Set defined by a left continuous function is open Let $f:[0,1] \rightarrow \mathbb{R}$ be monotone decreasing and left-continuous (also, the set of discontinuities of $f$ is dense in [0,1], although it is probably irrelevant for my question).
Define for $n \geq 0$, $A_n = \{ x \in [0,1] \ | \ f(x-h)-f(x) > nh \text{ for some } h \in (0,\frac{1}{n})\}$
I need to show $A_n$ is open. 
I tried to show $A_n^c$ is closed, if $f$ were continuous it would be obvious, but it is only continuous from the left. I can show this way that if $x_m \uparrow x$, $x_m \in A_n$, then $x \in A_n$. But I'm having difficulty with a general $x_m \rightarrow x$.
Any help would be appreciated..
Thanks!
 A: If $x\in A_{n}$, so $f(x-h)-f(x)>nh$ for some $h\in (0,1/n)$. 
I want to show that a small open around $x$ is also in $A_{n}$. First notice a couple of things


*

*$\exists\epsilon >0$ such that $h+\epsilon,h-\epsilon\in (0,1/n).$

*$\exists\delta>0$ such that $f(x-h)-f(x)-\delta>nh$.

*Because $f$ is left continuous there exists a $\epsilon'>0$ such that $f(y)-f(x)<\delta$ for $y\in(x-\epsilon',x) $.

*Since $f$ is a monotone decreasing function we have that $f(y)\leq f(x)$ whenever $y\geq x$.


We set 
$\varepsilon''=\min(\varepsilon,\varepsilon',\delta/n)$ 
Take 
$z\in(x-\varepsilon'',x)$.
Now we have 
$\gamma:=x-z>\epsilon''\Rightarrow h-\gamma \in (0,1/n)$.
There for
$$
\begin{eqnarray*}
f(z-(h-\gamma))-f(z)&=&f(x-h)-f(z)\\
&\overset{3}{>}& f(x-h)-f(x)-\delta\\
&=&nh\\
&>&nh-n\gamma=n(h-\gamma)
\end{eqnarray*}
$$
and so $z\in A_n$, we get that $(x-\epsilon'',x)\subset A_n$. 
Take $y\in(x,x+\varepsilon'')$ and set $\nu=y-x$, note we have $h+\nu\in(0,1/n)$ and $\nu\leq \delta/n\Rightarrow \delta- \nu n\geq0$. Recall $f$ is a monotone decreasing function we have that $f(y)\leq f(x)$ whenever $y\geq x$. We get
$$
\begin{eqnarray*}
f(y-(h+\nu))-f(y)&=& f(x-h)-f(y)\\
&\overset{4}{\geq}&f(x-h)-f(x)\\
&\geq&f(x-h)-f(x)-(\delta-\nu n)\\
&\overset{2}{>}&nh+n\nu=n(h+\nu)
\end{eqnarray*}$$ and so $y\in A_n$
So we have shown that $(x-\varepsilon'',x+\varepsilon'')\subset A_n$
