# Mortivation for the definition of measurable function?

1) We say that $f:\mathbb R^n\to \mathbb R$ is measurable if $$\{x\in\mathbb R^n\mid f(x)<\alpha \},\tag{D}$$ is measurable for all $\alpha$. What is the motivation for such definition ?

We can define for example continuity as : $f^{-1}(O)$ is open in $\mathbb R^n$ for all open set in $\mathbb R$. A possible definition for measurability could be for example $f^{-1}(O)$ is measurable for all open of $\mathbb R$, no ? So what is the motivation for the definition (D) ?

2) In more abstract measurable spaces, $f:(X,\mathcal F)\to (Y, \mathcal G)$ we say that $f$ is measurable if $$f^{-1}(U)\in \mathcal F\tag{D'}$$ for al $U\in \mathcal G$. Is there a correlation with the definition (D) ? For example, if $Y$ is a space with an order $\mathcal R$, would it be equivalent to $$\{x\in X\mid f(x)\mathcal R\alpha \}\in \mathcal F,$$ for all $\alpha \in Y$ ? Because I unfortunately don't really see the relation between (D) and (D').

• By combining (D) and the properties of the inverse image you get that the inverse images of the elements of the algebra generated by the sets $(-\infty,\alpha)$ also need to be measurable. – user574380 Jul 13 '18 at 10:26
• I would practicize (D') as definition and (D) as statement that can be shown to be equivalent with the definition in a particular case. – drhab Jul 13 '18 at 10:44
• The connection between $D$ and $D'$ is more clear if you write $D$ as $f^{-1}((-\infty,a)) \in \mathcal F$ for all $a$. – littleO Jul 13 '18 at 11:43
• A good motivation is indeed to extend the topological definition of continuity by saying that $f^{-1}(\mathcal O)$ is measurable for all open set $\mathcal O$. In fact : the function $f:(\mathbb R^n,\mathcal M)\longrightarrow (\mathbb R,\mathcal B)$ is measurable $\iff$ $f^{-1}(\mathcal O)$ is measurable for all open $\mathcal O\subset \mathbb R$. – Surb Jul 13 '18 at 11:47

$(D)$ is actually a necessary and sufficient condition for $f:\mathbb R^n\to\mathbb R$ to be measurable if $\mathbb R$ is equipped with the Borel-$\sigma$-algebra.

Especially the fact that it is sufficient is useful, and makes it easy to prove that functions are measurable.

In general if $X$ is a topological space with topology $\tau$ then by definition the corresponding Borel-$\sigma$-algebra is the $\sigma$-algebra generated by $\tau$ and denoted by $\sigma(\tau)$.

Here $\mathbb R$ is looked at as equipped with its usual topology.

If $\tau$ denotes this topology and $\mathcal M$ denotes the $\sigma$-algebra on $\mathbb R^n$ then it can be shown that $$f^{-1}(\sigma(\tau))\subseteq \mathcal M\text{ if and only if }f^{-1}((-\infty,\alpha))\in\mathcal M\text{ for every }\alpha\in\mathbb R$$

Here $f^{-1}(\sigma(\tau))$ serves as a notation for $\{f^{-1}(A)\mid A\in\sigma(\tau)\}$.

Note that $f^{-1}((-\infty,\alpha))=\{x\in\mathbb R^n\mid f(x)<\alpha\}$.

• Are you sure that $(D)$ is necessary and sufficient for $f$ to be measurable with $\mathbb R$ and $\mathbb R^n$ are both equipped with Borel $\sigma -$algebra ? Take for example $n=1$. If $B$ is a Borel set and $f:\mathbb R\longrightarrow \mathbb R$ is continuous, there is no reason for $f^{-1}(B)$ to be Borel although it's measurable... – Peter Jul 13 '18 at 10:53
• Yes, I am sure. In the situation you sketch continuity of $f$ means exactly that $f^{-1}(\tau)\subseteq\tau$. From this it follows that $\sigma(f^{-1}(\tau))\subseteq\sigma(\tau)$. But it can be shown in general that $\sigma(f^{-1}(\tau))=f^{-1}(\sigma(\tau))$ (for a proof of that see here) so we get $f^{-1}(\sigma(\tau))\subseteq\sigma(\tau)$. That means that - if $f$ is continuous - $f^{-1}(B)$ is Borel whenever $B$ is Borel. – drhab Jul 13 '18 at 11:04
• Ok for $f$ continuous, but for $f$ not continuous ? Because my teacher said that measurability if for $f:(\mathbb R,\mathcal M)\to (\mathbb R, \mathcal B)$ (and not $f:(\mathbb R,\mathcal B)\to (\mathbb R, \mathcal B)$, i.e. $f^{-1}(B)\in \mathcal M$ for all $B\in \mathcal B$ (where $\mathcal M$ is the lebesgue measurable function $\sigma -$algebra). See here – Peter Jul 13 '18 at 11:11
• Define $\mathcal A$ as collection of open sets $(-\infty,a)$. Let $(X,\mathcal M)$ be measurable space and let $f:X\to\mathbb R$ be a function that satisfies $f^{-1}(\mathcal A)\subseteq\mathcal M$. Then $\sigma(f^{-1}(\mathcal A))\subseteq\mathcal M$. But as I said: we have $\sigma(f^{-1}(\mathcal A))=f^{-1}(\sigma(\mathcal A))$ so that $f^{-1}(\sigma(\mathcal A))= \sigma(f^{-1}(\mathcal A))\subseteq\mathcal M$. Next to that it can be shown that $\tau\subseteq\sigma(\mathcal A)$ and consequently $\sigma(\tau)=\sigma(\mathcal A)$. So we end up with $f^{-1}(\sigma(\tau))\subseteq\mathcal M$. – drhab Jul 13 '18 at 11:25
• Under $15$? Not any more :-) Glad to help. – drhab Jul 13 '18 at 12:28

The two definitions are equivalent with the standard Borel sigma algebra.

The main trick is as follows: when checking the second definition, you don't need to check all $U\in\mathcal{G}$ - it is sufficient to check a generating set. For example, the open sets do generate the Borel sigma algebra so your suggested definition of measurability is also equivalent. But we can find an even smaller generating set, which is what the definition (D) is based on: all rays of the form $(-\infty,a)$.

And that's how you should read $(D)$. $f$ is measurable if $f^{-1}((-\infty,a))$ is measurable for each $a\in\mathbb{R}$.