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Although I have always known the definition of measurability in terms of pre images of measurable sets being measurable, I don't really conceptually understand the purpose of measurable functions or what it means.

More specifically, this dawned on me when I was looking at a paper talking about limits of a measurable function, $U:\mathbb{R_+} \rightarrow \mathbb{R_+}$ (the limit was $\lim_{t\rightarrow \infty}\frac{U(tx)}{U(t)}$), without specifying why the function had to be measurable at all. It's left me in a lot of doubt of my own knowledge of measure theory. Is anyone able to give me some intuition and a possible reason for measurability with regards to taking limits?

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    $\begingroup$ I always regarded this notion as extremely technical detail needed for Lebesgue integration. It is a very mild regularity assumption. I have never needed to deal with nonmeasurable functions in practice. These are horrible objects with place only in pure mathemathics. $\endgroup$
    – Blazej
    Feb 23, 2017 at 11:42
  • $\begingroup$ A measurable function (might need to be bounded or of bounded variation - not sure!) is approximately continuous i.e. continuous except on a set of measure 0. Measurability is quite a strong condition really, it stops the function being too wild. Limits involving measurable functions typically involve metrics which give 0 weight to sets of measure 0 e.g. integral norms. $\endgroup$
    – Paul
    Feb 23, 2017 at 11:45
  • $\begingroup$ @Paul I strongly disagree. $1_{\mathbb Q}$ is a Borel-measurable function and is nowhere continuous. In most contexts it is quite hard to find functions that are not measurable. It is not a stop for functions on being "wild" and does not ask for things like almost continuous or bounded. $\endgroup$
    – drhab
    Feb 23, 2017 at 12:06
  • $\begingroup$ So could my understanding be that the assumption of measurability ensures that Lebesgue integration can be understood? $\endgroup$
    – Necroticka
    Feb 23, 2017 at 12:28
  • $\begingroup$ @drhab Sorry, I was being overly vague - ${{1}_{\mathbb{Q}}}$ is nowhere continuous, but it is approximately continuous almost everywhere and even has a well defined approximate derivative of 0 almost everywhere. It is "like" the continuous function f(x) = 0 except at a set of measure 0. Measurable functions in general (modulo some boundedness criteria I think, but can't remember!) are like this, not worse, so not unimaginably exotic (wild..). Then limits involving (Lebesgue) integral norms will essentially "ignore" the measure 0 bits and make sense of a limit of measurable functions. $\endgroup$
    – Paul
    Feb 23, 2017 at 13:14

3 Answers 3

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I'm going to dissent from some of the comments and advise against thinking about measurability in terms of a topological property like continuity, even for heuristic purposes. The fact is that measurability is not a topological concept, and it's important to be conceptually clear about this, especially if you go on to study more abstract measure theory in which measurable spaces do not always come equipped with a natural topology (unlike the real numbers, for instance). There are, of course, important results linking measurability and continuity. I mentioned two in my comment: Lusin's theorem and Lebesgue's criterion for Riemann integrability. You should know these results and understand why they hold, but you should not conclude on their basis that, in general, the concept of measurability can be (approximately) reduced to the concept of continuity.

Now, as for intuitions, my advice is to always keep in mind that assertions about measurability are always relative to particular sigma-algebras. As you mention, a function $f: (X_1, \mathcal{F}_1) \to (X_2, \mathcal{F}_2)$ between two measurable spaces is $(X_1, \mathcal{F}_1, X_2, \mathcal{F}_2)$-measurable (notice the qualification of "measurable") provided $$f^{-1}(B_2) \in \mathcal{F}_1$$ for all $B_2 \in \mathcal{F}_2$. Sometimes mention of one or both sigma-algebras is suppressed when the context makes clear what measurable spaces one is working with. For example, in real analysis one often deals with functions from $(\mathbb{R}, \mathcal{L})$ into $(\mathbb{R}, \mathcal{B})$, where $\mathcal{L}$ is the collection of Lebesgue measurable sets and $\mathcal{B}$ is the Borel sigma-algebra.

