Uncountable infimum of measurable functions 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.
Thanks
 A: 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.
A: 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.
