Let $f$ be a real-valued continuous function of bounded variation on $[a,b]$. Suppose $f$ is absolutely continuous on $[a+\eta,b]$ for every $\eta\in(0,b-a)$. Show that $f$ is absolutely continuous on $[a,b]$.
The solution I can think of require Banach-Zarecki Theorem, which says absolutely continuous is equivalent to continuous, bounded variation and mapping null set to null set. For any null set $N\subset [a,b]$, write $\displaystyle N=(\bigcup_{k\in\mathbb{N}}(N\cap[a+\frac{1}{k},b]))\cup (N\cap\{a\})$, take image under $f$ for both sides, argue right hand side has measure $0$ by given condition, therefore left hand side has measure $0$.
However the given conditions look quite strong that I think there should be a more elementary solution, by just using the definition of absolute continuity and bounded variation. Does anyone have any idea?
Let $Tf$ be the variation function of $f$, that is $Tf(x)$ is the variation of $f$ on $[a,b]$. Since $f$ is continuous and has bounded variation, $Tf$ is continuous on $[a,b]$. Therefore, for every $\epsilon>0$ there exists $c\in (a,b)$ such that $Tf(b)-Tf(c)<\epsilon/2$, meaning that the variation of $f$ on $[c,b]$ is less than $\epsilon/2$.
Since $f$ is absolutely continuous on $[a,c]$, there exists $\delta>0$ such that for any finite sequence of disjoint subintervals $[s_j, t_j]$ in $[a,c]$ of total length $<\delta$ we have $\sum_j |f(s_j)-f(t_j)|<\epsilon/2$. Combining this with the first paragraph, we conclude that for any finite sequence of disjoint subintervals $[s_j, t_j]$ in $[a,b]$ of total length $<\delta$ we have $\sum_j |f(s_j)-f(t_j)|<\epsilon$, as required.