The intuition behind the formal definition is simply that the function $f$ can be measured. That is, if we now equip $(X_1, \mathcal{F}_1)$ with a measure $\mu$, the measurability of $f$ guarantees that any "reasonable statement" $B_1$ about the values that $f$ takes is in fact a set in $\mathcal{F}_1$, and hence $\mu(B_1)$ makes sense. Returning to the canonical real analysis setup, $B_1$ might be a statement like "$f(x) \leq 5$" or "$f(x) \in (0, \infty)$". Measurability guarantees that these rough "statements" are actually sets in $\mathcal{F}_1$ that can be measured. For example, the first statement corresponds to the set $B_1 = \{x \in \mathbb{R}: f(x) \leq 5 \}$.

With this in mind, let me explain why I recommend not conflating measurability and continuity. Consider a function $f: (\mathbb{R}, \mathcal{F}_1) \to (\mathbb{R}, \mathcal{B})$, where $\mathcal{F}_1$ is the trivial sigma-algebra $(\emptyset, \Omega)$. From the formal definition, we conclude that $f$ is measurable if and only if it is a constant function (verify this). But this rules out many "well-behaved",and, in particular, continuous, functions! In other words, there are many continuous functions that are not $(\mathbb{R}, \mathcal{F}_1, \mathbb{R}, \mathcal{B})$-measurable. We see that a lot turns on the choice of $\mathcal{F}_1$.

Another important example along these lines is that there are continuous functions $f: (\mathbb{R}, \mathcal{L}) \to (\mathbb{R}, \mathcal{L})$ that are not measurable. See the discussion here. So we see that a lot turns on the choice of sigma-algebra in the codomain as well.

Finally, in response to your question about limits, it's a useful exercise to prove, directly from the definition, that the pointwise limit of a sequence of real-valued measurable functions is measurable. (As PhoemueX pointed out in the comments, pointwise convergence doesn't make sense for functions between general measurable spaces, so we now consider functions taking values in $(\mathbb{R}, \mathcal{B})$.) This sort of exercise should help your intuitions quite a bit. Consider $f_n: (X_1, \mathcal{F}_1) \to (\mathbb{R}, \mathcal{B})$, where $(X_1, \mathcal{F}_1)$ is an arbitrary measurable space, each $f_n$ is measurable, and $f_n \to f$. First, it suffices to show that $f^{-1}((-\infty, x]) \in \mathcal{F}_1$ for all $x \in \mathbb{R}$ because sets of the form $(-\infty, x]$ generate $\mathcal{B}$ (verify). Now try to show that $$f^{-1}((-\infty, x]) = \cap_{m=1}^\infty \cup_{n=1}^\infty \cap_{k=n}^\infty \{ f_k^{-1}((-\infty, x + 1/m])\}.$$ (Just think about the definition of a limit and translate the quantifiers "for all" and "for some" to $\cap$ and $\cup$, respectively.) Do you know how to conclude from here?

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    $\begingroup$ At the very beginning of integration theory, one proves that a nonnegative function is the pointwise limit of an increasing sequence of simple functions. In fact that is a necessary and sufficient condition. It may be dismissed as a mere technical lemma on the way toward building up the general integral, but to me, it provides profound insight into the nature of measurability. $\endgroup$ Feb 23, 2017 at 15:19
  • $\begingroup$ That's a good point, and it can be generalized beyond functions on the reals, so is good for general intuitions. But which functions are simple depends on one's choice of sigma-algebra, which was the main point I was trying to communicate in my answer. $\endgroup$
    – aduh
    Feb 23, 2017 at 15:26
  • $\begingroup$ Since you are talking about measurable maps between any two measure spaces: In this generality, it does not make sense to talk about pointwise convergence. Maybe you should clarify this in the last part of the answer. Otherwise, great answer, +1. $\endgroup$
    – PhoemueX
    Feb 23, 2017 at 15:54
  • $\begingroup$ @PhoemueX Thanks, you're right. I'll try to clarify that. $\endgroup$
    – aduh
    Feb 23, 2017 at 16:33
  • $\begingroup$ @aduh Totally agreed. In case it wasn't clear, my comment was intended as a supplement, not a criticism, of your answer. $\endgroup$ Feb 23, 2017 at 20:15
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"Measurable" is the formal way of saying "Metaphysics-invariant".

