When is the infimum of an arbitrary family of measurable functions also measurable?

Question

Let $(X,\Sigma,\mu)$ be a measure space and consider a family of $\mu$-measurable functions $f_i:X \to \mathbb{R}$ for $i$ lying in some index set $I$. Define $$f(x) = \inf_{i \in I} f_i(x)$$

I think every good analysis book mentions or proves that if $I$ is countable, then $f$ is also $\mu$-measurable. What is not so clear is:

If $I$ has cardinality of the continuum, is $f$ still $\mu$-measurable?

Since I strongly suspect that the answer is "no" although a counterexample is not immediately coming to mind, here is the real question:

What are minimal conditions on $f_i$ that will make $f$ $\mu$-measurable even when $I$ is uncountably large?

By minimal conditions I am hoping for some weak properties, such as those that hold $\mu$-almost everywhere rather than those that require something strong from the global structure of $I$, such as a (partial or total) ordering.

If you know of a book or paper that deals with this, please let me know. I understand there is a good chance this type of thing is not considered research material by folks here; in this case I will delete the question.

Answer 1

As you expected, the infimum of continuum many measurable functions need not be measurable, even in the case where $X$ is the real line with Lebesgue measure. In fact, if $A$ is any subset of $\mathbb R$ (in particular not necessarily measurable), its characteristic function is the infimum of at most continuum many measurable functions, namely the characteristic functions of the sets $\mathbb R-\{x\}$ for all $x\notin A$.

A similar argument shows that even the infimum of $\aleph_1$ measurable functions won't be measurable if, in your measure space, the intersection of some $\aleph_1$ measurable sets is not measurable. And there are such measure spaces. For example, take $X$ to be a set of cardinality $\aleph_1$, take $\Sigma$ to be the $\sigma$-field of countable and co-countable sets, and take $\mu$ to give the countable sets measure 0 and the co-countable sets measure 1.

Whether the same thing happens for Lebesgue measure on $\mathbb R$ is independent of the usual ZFC axioms of set theory. It is consistent with ZFC that there is a non-Lebesgue-measurable set of cardinality $\aleph_1$, and it is also consistent that every union of $\aleph_1$ sets of Lebesgue measure 0 has measure 0, which implies that every union or intersection of $\aleph_1$ measurable sets is measurable.

The smallest cardinal $\kappa$ such that the union of some $\kappa$ sets of Lebesgue measure 0 does not have measure 0 is called the additivity of Lebesgue measure; it is one of the many well-studied cardinal characteristics of the continuum. It is also the smallest $\kappa$ such that the infimum of some $\kappa$ Lebesgue measurable functions $\mathbb R\to\mathbb R$ fails to be measurable.

Answer 2

If you want a (sometimes useful) positive result: suppose $f_i : \mathbb R \to \mathbb R$ are continuous. Then $g(x) := \inf_{i \in I} f_i(x)$ is upper semicontinuous, and therefore (Borel) measurable. More general domains are possible, of course.

Answer 3

Here is a set of of measurable functions with cardinality the continuum whose infimum is not (Borel) measurable:

Let $S\subset [0,1]$ be a non-measurable set. For $t\in[0,1]$ let $f_t(x)$ be the function defined as follows:

If $t \in S$ then $f_t(x) = 2$ for $t \not = x$ and $f_t(t) = 1$. If $t \not \in S$, then let $f_t(x) \equiv 2$. Then $f(x) := \inf_{t\in[0,1]} f_t(x)$ is $2$ on $S^c$ and $1$ on $S$, so is certainly not measurable.

Answer 4

The infimum of any family of measurable functions is measurable if we interpret the infimum as the lattice infimum. For details see: https://mathoverflow.net/a/316658/121665