More specifically, letting $I=[0,1]$, do there exist $f,E$ with $E$ a (necessarily nonseparable) Banach space and $f$ a bounded Lebesgue measurable function $I\to E$ such that $f$ is not equal almost everywhere to a pointwise limit of a sequence of simple Lebesgue measurable functions? Here "simple" means having finite range, and Lebesgue measurability of $f$ means that $f^{-1}[A]$ is a Lebesgue measurable set in $I$ for every Borel set $A$ in $E$ .

If such $f,E$ exist, and if one constructs $F=L^p(I,E)$ along the lines of Problem 4 on page 120 in Richard M. Dudley's book Real Analysis and Probability, Wadsworth 1989, one obtains $F$ where the set of (equivalence classes of) simple functions is not dense. If instead of Lebesgue measurable functions one here considers the Borel measurable ones, this pathology is not possible by Corollary 4.2.7 on page 97 and Problem 10 on page 99 loc. cit.

I guess that an explicit construction might not be possible, and one should (somehow) use something based on the axiom of choice.


1 Answer 1


No. In fact, every Lebesgue measurable function $f\colon I\to E$ is equal almost everywhere to a limit of simple Lebesgue measurable functions. As you hint at in the question, this is easy to show in the case where $E$ is separable. The general situation reduces to the separable case due to the following result. For a full proof, see Fremlin, Measure Theory, Volume 4 Part I, Lemma 451Q.

Theorem 1: Let $(X,\Sigma,\mu)$ be a finite compact measure space, $Y$ a metrizable space, and $f\colon X\to Y$ a measurable function. Then, there is a closed separable subspace $Y_0$ of $Y$ such that $f^{-1}(Y\setminus Y_0)$ is negligible.

That is, $f$ has essentially separable image. Restricting $f$ to the complement of a negligible set reduces the problem to the situation where the codomain is separable, in which case it is a limit of simple functions. Compactness of the space $(X,\Sigma,\mu)$ means that there is a family $\mathcal{K}\subseteq\Sigma$ such that any subset of $\mathcal{K}$ with the finite intersection property has nonempty intersection, and such that $\mu$ is inner-regular with respect to $\mathcal{K}$. That is, $\mu(E)=\sup\{\mu(K)\colon K\in\mathcal{K},K\subseteq E\}$ for every $E\in\Sigma$. In particular, the Lebesgue measure is compact by taking $\mathcal{K}$ to be the collection of compact sets under the standard topology.

The proof of Theorem 1 is rather tricky, involving what Fremlin describes as "non-trivial set theory". It rests on the following two results.

Theorem 2: Any metrizable space has a $\sigma$-disjoint base $\mathcal{U}$. That is, $\mathcal{U}$ is a base for the topology, and can be written as $\bigcup_{n=1}^\infty\mathcal{U}_n$ where each $\mathcal{U}_n$ is a disjoint collection of sets.

(Fremlin, Measure Theory, Volume 4 II, 4A2L (g-ii))

Theorem 3: Let $(X,\Sigma,\mu)$ be a finite compact measure space and $\{E_i\}_{i\in I}$ be a disjoint family of subsets of $X$ such that $\bigcup_{i\in J}E_i\in\Sigma$ for every $J\subseteq I$. Then, $\mu\left(\bigcup_{i\in I}E_i\right)=\sum_{i\in I}\mu(E_i)$.

(Fremlin, Measure Theory, Volume 4 I, 451P).

Theorem 3 is particularly remarkable, as it extends the countable additivity of the measure to arbitrarily large unions of sets.

Once these two results are known, the proof that $f$ has essentially separable image in Theorem 1 is straightforward. Let $\mathcal{U}=\bigcup_{n=1}^\infty\mathcal{U}_n$ be a $\sigma$-disjoint base for $Y$. Let $\mathcal{V}_n$ be the collection of $U\in\mathcal{U}_n$ such that $\mu(f^{-1}(U)) = 0$. By countable additivity, $\mathcal{U}_n\setminus\mathcal{V}_n$ is countable. Also, $\{f^{-1}(U)\colon U\in\mathcal{V}_n\}$ is a disjoint collection of negligible subsets of $X$ and, by measurability of $f$, any union of a subcollection of these is measurable. It follows from Theorem 3 that its union is negligible. That is, $f^{-1}\left(\bigcup\mathcal{V}_n\right)$ is negligible. Setting $Y_0=Y\setminus\bigcup_n\bigcup\mathcal{V}_n$ then, by countable additivity, $f^{-1}(Y\setminus Y_0)$ is negligible. Also, $\bigcup_n(\mathcal{U}_n\setminus\mathcal{V}_n)$ restricts to a countable base for the topology on $Y_0$, so it is separable (in fact, it is second-countable).

Finding a $\sigma$-disjoint base for the topology on $Y$ is easy enough. Following Fremlin, you can do this by well-ordering $Y$ and letting $(q_n,q^\prime_n)$ be a sequence running through the pairs $(q,q^\prime)$ of rationals with $0 < q < q^\prime$. Letting $\mathcal{U}_n$ be the collection of sets of the form $$ G_{ny}=\left\{x\in Y\colon d(x,y) < q_n, \inf_{z < y}\,d(x,z) > q_n^\prime\right\} $$ (over $y\in Y$) gives a $\sigma$-disjoint base.

