这是一份anu澳大利亚国立大学MATH2320/MATH3116/MATH6110的成功案例

Let $\varepsilon>0$ be given. Let $u \in X \backslash A$ with $d(u, x)<\varepsilon / 2$. (Draw pictures.) Then $a \in A$ implies
$$
d(x, a) \leq d(x, u)+d(u, a)<\frac{\varepsilon}{2}+d(u, a) .
$$
Multiplying the inequality by $g(a)$ and using the fact that $1 \leq g(a) \leq 2$, we get
$$
g(a) d(x, a)<\varepsilon+g(a) d(u, a)
$$
Taking the infimum as $a \in A$ yields
$$
h(x) \leq \varepsilon+h(u) .
$$
Similarly, we show that $h(u) \leq \varepsilon+h(x)$. Thus, $|h(u)-h(x)| \leq \varepsilon$ for $u \in(X \backslash A) \cap B(x, \varepsilon)$. This proves that $f$ is continuous at $x \in X \backslash A$.

MATH2320/MATH3116/MATH6110 COURSE NOTES ：

In fact, $a \notin A \cap B(x, \delta)$ implies
$$
d(u, a) \geq d(x, a)-d(x, u)>\delta-\frac{\delta}{4}=\frac{3 \delta}{4}
$$
so that
$$
\inf {g(a) d(u, a): a \in A \cap B(x, \delta)} \geq \frac{3 \delta}{4} \text { as } g \geq 1 \text {. }
$$
On the other hand, since $x \in A \cap B(x, \delta)$, we deduce
$$
g(x) d(x, u) \leq 2 d(x, u)<2 \cdot \frac{\delta}{4}=\frac{\delta}{2}<\frac{3 \delta}{4}
$$