Correct definition of measurable function

Question

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 $\alpha$.

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

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"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 $\forall E\in T \subseteq \mathcal{P}(\mathbb{R})$ $$f^{-1}(E) := \{x\in X \,\, | \,\, f(x)\in E\} \in \Sigma$$

Answer 1

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.

Answer 2

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.