Let $V$ be a bounded open set in $\mathbb{R}^n$ with $n>0$. A function $f:V\mapsto [-\infty, \infty)$ is called upper semicontinuous if for each $x\in V$ and $\lambda\in\mathbb{R}$ such that $f(x)<\lambda$, there exists an open set $W$ in $V$ containing $x$ such that $f(z)<\lambda$ for all $z\in W$. It is a well-known theorem that states: if $f$ is upper semicontinuous, then there exists a sequence of continuous functions $f_n$ that decreases pointwise to $f$.
My question: If my memory is correct, in case $f$ takes the value of $-\infty$, some $f_n$ may take the value of $-\infty$. In other words, if $f$ is not finite, $f_n$ is not necessarily finite. Is that correct? Can you give an example?
Best Answer
If $f(\xi) = -\infty$, then it is of course possible that $f_{n(\xi)}(\xi) = -\infty$ for some $n(\xi)$. Since $f_n$ decreases pointwise to $f$, we then have $f_n(\xi) = -\infty$ for all $n \ge n(\xi)$. As a simple example take $f(x) \equiv -\infty$ (which is continuous). Then you may take $f_n = f$ for all $n$. Another example is $f : [0,1] \to \mathbb R, f(x) = \dfrac{1}{x-1}$ for $x < 1$ and $f(1) = -\infty$. This is again a continuous function and you may take $f_n = f$ for all $n$.
However, you can avoid that the $f_n$ take the value $-\infty$. If you are given $(f_n)$ which decreases pointwise to $f$, then the functions $\bar f_n(x) = \max(f_n(x),-n)$ are continuous, do not take the value $-\infty$ and decrease pointwise to $f$.