Is a semicontinuous real function Borel measurable? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:30:12Z http://mathoverflow.net/feeds/question/90794 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90794/is-a-semicontinuous-real-function-borel-measurable Is a semicontinuous real function Borel measurable? kenneth 2012-03-10T04:16:09Z 2012-03-19T21:46:49Z <p>Let $f(x,u): [0,1]^2 \mapsto \mathbb{R}$ be a continuous function.</p> <p>[Q] Is $g(x) = \inf_{u\in [0,1]} f(x,u)$ always Borel measurable? If not, can one find a counter-example?</p> <p>Note that, for any $c$, we have $$(x: g(x) &lt; c) = \text{Proj}_x ((x,u): f(x,u) &lt; c),$$ where $\text{Proj}_x$ is a projection operator to $x$-axis. In the context of measurable selection theorem, the projection of Borel set $((x,u): f(x,u) &lt; c)$ of $\mathbb{R}^2$ is not necessarily a Borel set of $\mathbb{R}$. But, I can not find a counter-example.</p> <p>If there exists a proper counter-example, then it also implies that a semicontinuous real function is not necessarily Borel measurable.</p> <p>Thanks.</p> http://mathoverflow.net/questions/90794/is-a-semicontinuous-real-function-borel-measurable/90798#90798 Answer by Rami for Is a semicontinuous real function Borel measurable? Rami 2012-03-10T04:56:49Z 2012-03-10T05:03:23Z <p>I think that the answer is positive:</p> <p>It is enough to show that the set $( x | g(x) &lt; c )$ 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.</p> http://mathoverflow.net/questions/90794/is-a-semicontinuous-real-function-borel-measurable/90799#90799 Answer by GH for Is a semicontinuous real function Borel measurable? GH 2012-03-10T05:13:20Z 2012-03-10T05:18:50Z <p>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).</p> <p>In general, any upper semi-continuous function $g(x)$ is Borel, in fact Baire-1. To see this, note first that each level set <code>$\{x:g(x)\geq c\}$</code> is closed, hence <code>$\{x:g(x)&gt;c\}$</code> is an $F_\sigma$-set, <code>$\{x:a&lt;g(x)&lt;b\}$</code> 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 $F_\sigma$.</p> http://mathoverflow.net/questions/90794/is-a-semicontinuous-real-function-borel-measurable/91669#91669 Answer by Felipe Olmos for Is a semicontinuous real function Borel measurable? Felipe Olmos 2012-03-19T21:46:49Z 2012-03-19T21:46:49Z <p>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 $X$.</p> <p>Since you only have to check the measurability property for a generating class of the Borelians in $\mathbb{R}$ you are done.</p>