"Upper derivative" of indefinite integral of upper semicontinuous function The following problem is stated as Exercise 22.A(iii) in the book Van Rooij, Schikhof: A Second Course on Real Functions.

Let $f\colon [a,b]\to{\mathbb R}$ be Lebesgue integrable and upper semicontinuous. If $v$ is an indefinite integral of $f$, then $D^+v \le f$.

This serves an auxiliary result used a few times when defining and proving some basic properties of Perron integral. The same result can be found as Theorem 5.32 in Gordon: The integrals of Lebesgue, Denjoy, Perron, and Henstock.
Here indefinite integral means that $v(y)-v(x) = \int_x^y f(t) \; \mathrm{d}t$ for each $x,y\in[a,b]$ and $D^+v$ is defined as
$$D^+v(x) = \limsup_{y\to x} \frac{v(y)-v(x)}{y-x}.$$
This quantity is related to Dini derivatives, with the difference the the limit superior is not one-sided. (Some texts use the notation $\overline Dv$ instead of $D^+v$, but I used here notation from the text where I saw this problem.)
How can we show the above result?
I'll be grateful for comments on my approach posted below as an answer and also for other possible solutions.

Since $v$ is indefinite integral of $f$, we have $$v(y)-v(x) = \int_x^y f(t) \; \mathrm{d}t$$
for any $x,y\in[a,b]$.
We have
\begin{align*}
D^+ v(x)
&= \limsup_{y\to x} \frac{v(y)-v(x)}{y-x} \\
&= \limsup_{y\to x} \frac{\int_x^y f(t) \; \mathrm{d}t}{y-x}. \\
\end{align*}
From upper semicontinuity we know that for any given $x$ and $\varepsilon>0$ we have $f(y)<f(x)+\varepsilon$ for all $y$'s that are close enough to $x$. So for such $y$'s we get
$$\int_x^y f(t) \; \mathrm{d}t \le (f(x)+\varepsilon)(y-x)$$
and
$$D^+ v(x) \le f(x)+\varepsilon.$$
Since this is true for every $\varepsilon>0$, we finally get $D^+ v(x) \le f(x).$
