Are these two definitions of a real-valued, measurable function equivalent? ($(X, \Sigma, \mu)$ is the measure space.)

Definition 1: $f: X \to \mathbb{R}$ is said to be measurable if for all $\alpha \in \mathbb{R}$, $\{ x \mid f(x) > \alpha \} \in \Sigma$ (i.e., $f^{-1}(\alpha, \infty) \in \Sigma$)

Definition 2: $f: X \to \mathbb{R}$ is said to be measurable if, given the $\sigma$-algebra $\Sigma'$ on $\mathbb{R}$, $E \in \Sigma ' \implies f^{-1}(E) \in \Sigma$.

I am under the impression that Definition 1 is the definition of Lebesgue measurable, that is, $\Sigma '$ must be the $\sigma$-algebra of Lebesgue measurable sets, and $\mu = m$, Lebesgue measure, even though this is probably not true.

The reason I am asking this is that I am having trouble proving the statement that the pointwise limit of a sequence of Borel measurable functions is Borel measurable. I wasn't sure if I could just use the property we derived from Definition 1 that the pointwise limit of a sequence of measurable functions is measurable, or if Definition 1 only applies to Lebesgue measurable functions.

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    $\begingroup$ In definition1: $\Bbb R$ with Borel $\sigma$-algeba . $\endgroup$ – Hamou Aug 21 '14 at 19:41
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    $\begingroup$ Why consider $\Bbb R$ with the Borel $\sigma$-algebra by default? Just to win the measurability of continuous functions. $\endgroup$ – Hamou Aug 21 '14 at 19:50
  • $\begingroup$ @Hamou That makes sense. Thanks! $\endgroup$ – layman Aug 21 '14 at 19:52
  • $\begingroup$ @Hamou ... but this is only true if $X$ is endowed with the Borel $\sigma$-algebra $\Sigma = \mathcal{B}(X)$. $\endgroup$ – saz Aug 21 '14 at 19:56
  • $\begingroup$ @zas evidently, we talk about continuous functions, when there is a topology in $X$ and endowed with.... and in the practice $X$ is a topological space that is $X=\Bbb R^n$ or.... $\endgroup$ – Hamou Aug 21 '14 at 20:03

Definition 1 gives measurability of $f$ with respect to the Borel-$\sigma$-algebra on $\mathbb{R}$, i.e. the $\sigma$-algebra generated by the open sets. In contrast, Definition 2 defines the measurability of

$$f: (X,\Sigma) \to (\mathbb{R},\Sigma'),$$

i.e. we do not necessarily consider the Borel-$\sigma$-algebra on $\mathbb{R}$, but an arbitrary $\sigma$-algebra on $\mathbb{R}$. If $\Sigma' = \mathcal{B}(\mathbb{R})$, then both definitions are equivalent. This follows from the fact that the family $\{(a,\infty); a \in \mathbb{R}\}$ is a generator of $\mathcal{B}(\mathbb{R})$.

  • $\begingroup$ Is it normal to learn about Definition 1 long before learning about the Borel $\sigma$-algebra? We were just taught that a real-valued function is "measurable" (not Borel measurable) if Definition 1 holds. $\endgroup$ – layman Aug 21 '14 at 19:44
  • $\begingroup$ @user46944 From my point of view: No, it is not normal. Talking about measurability without defining $\sigma$-algebras doesn't make sense at all. $\endgroup$ – saz Aug 21 '14 at 19:46
  • $\begingroup$ So basically, if we say $\{ x \mid f(x) > \alpha \} \in \Sigma$ is the condition for measurability, then we are saying that the codomain has $\sigma$-algebra $\mathcal{B}(\mathbb{R})$ (the Borel $\sigma$-algebra). $\endgroup$ – layman Aug 21 '14 at 19:47
  • $\begingroup$ @user46944 Yes, exactly. A mapping is Borel-measurable if, and only if, this condition is satisfied. $\endgroup$ – saz Aug 21 '14 at 19:48
  • $\begingroup$ And this differs from a function being Borel measurable because if a function is Borel measurable, it is assumed that the $\sigma$-algebra in the domain is $\mathcal{B}(X)$ (if $X$ is the domain)? $\endgroup$ – layman Aug 21 '14 at 19:49

Definition 1 follows from definition 2 (assuming $\Sigma'$ is the Borel sigma algebra) because of the following lemma.