The really involved part of the proof is in establishing Theorem 3. I suggest you look in Fremlin for the details, but the idea is as follows. By countable additivity, only countably many $E_i$ can have positive measure so, removing these, we can suppose that every $E_i$ is negligible. Also, restricting $X$ to the union of the $E_i$ if necessary, we can suppose that $X=\bigcup_iE_i$. Then define the function $f\colon X\to I$ by $f(x)=i$ for $x\in E_i$. Using the power set $\mathcal{P}I$ for the sigma-algebra on $I$, $f$ will be measurable. Then let $\nu=\mu\circ f^{-1}$ be the image measure on $(I,\mathcal{P}I)$. Fremlin breaks this down into two cases.

a) $\nu$ is atomless. As with any finite atomless measure space, there will be a measure preserving map $g\colon I\to[0,\gamma]$ for some $\gamma\ge0$, with respect to the Lebesgue measure $\lambda$ on $[0,\gamma]$. Using compactness, it can be shown that the sets on which $\lambda$ and $\nu\circ g^{-1}$ are well-defined coincide (precisely, $\mu$ is compact, so it is perfect, so $\nu\circ g^{-1}$ is perfect and therefore is Radon). The existence of non-Lebesgue sets will then give a contradiction unless $\gamma=0$, so $\mu(X)=0$.

b) $\nu$ has an atom $M\subseteq I$: In this case, $\mathcal{F}=\{F\subseteq M\colon\nu(M\setminus F)=0\}$ is a non-principal ultrafilter on $M$ which is closed under countable intersections. Again making use of compactness of $\mu$, this can be used to derive a contradition, but it requires some tricky set theory. I refer you to Fremlin (451P) for the full details of this argument.

Update: I will, however, give a brief overview of the ideas involved in (b). It is possible to reduce the problem to the case where $M$ is a regular uncountable ordinal and $\mathcal{F}$ is a normal ultrafilter. Using $[S]^n$ to denote the collection of size-$n$ subsets of a set $S$ and $[S]^{ < \omega}=\bigcup_{n=0}^\infty[S]^n$ for the collection of finite subsets, normal ultrafilters have the following property.

If $\mathcal{S}\subseteq[M]^{ < \omega}$ then there exists an $F\in\mathcal{F}$ such that, for each $n\ge0$, $[F]^n$ is either a subset of $\mathcal{S}$ or disjoint from $\mathcal{S}$.

See, Frelim (4A1L). This contradicts compactness as follows. Set $G_i=\bigcup\{E_j\colon j\in M, j\ge i\}$. Then choose $K_i\in\mathcal{K}$ with $K_i\subseteq G_i$ and $\mu(K_i) > 0$. Let $\mathcal{S}$ consist of the finite subsets $S\subset M$ such that $\bigcap_{i\in S}K_i=\emptyset$. Choose $F\in\mathcal{F}$ as above. It is not possible for $[F]^n$ to be a subset of $\mathcal{S}$. Otherwise, every $x\in X$ would be contained in no more than $n$ of the sets $\mathcal{K}^\prime=\{K_i\colon i\in F\}$. So, $\sum_{i\in F}\mu(K_i)\le n\nu(M)$. But, as this sum is over an uncountably infinite set of positive numbers, it should be infinite. Therefore, $[F]^{ < \omega}\cap\mathcal{S}=\emptyset$, and $\mathcal{K}^\prime$ satisfies the finite intersection property. So, by compactness, $\bigcap_{i\in F}G_i\supseteq\bigcap\mathcal{K}^\prime\not=\emptyset$. This contradicts the fact that, as $F\in\mathcal{F}$ is an unbounded subset of $M$, this intersection is empty.

  • $\begingroup$ I gave you +1 just for the effort and would give more if I could. I thought about the question quite a bit, but couldn't come to grips with it, but I hadn't expected it to be that difficult. A very naive question (triggered by the phrase "non-trivial set theory"): Theorems 1,2 and 3 are theorems of ZFC and no trickery such as Martin's axiom or the continuum hypothesis is involved? $\endgroup$ Commented Feb 6, 2011 at 17:36
  • $\begingroup$ Yes, they are all theorems of ZFC. No additional trickery is required. I think, by the phrase "non-trivial set theory", Fremlin means that it requires rather more in-depth set theory than you would expect for a pure measure theory question. $\endgroup$ Commented Feb 6, 2011 at 17:43
  • $\begingroup$ +1 Nicely done, George. I was also interested in this, but couldn't answer it... $\endgroup$
    – user6096
    Commented Feb 6, 2011 at 18:05
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    $\begingroup$ In case (b) at the end, the existence of a non-principal ultrafilter closed under countable intersections is equivalent to the existence of a measurable cardinal, a large cardinal notion whose existence is not provable in ZFC. In particular, if there are no measurable cardinals, then this case does not occur. $\endgroup$ Commented Feb 6, 2011 at 22:16
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    $\begingroup$ The links to eom.springer.de are broken, but they can now be found at encyclopediaofmath.org/wiki/Perfect_measure and encyclopediaofmath.org/wiki/Radon_measure. $\endgroup$ Commented Jul 25, 2022 at 18:47

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