Let $f$ be continuous on $[a,b]$, and assume that $D^-f\geq 0.$ Show that $f$ increasing on $[a,b]$.(Hint: first show that for a function $g$ on $(a,b)$, then apply this to the function $g=f+x\epsilon.$ )
Here is my attempt for proving $g$ is increasing.
Let $g$ be a function on $(a,b)$, $a,b \in \mathbb{R}.$ Assume that $D^-f(x)\geq \epsilon>0$, for any $\epsilon>0.$ Let $x<y \in (a,b).$ Since $$\lim_{\delta \rightarrow 0}\inf_{0<h<\delta}\frac{g(x+h)-g(x)}{h}\geq \epsilon>0,$$ and $$\lim_{\delta \rightarrow 0}\inf_{-\delta<h<0}\frac{g(x+h)-g(x)}{h}\geq \epsilon>0,\ \ \ \delta>0$$ then $\frac{g(x+h)-g(x)}{h}\geq \epsilon>0$ for any $0<h<\delta,$ and $\frac{g(x+h)-g(x)}{h}\geq \epsilon > 0$ for any $-\delta<h<0$. Assume for the contrary that $g(y)<g(x)$. Let $\delta=|y|+1$. Then, we have that $$ \frac{g(y)-g(x)}{h}<0,\ \ \ \ \text{for any $h>0$}$$ but since $y>x$, we can find such an $0<h<\delta$ with $\delta=|y|+1$ such that $y=x+h,$ so we have $$ \frac{g(x+h)-g(x)}{h}<0$$ which is a contradiction since $\frac{g(x+h)-g(x)}{h}>0$.
I am not sure if my argument works properly, so I would appreciate any thought about that.