7
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Do you have an example where $g$ is not continues $\endgroup$ – Rami Mar 10 '12 at 5:27
14
$\begingroup$

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 $F_\sigma$.

$\endgroup$
0
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ In-fact the division of $U$ into 2 sets is unnecessary and the image of $U$ is just open. This dose not prove continuity yet since it is not enough to check continuity on sets like this. $\endgroup$ – Rami Mar 10 '12 at 5:23
  • $\begingroup$ I did not explained why it is enough to show that $(x|g(x) < c)$ is Borel. I now understand that is probably the point that you where interested in. But it is explained in the answer of GH so there is no point to repeat it $\endgroup$ – Rami Mar 10 '12 at 5:32
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.