Value bound for integrable periodic function

$\def\F{\mathscr{F}}\def\R{\mathbb{R}}\def\peq{\mathrel{\phantom{=}}{}}\def\ac{\text{ac}}\def\pc{\text{pc}}\def\emptyset{\varnothing}$Define\begin{align*} \F_{\ac} &= \{F: \R \to \R \mid F \text{ is absolutely continuous and increasing},\\ &\peq F(0) = 0,\ F(x + 1) = F(x) + 1\ (\forall x \in \R)\},\\ \F_{\pc} &= \{F: \R \to \R \mid F \text{ is piecewise continuous and increasing},\\ &\peq F(0) = 0,\ F(x + 1) = F(x) + 1\ (\forall x \in \R)\}, \end{align*} and for any $F: \R \to \R$ and $a \in (0, 1)$, define$$ A(F, a) = \{x \in [0, 1) \mid F(x + a) - F(x) \geqslant a\}. $$ Finally, for $B \subseteq \R$ and $c \in \R$, define $B + c = \{x + c \mid x \in B\}$ and $B - c = B + (-c)$. Now the original question can be phrased as finding $\inf\limits_{F \in \F_{\ac}} \left| A\left( F, \dfrac{3}{5} \right) \right|$, where $|\,·\,|$ is the Lebesgue measure, as $F' = f$.

Step 1: $\inf\limits_{F \in \F_{\ac}} \left| A\left( F, \dfrac{3}{5} \right) \right| \geqslant \inf\limits_{F \in \F_{\pc}} \left| A\left( F, \dfrac{3}{5} \right) \right| \geqslant \dfrac{1}{5}$.

Proof: For a fixed $F \in \F_{\pc}$ and any $x \in \left[ 0, \dfrac{1}{5} \right)$, if $A\left( F, \dfrac{3}{5} \right) \cap \left\{ x, x + \dfrac{1}{5}, \cdots, x + \dfrac{4}{5} \right\} = \emptyset$, then\begin{align*} F(x) &> F\left( x + \frac{3}{5} \right) - \frac{3}{5} > F\left( x + \frac{6}{5} \right) - \frac{6}{5} = F\left( x + \frac{1}{5} \right) - \frac{1}{5}\\ &> F\left( x + \frac{4}{5} \right) - \frac{4}{5} > F\left( x + \frac{7}{5} \right) - \frac{7}{5} = F\left( x + \frac{2}{5} \right) - \frac{2}{5}\\ &> F(x + 1) - 1 = F(x), \end{align*} a contradiction. Thus there exists $k \in \{0, 1, \cdots, 4\}$ such that $x + \dfrac{k}{5} \in A\left( F, \dfrac{3}{5} \right)$, which implies that$$ x \in A\left( F, \dfrac{3}{5} \right) \cap \left[ \dfrac{k}{5}, \dfrac{k + 1}{5} \right) - \dfrac{k}{5}. $$ Therefore $\left[ 0, \dfrac{1}{5} \right) \subseteq \bigcup\limits_{k = 0}^4 \left( A\left( F, \dfrac{3}{5} \right) \cap \left[ \dfrac{k}{5}, \dfrac{k + 1}{5} \right) - \dfrac{k}{5} \right)$ and\begin{align*} \left|\left[ 0, \dfrac{1}{5} \right) \right| &\leqslant \sum_{k = 0}^4 \left| A\left( F, \frac{3}{5} \right) \cap \left[ \frac{k}{5}, \frac{k + 1}{5} \right) - \frac{k}{5} \right| = \sum_{k = 0}^4 \left| A\left( F, \frac{3}{5} \right) \cap \left[ \frac{k}{5}, \frac{k + 1}{5} \right) \right|\\ &= \left| \bigcup_{k = 0}^4 \left( A\left( F, \frac{3}{5} \right) \cap \left[ \frac{k}{5}, \frac{k + 1}{5} \right) \right) \right| = \left| A\left( F, \frac{3}{5} \right) \right|, \end{align*} i.e. $\left| A\left( F, \dfrac{3}{5} \right) \right| \geqslant \dfrac{1}{5}$. Note that $\F_{\ac} \subseteq \F_{\pc}$, then$$ \inf\limits_{F \in \F_{\ac}} \left| A\left( F, \frac{3}{5} \right) \right| \geqslant \inf\limits_{F \in \F_{\pc}} \left| A\left( F, \frac{3}{5} \right) \right| \geqslant \dfrac{1}{5}. $$

Step 2: $\inf\limits_{F \in \F_{\ac}} \left| A\left( F, \dfrac{3}{5} \right) \right| = \min\limits_{F \in \F_{\pc}} \left| A\left( F, \dfrac{3}{5} \right) \right| = \dfrac{1}{5}$.

