# Correct definition of measurable function

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

• 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. – Vera Jul 6 '18 at 13:36
• @Vera could you please show me a proof? or direct me to a proof? – Euler_Salter Jul 6 '18 at 13:38
• 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. – littleO Jul 6 '18 at 13:39
• 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 $(-\infty,x]$. – Vera Jul 6 '18 at 13:49

## 2 Answers

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.

• 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? – Euler_Salter Jul 6 '18 at 13:08
• @Euler_Salter The first is a "measurable function from $X$ to $Y$". The second is a "measurable function from $X$ to $\mathbb{R}$." – Yanko Jul 6 '18 at 13:09
• But in the second definition the function starts from $E\subseteq X$ not from $X$! – Euler_Salter Jul 6 '18 at 13:12
• I am trying to prove that I can obtain the second definition – Euler_Salter Jul 6 '18 at 13:17
• @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 $\mathbb{R}$". – Yanko Jul 6 '18 at 16:58

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.