Is the set $\{ \int_0^x f\,\mathrm d\lambda\mid f(x)=0\}$ a Lebesgue-null set for $f\geq0$? We are given a non-negative, integrable function $f: [0, H]\rightarrow \mathbb R_{\geq0}$ and we define the absolutely continuous and non-decreasing function $F(x):=\int_0^x f\,\mathrm d\lambda$ where $\lambda$ denotes the Lebesgue measure.
I am interested in the set $$
M := \left\{ F(x) ~\middle\vert~ x\in[0,H], f(x)=0\right\}.
$$
Can we show that $M$ is in general a Lebesgue-null set or is there an $f$ such that $M$ has positive Lebesgue measure?
 A: By now, I have found the answer to this problem.
As the commentators stephenkk and GEdgar pointed out, my question is a special variant of Sard's theorem which states (as cited from Wikipedia):
Theorem 1. Let $f: \mathbb R^n\rightarrow\mathbb R^m$ be $k$ times continuously differentiable where $k\geq\max\{n-m+1, 1\}$. Let $X$ denote the set of points $x\in\mathbb R^n$ where the Jacobian matrix of $f$ has rank smaller $m$. Then $f(X)$ is a Lebesgue-null set.
In our case we have $m=n=1$, but $f$ is not necessarily continuously differentiable (not even everywhere differentiable).
Luckily, however, in this special case any arbitrary function has the following property as shown in this answer:
Theorem 2. Let $X$ denote the set of points $x$ where $f$ is differentiable with $f'(x)=0$. Then $f(X)$ is a Lebesgue-null set.
The corresponding proof uses the Infinite Version of Vitali's Covering Lemma (cf. Wikipedia).
I have found yet another answer, that proves the theorem in the special case of absolutely continuous functions using Vitali's Covering Theorem for the Lebesgue-measure (cf.  Wikipedia).
I want to give a slightly more in-depth proof in this setting following the ideas of this answer.
The version of Vitali's Covering Theorem used originates from [1, Exercise 13] where we say a closed interval $[a,b]$ is non-degenerate if $a < b$.
Definition.
A collection $\mathcal{V}$ of closed, bounded, non-degenerate intervals is called a Vitali-Covering of a set $A\subseteq\mathbb R$ if for all $x\in A$ and $\varepsilon >0$ there exists an interval $[a,b]\in\mathcal{V}$ with $x\in[a,b]$ and $b-a < \varepsilon$.
Vitali's Covering Theorem.
Let $A$ be a measurable set with $\lambda^*(A) < \infty$ and let $\mathcal{V}$ be a Vitali-Covering of $A$ consisting of closed, bounded intervals.
Then there exists a finite or countably infinite disjoint sequence $([a_i, b_i])_{i\in I}$ of intervals in $\mathcal{V}$ with $$
    \lambda^*\left( A\setminus\bigcup_{i\in I} [a_i, b_i] \right) = 0
.$$
Here, $\lambda^*$ is the Lebesgue outer measure.
Moreover, in the following proof it is important to know, that an absolutely continuous function $f$ has the Luzin N property, which means that $f$ maps Lebesgue-null sets to Lebesgue-null sets.
A simple proof of this can be seen eg. in this answer.
Theorem 3. Let $I$ be an interval (potentially unbounded) and let $f:I \rightarrow\mathbb R$ be absolutely continuous on any subinterval $[a,b]\subseteq I$. Let $X$ denote the set of points $x\in \mathbb R$ at which $f$ is differentiable with $f'(x)=0$ . Then $f(X)$ is a Lebesgue-null set.
Proof.
We can restrict $f$ to a bounded, closed interval $[a,b]$ and show the property on the restriction, because then we use the subadditivity of the Lebesgue-measure to show $$
  \lambda^*(f(X)) \leq \sum_{z\in\mathbb Z} \lambda^*(f|_{[z, z+1]}(X\cap[z,z+1]) = 0
.$$
Let $f:[a,b]\rightarrow \mathbb R$ absolutely continuous. For arbitrary $\varepsilon > 0$ we show $\lambda(f(A)) < \varepsilon$.
We define a Vitali-Covering as the set $$
\mathcal{V} := \left\{
    B_h(x) ~\middle\vert~ x\in A, h>0, \forall y\in B_h(x): |f(y)-f(x)| < \frac{\varepsilon}{2(b-a)} \cdot |y - x|
\right\}
,$$
where we denote $B_h(x) := [x - h, x+h]$.
This is indeed a Vitali-Covering for $A$ as for all $x\in A$ we have $$
    \lim_{y\rightarrow x} \frac{|f(y) - f(x|}{|y-x|} = 0
$$
and thus for all $\delta > 0$ there is an interval of length smaller than $\delta$ in $\mathcal V$ containing $x$.
By Vitali's Covering Theorem there is a finite and pairwise disjoint sequence $(B_{h_i}(x_i))_{i\in I}$ of intervals in $\mathcal V$ with $\lambda^*(A\setminus\bigcup_{i\in I} B_{h_i}(x_i)) = 0$.
We use this together with the Luzin N property to estimate \begin{align*}
    \lambda^*(f(A))
    &\leq \lambda^*\left( f\left(A\setminus\bigcup_{i\in I} B_{h_i}(x_i)\right) \cup f\left(\bigcup_{i\in I} B_{h_i}(x_i)\right)  \right) \\
    &\leq\sum_{i\in I} \lambda^*\left(f\left(B_{h_i}(x_i)\right)\right)
.\end{align*}
For all $i\in I$ we know that $f(B_{h_i}(x_i))$ is contained in an interval of length $h_i\cdot \varepsilon/(b-a)$.
Moreover, as all $B_{h_i}(x_i)$ are pairwise disjoint subsets of $[a,b]$, we have $\sum_{i\in I} 2\cdot h_i \leq b-a$.
This implies $$
    \lambda^*(f(A))
    \leq \sum_{i\in I} \lambda^*(f(B_{h_i}(x_i)))
    \leq \sum_{i\in I} \frac{h_i\cdot \varepsilon}{b-a}
    \leq \frac{\varepsilon\cdot(b-a)}{2\cdot(b-a)} < \varepsilon
.$$
$\square$
Let's answer the original question.
In our case $F: [0, H] \rightarrow \mathbb R$ is an absolutely continuous function (even if $f$ is not non-negative) by the fundamental theorem of Lebesgue integral calculus (cf. Wikipedia) with derivative $F'(x)=f(x)$ for almost all $x\in\mathbb R$.
Hence, the theorem above applies.

References
[1]: Halsey L. Royden and Patrick M. Fitzpatrick. Real Analysis. 4th ed. Prentice Hall, 2010. ISBN: 9780131437470.

Disclaimer: This is an enriched excerpt of my Master's thesis.
