I know that in general it does not hold, but there exists a positive result under some conditions for which the following claim holds?

Claim: given an uncountable set of measurable functions $f_i : U \rightarrow[0,1]$, then it holds that $\inf_i f_i$ is measurable.


  • $\begingroup$ Voted to close as too broad, but voted to close as off-topic by mistake. $\endgroup$ – YCor Dec 1 '18 at 17:02
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    $\begingroup$ @YCor This is actually a reasonable question because it has a reasonable and non-obvious answer (below) so I would not close it. $\endgroup$ – Piotr Hajlasz Dec 1 '18 at 17:16
  • $\begingroup$ @PiotrHajlasz this is my point with "too broad". "Give me a theorem" can also have a reasonable and non-obvious answer (certainly it's not as broad, but...). Anyway your answer is interesting; I have no idea if it's of interest to the OP since I don't know whether (s)he's willing to change the ordering to define supremum. $\endgroup$ – YCor Dec 1 '18 at 17:19
  • $\begingroup$ Sorry in the claim I wrote sup f_i instead of inf f_i... I correct the sentence $\endgroup$ – Ruggero Dec 1 '18 at 17:22
  • $\begingroup$ Possible duplicate of When is the infimum of an arbitrary family of measurable functions also measurable? $\endgroup$ – Gerald Edgar Dec 10 '18 at 11:35

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

My answer is for supremum, but the same holds for infimum since the corresponding results can be obtained from the equality $\inf A=-(\sup (-A))$.

The pointwise supremum of Borel functions need not be measurable.

Example. Let $I\subset[0,1]$ be a non-measureable set and for $i\in I$ we define $$ f_i(x)= \begin{cases} 1 & \text{if $x=i$},\\ 0 & \text{if $x\neq i$.} \end{cases} $$ Then all functions $f_i$ are Borel measurable, however, $\sup_{i\in I} f_i=\chi_I$ is the characteristic function of a non-measurable set and hence is non-measurable.

Instead of pointwise supremum you should consider the so called lattice supremum.

Definition. If $\mathcal{F}$ is a family of measurable functions on $\Omega$, we define the lattice supremum $\bigvee\mathcal{F}$ as a function that satisfies the following two properties: $$ \forall f\in \mathcal{F}\ \ \ \ f\leq \bigvee\mathcal{F} \ \ \text{a.e.} $$ $$ (\forall g\ \forall f\in \mathcal{F}, f\leq g \ \ \text{a.e.}) \quad \Rightarrow \quad (\bigvee\mathcal{F}\leq g \ \ \text{a.e.}) $$ Here of course, we consider measurable functions $g$. Similarly we introduce the lattice infimum $\bigwedge\mathcal{F}$.

If $\mathcal{F}$ is a countable family, then $\mathcal{F}$ can be obtained as the pointwise supremum of $\mathcal{F}$. However, if $\mathcal{F}$ is uncountable, we must distinguish between the lattice supremum $\bigvee\mathcal{F}$ and the pointwise supremum $$ \sup\mathcal{F}: x\to \sup \{u(x): u\in\mathcal{F}\}. $$ The latter one heavily depends on the choice of representatives as the above example shows.

The following result is well known. A similar result is true for the lattice infimum with a very similar proof.

Theorem. Let $\mathcal{F}$ be a class of measurable functions defined in a measurable set $E\subset\mathbb{R}^n$. Then $\bigvee\mathcal{F}$ exists and there is a countable subfamily $\mathcal{G}\subset\mathcal{F}$ such that $$ \bigvee\mathcal{F}=\bigvee \mathcal{G}=\sup \mathcal{G}. $$

Proof. First observe that we may assume that the family $\mathcal{F}$ is bounded in $L^\infty$ and consists of nonnegative functions, otherwise we replace $\mathcal{F}$ by a family of functions $\pi/2+\arctan u$, where $u\in\mathcal{F}$. We can also assume that the functions are defined in a set of finite measure, otherwise we make a diffeomorphic change of variables which maps $E$ onto a bounded set. Let $$ s = \sup \left\{ \int_{E} \max\{u_1,\ldots,u_k\} \,dx: \, u_1,\ldots,u_k\in\mathcal{F} \ \text{for some $k$}\right\}\, . $$ Obviously $s<\infty$. Now there exists a sequence $\{u_1, u_2, \ldots\}\subset\mathcal{F}$ such that $v_k=\max\{u_1,\ldots,u_k\}$ satisfies $\lim\int_E v_k\,dx=s$. Since $v_k$ is nondecreasing we have the a.e. convergence $\lim v_k=v$. Obviously $\int_E v\,dx=s$. This easily implies that $v=\bigvee\mathcal{F}$ and so we can take $\mathcal{G}=\{u_1,u_2\ldots\}$. $\Box$

This proof is taken verbatim from Lemma 2.6 in [1]. Fore more information see [2].

[1] P. Hajlasz, J. Maly, Approximation in Sobolev spaces of nonlinear expressions involving the gradient Ark. Mat. 40 (2002), 245-274.

[2] P. Meyer-Nieberg, Banach Lattices, Springer-Verlag, Berlin, 1991.

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  • $\begingroup$ Thanks, actually in the claim I wrote sup f_i instead of inf f_i... but I suppose that your answer is similar. $\endgroup$ – Ruggero Dec 1 '18 at 17:25
  • $\begingroup$ Your definition of lattice supremum does not have a quantifier for $g$. I assume that it is suppose to say "for all measurable $g$"? $\endgroup$ – Adam Smith Oct 22 '19 at 15:41
  • $\begingroup$ @AdamSmith You are correct. The quantifier for $g$ is not shown, but it is implied from the context. $\endgroup$ – Piotr Hajlasz Oct 22 '19 at 19:25
  • $\begingroup$ I agree it can be deduced. That said, I think it's worth specifying explicitly that $g$ is measurable —without that restriction on $g$, you just recover the usual notion of pointwise infimum. The whole point here is to change the definition, so it makes sense to emphasize the thing that actually changed. (More generally, "implied from the context" relies on a certain level of reader expertise...) $\endgroup$ – Adam Smith Oct 22 '19 at 21:39
  • $\begingroup$ @AdamSmith Is it better now? $\endgroup$ – Piotr Hajlasz Oct 22 '19 at 21:43

More a comment than an answer but I am not entitled. If, as is common, you do not consider actual functions but equivalence classes of a.e. equal functions with respect to a suitable measure, the situation is much more pleasant—— the formal statement is that the usual Banach lattices of Lebesgue integrable functions are Dedekind complete, i.e., order bounded sets have suprema and infima.

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    $\begingroup$ My answer is a detailed version of yours. $\endgroup$ – Piotr Hajlasz Dec 1 '18 at 17:17
  • $\begingroup$ If your index set is $[0,1]$ or $\mathbb{R}^+$ (a typical situation in the theory of stochastic processes), then it would help to know that the process has continuous paths, i.e., $i\mapsto f_i(\omega)$ is continuous for each $\omega$. In this case $\inf f_i$ is the same as the inf taken over countably many $i$ (say rational $i$). I'm not sure whether the setting of the question is close to such an assumption. $\endgroup$ – Dirk Werner Dec 2 '18 at 21:41

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