Let $f(x,u): [0,1]^2 \mapsto \mathbb{R}$ be a continuous
function.


[Q] Is $g(x) = \inf\_{u\in [0,1]} f(x,u)$ always Borel measurable?
If not, can one find a counter-example?


Note that, for any $c$, 
we have
$$(x: g(x) < c) = \text{Proj}\_x ((x,u): f(x,u) < c),$$
where $\text{Proj}\_x$ is a projection operator to $x$-axis.
In the context of measurable selection theorem,
the projection of Borel set $((x,u): f(x,u) < c)$ of
$\mathbb{R}^2$ is
not necessarily a Borel set of $\mathbb{R}$.
But, I can not find a counter-example.


If there exists a proper counter-example, then it also
implies that a semicontinuous real function is not necessarily Borel measurable.


Thanks.



Is a semicontinuous real function  Borel measurable?
We have that $g(x) = \inf\_{u\in [0,1]\cap\mathbb{Q}} f(x,u)$, because $f(x,u)$ is continuous. This shows immediately that $g(x)$ is Borel, in fact Baire-1 because it is the pointwise limit of continuous functions (since $\mathbb{Q}$ is countable).


In general, any upper semi-continuous function $g(x)$ is Borel, in fact Baire-1. To see this, note first that each level set $\{x:g(x)\geq c\}$ is closed, hence $\{x:g(x)>c\}$ is an $F\_\sigma$-set, $\{x:a<g(x)<b\}$ is the intersection of two $F\_\sigma$'s which is $F\_\sigma$, hence the inverse image of any open set is a countable union of $F\_\sigma$'s which is $F\_\sigma$.