Suppose $(X,\Sigma)$ and $(X',\Sigma')$ are two measurable spaces, and suppose that the $\sigma$-algebra $\Sigma'$ is generated by the family of sets $\Pi$. Then $f : X \rightarrow X'$ is $\Sigma/\Sigma'$ measurable if (and only if, trivially) $f^{-1}(E) \in \Sigma$ for all $E \in \Pi$.

(Proof hint: $\{ E : f^{-1}(E) \in \Sigma\}$ is a $\sigma$-algebra containing $\Pi$.) Apply the lemma to the family $\{(-\infty,a) : a \in \mathbb{R}\}$, which generates the Borel $\sigma$-algebra. The Lebesgue $\sigma$-algebra is slightly larger; it is the completion of the Borel $\sigma$ algebra.

As for your question, here are two hints:

  1. $\{\sup_j f_j \leq a\} = \cap_{j=1}^\infty \{f_j \leq x\}$.

  2. $\limsup_{j\rightarrow\infty} f_j = \inf_n \sup_{j \geq n} X_j.$

Prove the same relation for $\liminf$ and conclude that both $\limsup f_j$ and $\liminf f_j$ are Borel-measurable. If $\lim_{j\rightarrow\infty} f_j$ exists, then ...

  • $\begingroup$ Thanks, this helped a lot! $\endgroup$ – layman Aug 21 '14 at 20:07

Definition 2 doesn't really make sense. Or, more properly it is not equivalent to def 1 nor useful.

In measurability, you want a sigma algebra (i.e., a measurable space) in your domain, but a topological one in your codomain. Otherwise, if you don't have a notion of convergence in your codomain, you cannot really define a useful integral.

Definition 2, to be equivalent with definition 1, should be that given $E$ open, $f^{-1}(E)\in \Sigma$.

  • $\begingroup$ You mean a topological one in the codomain, so that all open sets are measurable? And hence, it makes sense to talk about $f_{n} \to f$? $\endgroup$ – layman Aug 21 '14 at 19:46
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    $\begingroup$ Typo corrected. For integration, which is the point of measurability, you need the preimages of open sets to be measurable. You don't need a notion of measurability in the codomain: an integral is a sum of "value of the function times size of the region where you approximate the value". $\endgroup$ – Martin Argerami Aug 21 '14 at 21:26
  • $\begingroup$ Thanks for the explanation. I will try to internalize it. $\endgroup$ – layman Aug 21 '14 at 21:43
  • $\begingroup$ If it helps, think of the "generalized Riemann sums" (or, the same, integrals of simple functions) $$\sum_k\,f(t_k)\,\mu(f^{-1}([t_k,t_{k+1}))).$$ $\endgroup$ – Martin Argerami Aug 21 '14 at 21:58
  • $\begingroup$ You're totally right. Except, I'm not sure where topology comes into play when we have $\int \limits_{X} s \,d\mu = \sum \limits_{ i = 1}^{n} \alpha_{i} \mu(E_{i})$. $E_{i}$ is not necessarily in the topology, but we do need a topology to talk about convergence almost everywhere, so I think I understand your point. :) $\endgroup$ – layman Aug 21 '14 at 22:20

I wanted to answer my own question in addition to the other answers given, to make extra clear what is going on here (this will be useful for me as I reference this in the future, and hopefully it will be useful for others, too).

As others have said, our definition of a real-valued function $f: X \rightarrow \mathbb{R}$ being measurable is that $\{x \mid f(x) > \alpha \} \in \Sigma$, where $\Sigma$ is a $\sigma$-algebra of subsets of $X$ $((X, \Sigma)$ is a measure space).

But this definition seems so different from the definition that if $(X, \Sigma_{1})$ and $(Y, \Sigma_{2})$ are measure spaces and we have a function $f: X \to Y$, then $f$ is measurable if $E \in \Sigma_{2} \implies f^{-1}(E) \in \Sigma_{1}$.

