Let $f$ be increasing on $[a,b]$, i.e. for all $x<y$ in $[a,b]$, $f(x)\leq f(y)$. Also assume $f$ satisfies the intermediate value property. Show that $f$ is continuous on $[a,b]$.

My attempt: Let $f$ be increasing on $[a,b]$ and satisfy IVP, i.e. $\forall x<y$ on $[a,b]$ and $\forall L$ between $f(x)$ and $f(y)$, $\exists c\in(x,y)$ with $f(c)=L$.

Let $\epsilon>0$. Let $c\in(a,b)$. To show $f$ continuous on $[a,b]$, we need to choose a $\delta>0$ such that whenever $|x-c|<\delta$, $|f(x)-f(c)|<\epsilon$.

Since $f$ is increasing and $a<c$, then $f(a)\leq f(c)$. I want to show that $f(c)-\epsilon\leq f(x)\leq f(c)+\epsilon$ but don't know how.

  • $\begingroup$ The one-sided limits for $f$ at any $x_0\in [a,b]$ should exist, which can be proved by monotone convergence theorem. From there, you can show that the left and right hand limits are equal (this is where the IVP comes in handy). $\endgroup$ – Trevor Norton Mar 27 '18 at 15:24
  • $\begingroup$ You can just prove that the preimage of an open interval is an open interval. $\endgroup$ – Michael Greinecker Mar 27 '18 at 16:36

Consider a point $c\in[a,b]$ where $f$ is supposed not continous. So $\exists\epsilon_0>0$ such that there is NO $\delta>0$ with the property that:

if |$x-c$|<$\delta$, then |$f(x)-f(c)$|<$\epsilon_0$.

Now, since $f$ is increasing on [a,b], it is one-to-one on [a,b]. This allows us to look at the expression

$f(c)-\epsilon_0<f(c)<f(c)+\epsilon_0$, and realize that there must be an $x_1,x_2\in[a,b]$ such that $x_1<x_2$ and,



So we have $f(x_1)<f(c)<f(x_2)$, and since $f$ is increasing, $x_1<c<x_2$.

So now choose $\delta$=min{$x_2-c,c-x_1$}

Now with this $\delta$ in mind, if |$x-c$|<$\delta$, it must be the case that $f(x_1)=f(c)-\epsilon_0<f(x)<f(c)+\epsilon_0=f(x_2)$.

But we suppose that there wasn't any $\delta$ that would work for this $\epsilon_0$.


Let $f^-(a)=f(a)$ and let $f^-(x)=\lim_{x'\to x^-}f(x')=\lim_{x'\to x\land x'<x}f(x')$ for $x\in (a,b].$

Similarly $f^+(b)=f(b)$ and let $f^+(x)=\lim_{x'\to x^+}f(x')=\lim_{x'\to x\land x'>x}f(x')$ for $x\in [a,b)$.

$f^-$ and $f^+$ exist because $f$ is monotonic. And because $f$ is increasing we have $f^-(x)\leq f(x)\leq f^+(x)$ for all $x\in [a,b].$

Therefore $f$ is continuous iff $f^-(x)=f^+(x)$ for all $x\in [a,b].$

Because $f$ is increasing we have

(I). $f^-(x)=\sup_{x'\in [a,x)}f(x')$ when $x\in (a,b].$

(II). $f^+(x)=\inf_{x'\in (x,b]}f(x')$ when $x\in [a,b). $

By contradiction suppose $f^+(x)-f^-(x)=r>0$ then at least one $y\in \{f^+(x)-f(x), f(x)-f^-(x)\}$ is positive.

If $f(x)<y<f^+(x)$ then by the def'n of $f^+(x)$ we must have $x<b$ so by (II) we have $\forall x'\in (a,b]\; (y<f^+(x)\leq f(x'))$ . So $x<b$ and $f(x)<y<f(b)$ but no $x'\in (x,b)$ can satisfy $f(x')=y$.

Similarly if $f^-(x)<y<f(x)$ then $a<x$ and (I) applies, implying that $f(a)<y<f(x)$ but no $x'\in (a,x)$ satisfies $f(x')=y.$


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