Let $f : \mathbb{R} \to \mathbb{R}$ be a measurable function and $g: \mathbb{R} \to [0,\infty)$ be a locally-integrable function such that \begin{equation} f(b)-f(a) \leq -\int_a^b g(x)dx \end{equation} holds for almost every $a \in \mathbb{R}$ and all $b \in (a,\infty)$.
This seems to me in a sense that $f(x)$ is monotone-decreasing "almost everywhere".
Now, my question:
is it possible to modify $f$ on a set of measure zero such that the above inequality holds for "all" $a, b \in \mathbb{R}$ such that $a<b$?
This issues looks more subtle than expected.. Could anyone please help me?
Let $\mathcal{A} = \{ a \in \mathbb{R}: \text{ for all } b > a, f(b) - f(a) \le - \int_a^b g(x) dx\}.$
For $a \not \in \mathcal{A}$ and $b > a$ with $b \in \mathcal{A}$, define $$ f_b^*(a) = f(b) + \int_a^b g(x) dx. $$ Note that for $c \ge b$ with $c \in \mathcal{A}$, $$f_c^*(a) - f_b^*(a) = f(c) - f(b) + \int_b^c g(x) dx \le 0.$$
As a result, on $\mathcal{A} \cap (a, \infty)$ the map $b \mapsto f_b^*(a)$ is decreasing.
As a result, we can define $$ f^*(a) = \begin{cases} f(a), a \in \mathcal{A}\\ \sup_{b \in \mathcal{A}, b> a} f_b^*(a). \end{cases} $$ Checking that this definition satisfies the desired inequality is essentially a lot of simple case checking. In what follows, I will always have $a < b$.
If both $a,b$ are in $\mathcal{A}$, the desired inequality is true by assumption.
If $b \in \mathcal{A}$ and $a \not \in \mathcal{A}$ then choose $c \in (a,b)$ such that $c \in \mathcal{A}$ (which is possible since the complement of $\mathcal{A}$ has measure $0$). Then $$f^*(b) - f^*(a) = f(b) - f^*(a) \le f(b) - f_c^*(a) = f(b) - f(c) - \int_a^c g(x) dx \le - \int_a^b g(x) dx.$$
If $a \in \mathcal{A}$ and $b \not \in \mathcal{A}$ then $$f^*(b) - f^*(a) = \sup_{c > b, c \in \mathcal{A}} f(c) - f(a) + \int_b^c g(x) dx \le \sup_{c > b} \Bigg ( - \int_a^c g(x) dx + \int_b^c g(x) dx \Bigg )$$ which implies the desired inequality.
Finally if $a \not \in \mathcal{A}$, $b \not \in \mathcal{A}$, $$f^*(b) - f^*(a) = \sup_{c > b, c \in \mathcal{A}} \Bigg ( f(c) + \int_b^c g(x) dx \Bigg ) - \sup_{d > a, d \in \mathcal{A}} \Bigg ( f(d) + \int_a^d g(x) dx \Bigg ).$$
Pick $d \in (a,b) \cap \mathcal{A}$. Then for each $c$ in the set over which we take the first supremum on the right hand side, we have $$ f(c) + \int_b^c g(x) dx - f(d) - \int_a^d g(x) dx \le - \int_d^c g(x) dx + \int_b^c g(x) dx - \int_a^d g(x) dx.$$
Since the right hand side is nothing but $-\int_a^b g(x) dx$ this implies the desired bound.