Is a semicontinuous real function  Borel measurable? 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.
 A: I think that the answer is positive:
It is enough to show that the set  $( x | g(x) < c )$ is Borel. as you saed it is an image under $Proj_x$ of an open set $U$. divide $[0,1]^2$ to a union of its interior $(0,1)^2$ and the boundary. Correspondingly divide $U$ into $U_0:= (0,1)^2 \cap U$ and its complement $Z$. it is enough  to show the the image of each of them under $Proj_X$ is Borel. Which is evident.
A: I think every (lower) semicontinuous function $f:X \to \mathbb{R}$ is Borel measurable, since you have the following characterization: for every $a \in \mathbb{R}$ the set
$$ f^{-1}((-\infty, a])$$
is closed in the topology that you are considering in $X$.
Since you only have to check the measurability property for a generating class of the Borelians in $\mathbb{R}$ you are done.
A: 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 $F_\sigma$.
