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Suppose $f_n$ are absolutely continuous on $[a,b], n\in\mathbb{N}$, and $f:[a,b]\rightarrow\mathbb{R}$ is such that $\lim\limits_{n\rightarrow\infty}T(f-f_n)=0$. Here $T$ denotes total variation on $[a,b]$. Prove that $f$ is also absolutely continuous.

Let $\epsilon>0$. Then there exists $N$ such that for $n\geq N$, $T(f-f_n)<\epsilon$. We need $\delta>0$ such that for any collection $\{[a_i,b_i]\}$ of nonoverlapping subintervals of $[a,b]$, $\sum|f(b_i)-f(a_i)|<\epsilon$ if $\sum(b_i-a_i)<\delta$. I don't know what to do from here.

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Choose $k$ such that $T(f-f_k) <\epsilon /2$. For this $k$ there exists $\delta >0$ such that for any finite collection of non-overlapping intervals $(a_i,b_i)$ with $\sum (b_i-a_i) <\delta $ we have $\sum |f_k(b_i)-f_k(a_i)| <\epsilon/2$. Now there is an obvious way of adding points to the collection $\{a_1,b_1,a_2,b_2,...,a_n,b_n\}$ to get a partition of $[0,1]$ so that each $(a_i,b_i)$ is a sub-interval in the partition. [You only have to arrange the intervals in a 'natural' order and add the points $0$ and $1$]. Hence $\sum |[f(a_i)-f_k(a_i)]-[f(b_i)-f_k(b_i)]| \leq T(f-f_k) <\epsilon/2$. It follows by triangle inequality that $\sum |f(b_i)-f(a_i)| <\epsilon$.

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