# Monotonic function satisfying darboux property $\Rightarrow$ continuous

Assume $$f : I \rightarrow \mathbb{R}$$ is a non-decreasing on an open interval $$I$$ and that $$f$$ satisfies the Intermediate value property or Darboux's property on $$I$$ (that is, for any $$a < b$$ in $$I$$ and any $$L$$ between $$f(a)$$ and $$f(b)$$, there exists $$c \in [a, b]$$ such that $$f (c) = L)$$.

Then, prove that $$f$$ is continuous.

However, I know that a function can be discontinuous and also satisfy the IVT at the same time. Could someone point me in the right direction?

• Obviously, you must use the fact that $f$ is non-decreasing.
– 5xum
May 27, 2014 at 12:10
• That is, if $f$ is non-decreasing, it can only have jump discontinuities. May 27, 2014 at 12:37
• $f(x) = \begin{cases}\sin(\frac{1}{x}) & ,\text{if}\ x\neq 0 \\0& ,\text{if}\ x = 0 \end{cases}$ satisfies Intermediate Value Property, but is not continuous at $0$
– OBDA
May 27, 2014 at 13:25

A monotonic function can have only discontinuities of first kind or simple discontinuity.

The idea of the proof is that if $p$ is a point of discontinuity, then $f(p-)$ and $f(p+)$ exists and $f(p-)<f(p+)$. We can easily show that $$\color{red}l=\sup\limits_{a<t<p}f(t)=f(p-)<f(p+)=\inf\limits_{p<t<b}f(t)=\color{red}L.$$

Now using definition of $\sup$ and $\inf$, choose some $\varepsilon>0$ and there exists $r$ and $s$ such that $$r<x_1<p\Rightarrow l-\varepsilon<f(x_1)\le l\\ p<x_2<s \Rightarrow L\le f(x_2)<L+\varepsilon$$

Observe that $f(x_1)<\dfrac{l+L}{2}<f(x_2)$ then by intermediate value property, there should be some $c\in (x_1,x_2)$ such that $f(c)=\dfrac{l+L}{2}$. A contradiction!

• @Jack D'Aurizio:How this is a contradiction?? Nov 10, 2017 at 15:26
• @Jack D'Aurizio:Actually yours name is first that clicked to my mind so, i tagged you here.Sorry!!,for the discomfort Nov 10, 2017 at 15:30

$$f$$ is non decreasing on $$[a,b]$$ then for any $$x_1\leq x_2 \implies f(x_1)\leq f(x_2)$$.

for $$c\in [a,b]$$, we want to prove that $$f$$ is continuous at $$c$$, so consider $$\epsilon>0$$. We want to find a delta such that $$x\in (c-\delta,c+\delta) \cap [a,b] \implies |f(x)-f(a)|<\epsilon$$.

So look in the interval $$(f(a)-\epsilon,f(a)+\epsilon)$$. We know that there exists $$f(y)$$ and $$f(z)$$ in $$(f(a)-\epsilon,f(a)+\epsilon)$$ such that $$f(y)\leq f(a) \leq f(z)$$ [IVT gurantees this]

So again by applying IVT, we get for all $$x \in (y,z) \implies f(x) \in (f(a)-\epsilon,f(a)+\epsilon)$$. choose $$\delta = \min [\dfrac{z-a}{2},\dfrac{a-y}{2}]$$

• You should mention the two cases where $f(a)$ is not an interior point. Feb 21, 2021 at 17:02