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'semicontinuous'
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| Title of object: |
semicontinuous |
| Canonical Name: |
LowerSemicontinuous |
| Type: |
Definition |
| Created on: |
2003-10-15 01:05:31 |
| Modified on: |
2005-12-14 23:17:59 |
| Classification: |
msc:26A15 |
| Defines: |
lower semicontinuous, upper semicontinuous, lower semi-continuous, upper semi-continuous |
Revision comment (for changes between this and next version):
| fix spelling of "continuous" |
Preamble:
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Content:
Suppose $X$ is a topological space, and $f$ is a function
from $X$ into the extended real numbers $\overline{\R}$; $f:X\to \overline{\R}$.
Then:
\begin{enumerate}
\item
If
$f^{-1}((a,\infty])=\{x\in X \mid f(x) >\alpha\}$
is an open set in $X$ for all $\alpha\in \R$,
then $f$ is said to be {\bf lower semicontinuous}.
\item
If
$f^{-1}([-\infty,\alpha))=\{x\in X \mid f(x) <\alpha\}$
is an open set in $X$ for all $\alpha\in \R$,
then $f$ is said to be {\bf upper semicontinuous}.
\end{enumerate}
In other words, $f$ is lower semicontinuous, if $f$ is continuous with
respect to the topology for $\overline{\R}$ containing $\emptyset$ and
open sets
$$
U(\alpha) = (\alpha,\infty], \quad \quad \alpha\in \R\cup \{-\infty\}.
$$
It is not difficult to see that this is a topology. For example,
for a union of sets $U(\alpha_i)$ we have $\cup_i U(\alpha_i)=U(\inf \alpha_i)$.
Obviously, this topology is much coarser than
the usual topology for the extended numbers.
However,
the sets $U(\alpha)$ can be seen as neighborhoods of infinity, so
in some sense, semicontinuous functions are "continous at infinity"
(see example 3 below).
\subsubsection{Examples}
\begin{enumerate}
\item A function $f\colon X\to \overline{\R}$ is continuous if and only if
it is lower and upper semicontinuous.
\item Let $f$ be the characteristic function of a set $\Omega\subseteq X$.
Then $f$ is lower (upper)
semicontinuous if and only if $\Omega$ is open (closed).
This also holds for the function that
equals $\infty$ in the set and $0\,$ outside.
It follows that the characteristic function of $\Q$ is not
semicontinuous.
\item On $\R$, the function $f(x)=1/x$ for $x\neq 0$ and $f(0)=0$, is not
semicontinuous. This example illustrate how semicontinuous "at infinity".
\end{enumerate}
\subsubsection{Properties}
Let $f\colon X\to \overline{R}$ be a function.
\begin{enumerate}
\item Restricting $f$ to a subspace preserves semicontinuity.
\item Suppose $f$ is upper (lower) semicontinuous, $A$ is a topological space, and
$\Psi\colon A\to X$ is a homeomorphism. Then $f\circ\Psi$ is upper (lower) semicontinuous.
\item Suppose $f$ is upper (lower) semicontinuous, and
$S\colon \overline{\R}\to \overline{\R}$ is a sense preserving homeomorphism.
Then $S\circ f$ is upper (lower) semicontinuous.
\item $f$ is lower semicontinuous if and only if
$-f$ is upper semicontinuous.
\end{enumerate}
\begin{thebibliography}{9}
\bibitem{rudin_real}
W. Rudin, \emph{Real and complex analysis}, 3rd ed., McGraw-Hill Inc., 1987.
\bibitem{cohn}
D.L. Cohn, \emph{Measure Theory}, Birkh\"auser, 1980.
\end{thebibliography} |
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