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I am new to measure theory and here is the definition I have :

(1) A function $f:(X, \mathcal{A}) \to (Y, \mathcal{B})$ is mesurable iff : $\forall B \in \mathcal{B}, f^{-1}(B) \in \mathcal{A}$

Why this definition and not this one ?

(2) A function $f:(X, \mathcal{A}) \to (Y, \mathcal{B})$ is mesurable iff : $\forall A \in \mathcal{A}, f(A) \in \mathcal{B}$

Thus with (2) a function is measurable iff it maps measurable sets to measurable sets. It's seems more natural to me. I know that the first definition extend the notion of continuity, but this explanation still doesn't convince me that (1) should be the most natural definition.

So are functions that respect (2) have a name ? And why (2) is not the definition of measurable functions ?

anomaly
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NickolasB
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    It's funny that the definition of continuous map is that $f:X\to Y$ is continuous if $f^{-1}(V)$ is open in $X$ for all $V$ open in $Y$. Why isn't it that $f(U)$ is open in $V$ for all $U$ open in $X$? – Angina Seng May 24 '20 at 18:28
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    @AnginaSeng ? I don't really see what is funny, but for your question it's just because there are continuous function that doesn't map open sets to open sets, for example $x^2$ maps $(-1,1)$ to $[0,1[$ – NickolasB May 24 '20 at 18:31
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    To me I guess it is natural because you alredy have a notion of continuity in hand and you might want to carry that notion onto measure theory so that the continuous functions you have are good fellows in measure theory . – Noob mathematician May 24 '20 at 18:39
  • The more natural form of "measurability" in my opinion follow the Lebesgue measure type so:

    Given the measure space $(X, \mathcal M)$ and $f$ an extended real valued function defined on $X$ then $f$ is measurable iff the set ${x \in X | f(x) > c }$ is measurable for each real number $c$.

     – Alessandro Romancino May 24 '20 at 19:42

2 Answers2

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Inverse images behave much more nicely than direct images. Indeed, one has the identities $$ f^{-1}(B\cap C)=f^{-1}(B)\cap f^{-1}(C) $$ and $$ f^{-1}(B^c)=(f^{-1}(B))^c $$ where ${}^c$ is the complement (relative to $X$ or $Y$). Neither of these identities hold for direct images, we only have $\subset$ in the first one. The first identity also generalizes to arbitrary unions.

If you're categorically minded, taking inverse images defines a function from the category of sets into the category of (complete) Boolean algebras, while taking direct images is merely a morphism of ordered sets.

When trying to explain where this comes from, one notices that the definition of a direct image is more complicated, in terms of the syntax. Indeed, compare $$ f^{-1}(B):=\{x\in X: f(x)\in B\} $$ with $$ f(A):=\{y\in Y: \exists x\in A(f(x)=y)\} $$ The condition that determines whether something is in $f(A)$ involves an existential quantifier, which is not the case for $f^{-1}(B)$, where the defining condition is quantifier free. And, as always, quantifiers are hard. Indeed, think of evaluating the truth value of any propositional formula. This can be done with a truth table, given enough time. Once quantifiers get into the mix, things become much harder, as they involve checking infinitely many objects.

Regarding the question of whether $f(A)$ is measurable for measurable $A$, the answer is, in general, no. This famously tripped up even Lebesgue himself, who published a faulty proof that the projection (which is continuous, even open!) of a Borel set in the plane is Borel in the line. This mistake, which involved an incorrect manipulation of $\exists$, went unnoticed for 10 years, until it was discovered by Souslin, marking the start of descriptive set theory. See here for more details.

However, not all is lost, as we have the following positive result:

Theorem: Let $X$ and $Y$ be standard Borel space (this means that the $\sigma$-algebras are induced by Polish topologies on the corresponding spaces). Suppose that $f:X\to Y$ is Borel measurable. Let $A\subset X$ be Borel measurable. If the restriction $f\mid A$ is injective, then $f(A)$ is Borel measurable.

Reveillark
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It is nice to play with definitions who knows it might give rise to new branches and new ways of looking at things.

Warning: This is my personal opinion .

You see we already have a notion of continuity. And initially (before proceeding towards abstract measure space) your measure spaces would arise from the topological spaces you already know and there the $\sigma$-algebra you consider would be generated by the open sets of the topology. Now you want your continuous functions you have on topological spaces to be your good guys not the nasty ones. And that is why in my opinion it is a natural definition for measurable function.

Indicator functions of non-measurable sets, such as: $$\chi_A(x) = \begin{cases} 1 & \text{if $x \in A$} \\ 0 &\text{elsewhere}\end{cases}$$ Where $A$ is a non-measurable set is now a measurable function for definition (2).

While this might seem trivial, it will for sure be of importance for basics properties of measurable functions such as:

  • Linearity
  • Measurability of the product of two measurable functions
  • Measurability of inf, sup, max and min of a sequence of measureable functions
  • etc.

Also we might go under problems against another very well used definition of measurability (which is most often used for Lebeasgue measure):

Given the measure space $(X, \mathcal M)$ and $f$ an extended real valued function defined on $X$ then $f$ is measurable iff the set $\{x \in X | f(x) > c \}$ is measurable for each real number $c$.

These problems will for sure translate to other problems into integration, functional spaces such as the $L^0$ space or even $L^p$ spaces in general. For example, Lusin's theorem will for sure not work with this definition.

Alessandro Romancino
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Noob mathematician
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