is a "measurable function from $X$ to $Y$". The second is a "measurable function from $X$ to $\mathbb{R}$." $\endgroup$ – Yanko Jul 6 '18 at 13:09
  • $\begingroup$ But in the second definition the function starts from $E\subseteq X$ not from $X$! $\endgroup$ – Euler_Salter Jul 6 '18 at 13:12
  • $\begingroup$ I am trying to prove that I can obtain the second definition $\endgroup$ – Euler_Salter Jul 6 '18 at 13:17
  • $\begingroup$ @Euler_Salter You are right, but if $X$ is a measure space so is every subset of $E$. Well then, a function which satisfies definition 2 is a "measurable function from $E$ to $\mathbb{R}$". $\endgroup$ – Yanko Jul 6 '18 at 16:58
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You should go for the first as definition.

The second is not more than a (handsome) sufficient condition for $f:X\to\mathbb R$ to be measurable under the extra condition that $\mathbb R$ is equipped with the Borel $\sigma$-algebra, which is the smallest $\sigma$-algebra that contains all open sets if $\mathbb R$ is equipped with its usual (order)-topology.

The condition is also necessary.

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