Example 2

Define a real-valued function $f$ on $\mathbf{R}$ by putting $f(x) = 0$ if $x$ is irrational, and $f(x) = 1/q$ if $x$ is rational and equal to the irreducible fraction $p/q$ ($q > 0$). For each integer $n > 0$, the set of rational numbers $p/q$ such that $q < n$ is closed, and its points are isolated; hence for every irrational $x$ there is a neighbourhood $V$ of $x$ such that $f(y) \leq 1/n$ for all $y \in V$, which shows that $f$ is continuous at $x$; and on the other hand, $f$ has a relative maximum at every rational point $x$. Hence $f$ is upper semi-continuous on $\mathbf{R}$.

The condition for $f$ to be lower semi-continuous at $a$ may be expressed by saying that, for each $h < f(a)$, the set $\overline{f}([h, +\infty])$ must be a neighbourhood of $a$.

It is enough that this condition should be satisfied for an increasing sequence $(h_n)$ of real numbers $< f(a)$ which tend to $f(a)$.

Endow $\overline{\mathbf{R}}$ with the topology in which the open sets are $\varnothing$ and all open intervals of $\overline{\mathbf{R}}$ unbounded on the right (that is, all intervals $]a, +\infty[$ for finite $a$, and the interval $[--\infty, +\infty] = ]\leftarrow, \rightarrow[]$). Then the real-valued function $f$ is lower semi-continuous at $a$ if and only if it is continuous at $a$ when considered as a mapping into $\overline{\mathbf{R}}$ endowed with this topology.