Increasing bounded function defined in a closed interval [closed]

Every increasing function defined on a closed interval such that the imagem is a interval, is continuos and integrable if this interval is bounded.

I dont know if this is true, I just thought that any increasing function that has a jump its not continuos, so should be false.

Solution 1:

The statement is true. Consider first an increasing function $f:I\to\mathbb{R}$ defined on an interval $I\subseteq\mathbb{R}$. Let $x_0\in I$ be arbitrary. We want to show that $f$ is continuous in $x_0$.

Hence, fix $\varepsilon>0$.

Suppose there are $x_1,x_2\in I$ such that $f(x_1)<f(x_0)<f(x_2)$. (Otherwise $f$ is constant or at least it is constant on the left/right hand side of $x_0$ and we can carry out the following construction only on one side.)

Define $y_1:=\max\{f(x_1),f(x_0)-\varepsilon\}$ and $y_2:=\min\{f(x_2),f(x_0)+\varepsilon\}$ Since $f(I)$ is an interval, we have $[y_1,y_2]\subseteq[f(x_1),f(x_2)]\subset I$. Therefore, there are $z_1,z_2\in I$ such that $f(z_1)=y_1$ and $f(z_2)=y_2$. Then, we can define $\delta:=\min\{z_2-x_0,x_0-z_1\}$.

Now, for any $x\in I$ with $\vert x-x_0\vert<\delta$, we have $$z_1<x<z_2$$ and because $f$ is increasing $$f(x_0)-\varepsilon\leq y_1=f(z_1)\leq f(x)\leq f(z_2)=y_2\leq f(x_0)+\varepsilon.$$ Thus, $\vert f(x)-f(x_0)\vert\leq\varepsilon$.

We have to be a bit careful about the question of integrability. The way your question is posed, I understand that the image interval is bounded. This does not guarantee integrability (consider e.g. $f\equiv 1$ on $\mathbb{R}$). However, if (what you probably meant) $I$ is a bounded, closed interval, then $f$ attains a maximum and minimum in I and thus $f$ is indeed integrable with $$\int_I\vert f(x)\vert\ \mathrm{d}x\leq\vert I\vert\max\{\vert\max f\vert,\vert\min f\vert\}.$$