What is going on here is that we automatically assume we are dealing with the Borel $\sigma$-algebra in the codomain when we talk about real-valued functions. Specifically, our function is going from $X$ to $\mathbb{R}$ with measure space $(X, \Sigma)$ as the domain, and $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$ as the codomain.

But if this is what we are assuming about the codomain, then by the second definition of measurability, shouldn't we be checking that $E \in \mathcal{B}(\mathbb{R}) \implies f^{-1}(E) \in \Sigma$? Yes, that is the condition that needs to be satisfied for a real-valued function to be measurable. So why are we only checking that $\{x \mid f(x) > \alpha \} \in \Sigma$?

Here is the reason: First, when we assume the codomain has the Borel $\sigma$-algebra, we need to check that the preimage of a Borel measurable set is measurable. (i.e., if $f: X \to Y$ is a function and $\Sigma$ is the $\sigma$-algebra on $X$, we need to check that $E \in \mathcal{B}(Y) \implies f^{-1}(E) \in \Sigma$). But actually, it suffices to show that only the preimage of open sets is measurable, because the set $\{ E \mid E \subseteq Y \, \& \, f^{-1}(E) \in \Sigma \}$ is a $\sigma$-algebra, and so if it contains the open sets, then it must contain the smallest $\sigma$-algebra containing the open sets, which is $\mathcal{B}(Y)$. So that would imply $\mathcal{B}(Y) \subseteq \{ E \mid E \subseteq Y \, \& \, f^{-1}(E) \in \Sigma \}$, and thus the preimage of all Borel measurable sets are measurable (and if the domain is specifically equipped with Borel $\sigma$-algebra, then the preimage of all Borel measurable sets are Borel measurable). So, it suffices to just check that the preimage of an open set is measurable to show that a function is measurable with respect to an arbitrary $\Sigma$ on the domain $X$.

But, we can take it a step further. We really only need to check that the preimages of the intervals $(\alpha, \infty)$ are measurable for all $\alpha \in \mathbb{R}$. This is because we have the following equivalence:

$O \subseteq \mathbb{R}$ is open and $f^{-1}(O) \in \Sigma$ $\iff$ $f^{-1}((\alpha, \infty)) \in \Sigma$ for all $\alpha \in \mathbb{R}$.

To prove this equivalence is easy. For the $\implies$ direction, suppose every open subset of $\mathbb{R}$ has a measurable preimage. Then since for any $\alpha \in \mathbb{R}$, $(\alpha, \infty)$ is an open subset of $\mathbb{R}$, by assumption $f^{-1}((\alpha, \infty)) \in \Sigma$, as desired.

For the $\impliedby$ direction, suppose for every $\alpha \in \mathbb{R}$, $\{ x \mid f(x) > \alpha \} = f^{-1}( (\alpha, \infty) ) \in \Sigma$. This directly implies that for any interval $I \subseteq \mathbb{R}$, $f^{-1}(I) \in \Sigma$. But every open subset of $\mathbb{R}$ can be written as a countable, disjoint union of open intervals. So we have if $O \subseteq \mathbb{R}$ is open, then since we can write $O = \bigcup \limits_{n = 1}^{\infty} I_{n}$, and $f^{-1}(O) = f^{-1}(\bigcup \limits_{n = 1}^{\infty} I_{n}) = \bigcup \limits_{n = 1}^{\infty} f^{-1}(I_{n})$, and because $f^{-1}(I_{n}) \in \Sigma$ (since $I_{n}$ is an interval), then $f^{-1}(O) \in \Sigma$, as desired.

So at the end of the day, to check that a real-valued function is measurable, by definition we must check that the preimage of a Borel measurable set is measurable. But this boils down, as shown above, to proving that $\{x \mid f(x) > \alpha \} = f^{-1}( (\alpha, \infty)) \in \Sigma$ for all $\alpha \in \mathbb{R}$, since this implies that the preimage of Borel measurable sets are measurable.

Note that to check if a real-valued function is Borel measurable, it also suffices to check that $f^{-1}((\alpha, \infty)) \in \mathcal{B}(X)$ for all $\alpha \in \mathbb{R}$, since this implies that the preimage of an open set is Borel measurable, which implies that the preimage of a Borel measurable set is Borel measurable.


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