12.1 Continuity

12.1   Definition (Continuity at a point.) Let $f$ be a real valued function such that $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}}$. Let $a\in\mbox{{\rm dom}}(f)$. We say that $f$is continuous at $a$ if and only if

$$\lim_{x\to a}f(x)=f(a).$$

Remark: According to this definition, in order for $f$ to be continuous at $a$ we must have

$$a\in\mbox{{\rm dom}}(f)$$

and

$$a \mbox{ is approachable from } \mbox{{\rm dom}}(f).$$

The second condition is often not included in the definition of continuity, so this definition does not quite correspond to the usual definition.

Remark: The method we will usually use to show that a function $f$ is not continuous at a point $a$, is to find a sequence $\{x_n\}$ in $\mbox{{\rm dom}}(f)\setminus\{a\}$ such that $\{x_n\}\to a$, but $\{f(x_n\}$ either diverges or converges to a value different from $f(a)$.

12.2   Definition (Continuity on a set.) Let $f$ be a real valued function such that domain $(f)\subset\mbox{{\bf R}}$, and let $S$ be a subset of domain$(f)$. We say that $f$ is continuous on $S$ if $f$ is continuous at every point in $S$. We say that $f$ is continuous if $f$ is continuous at every point of domain$(f)$.

12.3   Example ($\sin$, $\cos$, $\ln$ and power functions are continuous.) We
proved in lemma 11.17 that a function is continuous at every point at which it is differentiable. (You should now check the proof of that lemma to see that we did prove this.) Hence $\sin$, $\cos$, $\ln$, and $x^n$ (for $n\in\mbox{{\bf Z}}$) are all continuous on their domains, and if $r \in \mbox{{\bf Q}}\setminus \mbox{{\bf Z}}$, then $x^r$ is continuous on $\mbox{${\mbox{{\bf R}}}^{+}$}$.

12.4   Example. Let

$$f(x)=\cases{ 0 &if $x\leq 0$\cr 1 &if $x>0$.\cr}$$

Then $f$ is not continuous at $0$. For the sequence $${ \{ {1\over n}\}}$$ converges to $0$, but $${ \{ f( {1\over n})\}=\{1\} \to 1\neq f(0)}$$.

Our limit rules all give rise to theorems about continuous functions.

12.5   Theorem (Properties of continuous functions.) Let $f,g$ be real valued functions with $\mbox{{\rm dom}}(f)\subset\mbox{{\bf R}},\; \mbox{{\rm dom}} (g)\subset\mbox{{\bf R}}$, and let $c,a\in\mbox{{\bf R}}$. If $f$ and $g$ are continuous at $a$ and if $a$ is approachable from $\mbox{{\rm dom}} (f)\cap \mbox{{\rm dom}}(g)$, then $f+g,\; f-g,\;fg,\;$ and $cf$ are continuous at $a$. If in addition, $g(a)\neq 0$ then $${ {f\over g}}$$ is also continuous at $a$.

Proof: Suppose $f$ and $g$ are continuous at $a$, and $a$ is approachable from $\mbox{{\rm dom}} (f)\cap \mbox{{\rm dom}}(g)$. Then

$$\lim_{x\to a} f(x) = f(a) \mbox{ and }\lim_{x\to a} g(x) = g(a).$$

By the sum rule for limits (theorem 10.15) it follows that

$$\begin{eqnarray*} \lim_{x \to a} (f+g)(x) &=& \lim_{x \to a} f(x) + \lim_{x\to a} g(x)\\ &=& f(a) + g(a) = (f+g)(a). \end{eqnarray*}$$

Thus $f+g$ is continuous at $a$. The proofs of the other parts of the theorem are similar.

12.6   Example (An everywhere discontinuous function.) Let $D$ be the example of a non-integrable function defined in equation (8.37). Then $D$ is not continuous at any point of $[0,1]$. Recall

$$D(x)=\cases{ 1 &if $x\in S$\cr 0 &if $x\neq S$\cr}$$

where $S$ is a subset of $[0,1]$ such that every subinterval of $[0,1]$ of positive length contains a point in $S$ and a point not in $S$. Let $x\in [0,1]$.

Case 1.

If $x \in S$ we can find a sequence of points $\{t_n\}$ in $[0,1]\setminus S$ such that $\{t_n\}\to x$. Then

$$\{D(t_n)\}=\{0\}\to 0\neq D(x)$$

so $D$ is not continuous at $x$.

Case 2.

If $x \not\in S$ we can find a sequence of points $\{s_n\}$ in $S$ such that $\{s_n\}\to x$. Then

$$\{D(s_n)\}=\{1\}\to 1\neq D(x)$$

so $D$ is not continuous at $x$.

12.7   Example. Let

$$h(x)=\sqrt x \mbox{ for } x\in\mbox{{\bf R}}_{\geq 0}.$$

I claim that $h$ is continuous. We know that $h$ is differentiable on $\mbox{${\mbox{{\bf R}}}^{+}$}$, so $H$ is continuous at each point of $\mbox{${\mbox{{\bf R}}}^{+}$}$. In example 10.13 we showed that $${\lim_{x\to 0}h(x)=0=h(0)}$$ so $h$ is also continuous at $0$.

12.8   Example. Let

$$\begin{eqnarray*} f(x) &=& -x^2,\\ g(x) &=& \sqrt x. \end{eqnarray*}$$

Then $f$ and $g$ are both continuous functions. Now

$$(g\circ f)(x)=\sqrt{-x^2}$$

and hence $g \circ f$ is not continuous. The domain of $g \circ f$ contains just one point, and that point is not approachable from $\mbox{{\rm dom}}(g\circ f)$.

12.9   Theorem (Continuity of compositions.) Let $f,g$ be functions with domains contained in $\mbox{{\bf R}}$ and let $a\in\mbox{{\bf R}}$. Suppose that $f$ is continuous at $a$ and $g$ is continuous at $f(a)$. Then $g \circ f$ is continuous at $a$, provided that $a$ is approachable from $\mbox{{\rm dom}}(g\circ f)$.

Proof: Suppose $f$ is continuous at $a$ and $g$ is continuous at $f(a)$, and $a$ is approachable from $\mbox{{\rm dom}}(g\circ f)$. Let $\{x_n\}$ be a sequence in $\mbox{{\rm dom}}(g\circ f)\setminus \{a\}$ such that $\{x_n\}\to a$. Then $\{f(x_n)\}\to f(a)$ since $f$ is continuous at $a$. Hence $\{g\Big( f(x_n)\Big)\}\to g\Big( f(a)\Big)$ since $g$ is continuous at $f(a)$; i.e.,

$$\{ (g\circ f)(x_n)\}\to (g\circ f)(a).$$

Hence $g \circ f$ is continuous at $a$.