Proof: Defining$$ F_{\tfrac{3}{5}}(x) = \begin{cases} 0; & x \in \left[ 0, \dfrac{3}{5} \right)\\ \dfrac{1}{2}; & x \in \left[ \dfrac{3}{5}, 1 \right) \end{cases} $$ and extending $F_{\tfrac{3}{5}}$ on $\R$ so that $F_{\tfrac{3}{5}} \in \F_{\pc}$, it is easy to see that $\left| A\left( F_{\tfrac{3}{5}}, \dfrac{3}{5} \right) \right| = \dfrac{1}{5}$, thus $\min\limits_{F \in \F_{\pc}} \left| A\left( F, \dfrac{3}{5} \right) \right| = \dfrac{1}{5}$. Although $F_{\tfrac{3}{5}} \not\in \F_{\ac}$, it can be modified in the following way: For $0 < ε < \dfrac{1}{5}$, define$$ F_{\tfrac{3}{5}, ε}(x) = \begin{cases} 0; & x \in \left[ 0, \dfrac{3}{5} - ε \right)\\ \dfrac{1}{2ε} \left( x - \dfrac{3}{5} \right) + \dfrac{1}{2}; & x \in \left[ \dfrac{3}{5} - ε, \dfrac{3}{5} \right)\\ \dfrac{1}{2}; & x \in \left[ \dfrac{3}{5}, 1 - ε \right)\\ \dfrac{1}{2ε} (x - 1) + 1; & x \in [1 - ε, 1) \end{cases}, $$ then $F_{\tfrac{3}{5}, ε} \in \F_{\ac}$ and$$ A\left( F_{\tfrac{3}{5}, ε}, \dfrac{3}{5} \right) \cap \left( \left[0, \dfrac{2}{5} - ε \right] \cup \left[ \dfrac{3}{5}, 1 \right) \right) = \emptyset \Longrightarrow \left| A\left( F_{\tfrac{3}{5}, ε}, \dfrac{3}{5} \right) \right| \leqslant \dfrac{1}{5} + ε. $$ Therefore $\inf\limits_{F \in \F_{\ac}} \left| A\left( F, \dfrac{3}{5} \right) \right| = \dfrac{1}{5}$.

For general $a = \dfrac{m}{n} \in \mathbb{Q}$ where $m, n \in \mathbb{N}_+$, $m < n$ and $(m, n) = 1$, it can be proved analogously that$$ \inf_{F \in \F_{\ac}} |A(F, a)| = \min_{F \in \F_{\pc}} |A(F, a)| = \frac{1}{n} $$ with $F_a$ defined in the following way: Suppose $\widetilde{m} \in \{1, 2, \cdots, n - 1\}$ satisfies that $m \widetilde{m} \equiv 1 \pmod{n}$. Define$$ J_a = \left\{ k \in \{1, 2, \cdots, n - 1\} \,\middle|\, \left[ \frac{k \widetilde{m}}{n} \right] > \left[ \frac{(k - 1) \widetilde{m}}{n} \right] \right\} $$ and arrange the numbers in $J_a$ as $k_1 < \cdots < k_{\widetilde{m} - 1}$ (it is not hard to prove that $J_a$ has $\widetilde{m} - 1$ elements), and define$$ F_a(x) = \frac{1}{\widetilde{m}} \sum_{j = 1}^{\widetilde{m} - 1} I_{\left[ \tfrac{k_j}{n}, 1 \right)}(x). $$ Basically $F_a$ is a staircase function with jumps at numbers in $J_a$ of size $\dfrac{1}{\widetilde{m}}$ and can be extended on $\R$ so that $F_a \in \F_{\pc}$. In addition, it can be proved that $A(F_a, a) = \left[ \dfrac{n - m}{n}, \dfrac{n - m + 1}{n} \right)$.

Since $F_{a, ε} \in \F_{\ac}$, then$$ f_{a, ε} := F'_{a, ε} = \frac{1}{ε \widetilde{m}} \sum_{j = 1}^{\widetilde{m}} I_{\left[ \tfrac{k_j}{n} - ε, \tfrac{k_j}{n} \right]},\quad\forall 0<ε<\frac{1}{n} $$ where $k_{\widetilde{m}} := m$. In particular, $f_{\tfrac{3}{5}, ε} = \dfrac{1}{2ε} I_{\left[ \tfrac{3}{5} - ε, \tfrac{3}{5} \right] \cup [1 - ε, 1]}$.

Remark 2: So far I have failed to prove that the infimum is unattainable in $\F_{\ac}$. If anyone has proved/disproved it and post the proof as an answer to this question, I would be glad to offer a bounty on that answer.

Upper bound: $0.40$

Proof: Let $f = 2\cdot1_E$ where $E = [0,\frac{1}{8}]\cup[\frac{2}{8},\frac{3}{8}]\cup[\frac{5}{8},\frac{7}{8}]$. Then $A^c = [0,0.225]\cup[0.300,0.475]\cup[0.800,1]$, which has measure $0.60$.

Lower bound: $0.20$

Proof: For any $y$, at least one of $y,y+\frac{1}{5},y+\frac{2}{5},y+\frac{3}{5},y+\frac{4}{5}$ must be in $A$, as $$\sum_{j=0}^4 \int_{y+\frac{j}{5}}^{y+\frac{j}{5}+0.6} f(t)dt = 3\int_0^1 f(t)dt = 3.$$

It's very likely the answer for general $a$ is $\frac{1}{n}$ if $a =\frac{m}{n}$ reduced, with answer being $0$ for $a$ irrational. My lower bound proof gives the lower bound for this claim, so you just need examples to get the upper bound. I don't think this is too hard; I'm just lazy and lost interest in this problem.