Upper semi continuity approximation continuous functions

Question

I am currently studying Potential Theory of the Complex Plane written by Ransford. In his early section of his second chapter, he introduces the notion of upper semi continuous functions (u.s.c) and he proves a theorem of approximating such functions by continuous ones, Theorem 2.1.3, having the assumption that the u.s.c function $u$ should be bounded above.

Next, he presents an exercise which states that this condition can be removed by using a distinct test function, defined as follows.

Let $(X,d)$ be a metric space and $u\colon X\rightarrow [-\infty,+\infty)$ be an u.s.c. For $n\geq 0$, define the $$F_n\doteqdot \{x\in X\colon u(x)\geq n\}$$ $$\psi_n(x)\doteqdot \max(0,1-n dist(x,F_n))\qquad (x\in X)$$

Then one can show that the sum $\sum_{n=0}\psi_n$ converges locaaly uniformly to a continous function $\psi$ with $\psi\geq u$.

Now, the problem is first I cannot even vizualize how this series to such function. I thought of concrete examples of upper semi continuous functions, like $$u(x)=x \text{ for }0\leq x<1\qquad u(x)=2 \text{ for }1\leq x<+\infty$$ Then, computing easily the $F_n$ and $\psi_n$, the result is validated. I cannot think of a general way to approach its proof however. Any help shall be appreciated.

Answer 1

The fact lies here to see that the exercise desires locally uniform convergence. In particular, let $K$ be a compact subset of $X$. By a Proposition it is true that $u$ is bounded above on $K$, so there exists an index $N_0\in\mathbb{N}$, so that $$u(x)< N_0\qquad\forall x\in K\,.$$ In particular, we have that $K\cap F_n=\emptyset$ for every $n\geq N_0$,. Since the definition of distance says that $\operatorname{dist}(x,\emptyset)=+\infty$ then it is true that by definition of $\phi_n$ $$\phi_n(x)=0 \qquad\forall x\in K\qquad \forall n\geq N_0.$$ In particular, the above series is apparently a finite sum of functions or the partial sums of the series is an eventually constant sequence of functions. Hence it trivially converges uniformly on $K$ and therefore converges in $X$ to a function $\psi\colon X\rightarrow \mathbb{R}$. We have to prove that $\psi$ is continuous. Since the distance function is continuous, it is clear that $\psi_n$ is a maximum of continuous functions, and henceforth continuous. See a similar question answered in here.

So for each $n$, the function $\psi_n$ is continuous. Since $\psi$ consists of finite sum of continuous functions, it is also continuous. \par Finally $\psi(x)\geq u(x)$ for every $x\in X$. Fix an $x\in X$. Since $u$ does not take $+\infty$ value, then there exists a $n_x\in\mathbb{N}$ so that $$n_x-1\leq u(x)< n_x\,.$$ In particular, $x\in F_n$ for all $n=0,\dots n_x-1$ while $x\notin F_n$ for all $n\geq n_x$. So $$\operatorname{dist}(x,F_n)=0\qquad\forall n=0,\dots,n_x-1$$ and therefore $$\psi_n(x)\equiv 1\qquad \forall n=0,\dots,n_x-1\,.$$ However, by the definition of $\psi$, $$\psi(x)\geq n_x>u\,.$$ As $x\in X$ was arbitrary the result is proven.