Upper and lower semicontinuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$

Let $A \subseteq \mathbb{R}$, let $f : A \to \mathbb{R}$ and let $c \in A$, with neighbourhoods as in The $\varepsilon$-neighbourhood and the punctured $\varepsilon$-neighbourhood of a point of $\mathbb{R}$.

• $f$ is upper semicontinuous at $c$ when for every real $\varepsilon > 0$ there is a real $\delta > 0$ with $f(x) \;<\; f(c) + \varepsilon \qquad \text{for every } x \in A \cap N_\delta(c).$
• $f$ is lower semicontinuous at $c$ when for every real $\varepsilon > 0$ there is a real $\delta > 0$ with $f(x) \;>\; f(c) - \varepsilon \qquad \text{for every } x \in A \cap N_\delta(c).$
• $f$ is upper semicontinuous on $A$, respectively lower semicontinuous on $A$, when it is so at every point of $A$.

In words: an upper semicontinuous function cannot jump up in the limit, and a lower semicontinuous one cannot jump down. Both conditions are pointwise, both quantify over the same unpunctured neighbourhoods as Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point, and at $x = c$ each holds automatically, since $f(c) < f(c) + \varepsilon$ and $f(c) > f(c) - \varepsilon$.

Continuity is exactly the conjunction

$f$ is continuous at $c$ if and only if it is both upper and lower semicontinuous at $c$.

If $f$ is continuous at $c$, a $\delta$ witnessing $|f(x) - f(c)| < \varepsilon$ on $A \cap N_\delta(c)$ witnesses both displayed conditions, since $|f(x) - f(c)| < \varepsilon$ gives $-\varepsilon < f(x) - f(c) < \varepsilon$ (Basic properties of the absolute value).

Conversely, given $\varepsilon > 0$ take $\delta_1$ for the upper condition and $\delta_2$ for the lower one and put $\delta := \min\{\delta_1, \delta_2\} > 0$. For $x \in A \cap N_\delta(c)$ both $f(x) < f(c) + \varepsilon$ and $f(x) > f(c) - \varepsilon$ hold, that is $|f(x) - f(c)| < \varepsilon$ (Basic properties of the absolute value). So $f$ is continuous at $c$.

Consequently $f$ is continuous on $A$ exactly when it is both upper and lower semicontinuous on $A$.

Negation exchanges the two

$f$ is upper semicontinuous at $c$ if and only if $-f$ is lower semicontinuous at $c$, since $f(x) < f(c) + \varepsilon$ says the same thing as $-f(x) > -f(c) - \varepsilon$ (Complete ordered field (least-upper-bound property)). Every statement about one notion below is therefore proved for one of them and transferred to the other by this substitution, never proved twice.

Neither notion implies the other, and neither implies continuity. The indicator of a closed set is upper semicontinuous and the indicator of an open set is lower semicontinuous, and neither is continuous unless the set is clopen; the companion page uses an upper semicontinuous function on $[0,1]$ that attains no minimum.