A sigma-algebra $\mathcal{F}$ defines a "language" for events, where each event $A\in\mathcal{F}$ is a "sentence". The axioms of a sigma-algebra then define the logical operators $\lnot$, $\land$, $\lor$, $\forall$, $\exists$ for this language.

(Note that this is, despite appearances, not first-order-logic, not even any sub-$\omega_{CK}$ logic, but rather second-order logic -- because we allow arbitrary countable unions, not necessarily only of computable lists.)

The most common use for a sigma-algebra is to define what is observable. E.g. say $\Omega=\{HH,HT,TH,TT\}$ then $\mathcal{F}=\langle\{HH,TH\},\{HT,TT\}\rangle$ means that we observe the second coin toss while the first coin toss, not being directly observable, is basically metaphysics.

The notion that "the rules of the world are metaphysics-invariant" are made formal by the following fact:

  1. The conditional expectation $E(X\mid \mathcal{H})$ is $\mathcal{H}$-measurable for $\mathcal{H}\subseteq\mathcal{F}$.

  2. If you let $\mathcal{F}$ include "metaphysical" language, then metaphysics can still "give you information" on observable things ($\mathbf{P}$ can have correlations between them), but this information can always be captured by a sub-algebra $\mathcal{H}$, i.e. $E(X|\mathcal{H})$ exists, and you can condition on more and more things indefinitely.

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Taking an abstract measurable space $(X, \mathcal{S})$, without any topological structure necessarily, measurability of functions is directly linked to the structure of a $\sigma$-field $\mathcal{S}$, and to the need to define abstract integration of that function with respect to some measure $\mu$ on the space, defined on the measurable sets.

Non-measurable sets give non-measurable functions and vice versa.

If you have a non-measurable set, you have a non-measurable function by taking the indicator of that set.

Working in the other direction, you can write any real-valued measurable function as the pointwise almost-everywhere (w.r.t. $\mu$) limit of a sequence finite linear combinations of indicator functions of measurable sets.

When you come to defining an integral, at least in this abstract setting, you can start by defining it for such linear combinations of indicators, via $$\sum_1^n a_i \mu(A_i) = \int_X f \, d\mu $$ where $A_i$ is a finite partition of $X$ and $a_i$ are real numbers for $i = 1 \ldots n$, when $f$ is a function of that form.

That definition with the basic properties of measures let you define the abstract integral for any measurable function (with the caveat that not both of a function's integral's negative and positive parts can be infinite).

So again, the need for measurability is directly related to the sets for which a measure is defined, the sets in $\mathcal{S}$. A non-measurable set will will give you a non-measurable function for which the integral above cannot be defined.

When you have certain special topological structure on $X$, and certain properties for $\mu$, you get some comforting connections to Riemann integration. For example, Lusin's theorem on second-countable spaces with Radon measures allows you to think of measurable functions as 'almost' continuous.

But even there the statement 'almost' is related to properties of the measure and integral: Since it is discontinuous on a null set with respect to $\mu$, you can ignore that inconvenience when taking the integral.

Finally, everything above was given with real-valued functions and a Borel $\sigma$-field on the codomain in mind. Measurability depends as much on the codomain's $\sigma$-field as it does on the domain's, but the intuition is the same: follow the measurable and non-measurable sets to understand measurable and non-measurable functions.

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