# $\lim\limits_{n\rightarrow\infty}T(f-f_n)=0$.($T$:total variation, $f_n$: absolutely continuous). Then $f$ is absolutely continuous.

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.

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 $$\sum |f(b_i)-f(a_i)| <\epsilon$$.