Let $\lambda$ be the Lebesgue measure on $[a, b]$. I'm trying to prove below result mentioned as Theorem 5.2 in my lecture note of gradient flows.
Theorem A map $F:[a, b] \to \mathbb R$ is absolutely continuous if and only if there is a $\lambda$-integrable $f:[a, b] \to \mathbb R$ such that $$ \int_a^x f \mathrm d \lambda = F(x)-F(a) \quad \forall x \in [a, b]. $$ In that case, $F$ is differentiable $\lambda$-a.e. and $F'=f$ $\lambda$-a.e.
Could you confirm if my below attempt is fine?
Proof
Let $F$ be absolutely continuous. Then $F$ is of bounded variation. Then $F$ is differentiable $\lambda$-a.e. and its derivative $F'$ is $\lambda$-integrable. Let $G(x) := \int_a^x F' \mathrm d \lambda$. Then $G$ is absolutely continuous. So $G-F$ is absolutely continuous. By Lebesgue differentiation theorem, $G$ is differentiable $\lambda$-a.e. and $G' = F'$ $\lambda$-a.e. It follows that $(G-F)' = 0$ $\lambda$-a.e. Then $G-F$ is a constant. On the other hand, $G(a)=F(a)=0$. So $G=F$.
Let $f:[a, b] \to \mathbb R$ be $\lambda$-integrable such that $$ \int_a^x f \mathrm d \lambda = F(x)-F(a) \quad \forall x \in [a, b]. $$ Let $G(x) := \int_a^x f \mathrm d \lambda$. Then $G$ is absolutely continuous. Then $F =G + F(a)$ is absolutely continuous. By Lebesgue differentiation theorem, $G$ is differentiable $\lambda$-a.e. and $G' = f$ $\lambda$-a.e. On the other hand, $F'=G'$ $\lambda$-a.e.
