I don't understand how the proof of the theorem below works. (Theorem 13.26, Real Analysis, J. Yeh, 2nd ed.)
"Let $f$ be a real-valued continuous function on [a,b] such that $f'$ exists almost everywhere (a.e.) and $f'$ is $\mu_L$-integrable on [a,b]. If the upper-right Dini derivative $D^+f > -\infty$, on [a,b), then we have $\int_{[a,x]}f'd\mu_L=f(x)-f(a)$ for every $x\in[a,b]$."
The proof goes like this.
The author defines additional functions.
$g_n(x)=min\{f'(x),n\}$, $G_n=\int_{[a,x]}g_nd\mu_L$ $\;$ for $x\in[a,b]$
($\lim\limits_{n\rightarrow\infty}G_n=\lim\limits_{n\rightarrow\infty}\int_{[a,x]}g_nd\mu_L=\int_{[a,x]}f'd\mu_L$)
$h_n(x)=max\{f'(x),-n\}$, $H_n=\int_{[a,x]}h_nd\mu_L$ $\;$ for $x\in[a,b]$
($\lim\limits_{n\rightarrow\infty}H_n=\lim\limits_{n\rightarrow\infty}\int_{[a,x]}h_nd\mu_L=\int_{[a,x]}f'd\mu_L$)
Then,
$D^+G_n(x)=\limsup\limits_{h\downarrow0}h^{-1}\int_{[x,x+h]}g_nd\mu_L\le\limsup\limits_{h\downarrow0}h^{-1}\int_{[x,x+h]}n\mu_L=n$, and
$D^+(f-G_n)\ge D^+f-D^+G_n\gt -\infty$ (in addition to $\ge 0$ a.e.), because $D^+f > -\infty$ by supposition, and $D^+G_n\le n$.
Thus,
A real valued continuous fucntion $f-G_n$ is an increasing function, and $f(x)-G_n(x)-(f(a)-G_n(a))\ge0,\;f(x)-f(a)\ge G_n(x)\,for\,x\in[a,b)$.
Since this holds for every $n\in\mathbb{N}$, $f(x)-f(a)\ge\lim\limits_{n\rightarrow\infty}G_n(x)=\int_{[a,x]}f'd\mu_L$.
Also,
$D^+H_n(x)=\limsup\limits_{h\downarrow0}h^{-1}\int_{[x,x+h]}h_nd\mu_L\ge\limsup\limits_{h\downarrow0}h^{-1}\int_{[x,x+h]}-n\mu_L=-n$, and
$D^+(H_n-f)\ge D^+H_n-D^+f\gt -\infty$ (and $\ge 0$ a.e.), because $D^+f > -\infty$, and $D^+H_n\ge -n$.****
Thus,
$H_n-f$ is an increasing function, and $H_n(x)-f(x)-(H_n(a)-f(a))\ge0,\;H_n(x)\ge f(x)-f(a)\,for\,x\in[a,b)$.
$\lim\limits_{n\rightarrow\infty}H_n(x)=\int_{[a,x]}f'd\mu_L\ge f(x)-f(a)$.
Finally,
$\int_{[a,x]}f'd\mu_L\ge f(x)-f(a)\ge\int_{[a,x]}f'd\mu_L$,
$\int_{[a,x]}f'd\mu_L= f(x)-f(a)$.
I'm stuck at the line marked by "****."
How does $D^+f > -\infty$ and $D^+H_n\ge -n$ mean $D^+H_n-D^+f\gt -\infty$?