2a.
$ ( \Rightarrow ) $ Say $ f $ is A.C. Then $ f $ is of bounded variation, and since $ f $ is clearly nondecreasing, $ f $ must be bounded.
In particular, $ \infty > f(1)=\sum_{n=1}^\infty m([0,1]\cap G_n) \Rightarrow \sum_{n=1}^\infty m(G_n)<\infty $.
$ ( \Leftarrow ) $
Now say $ \sum_{n=1}^\infty m(G_n)<\infty $.
I claim $ f'(x) = \sum_{n=1}^\infty \chi_{G_n}(x) $ a.e. as follows:
For $ h>0 $, $ \frac{f(x+h)-f(x-h)}{2h}=\frac{1}{2h}\sum_{n=1}^\infty m([x-h,x+h]\cap G_n)=\frac{1}{2h}\sum_{n=1}^\infty \int_{x-h}^{x+h} \chi_{G_n}=\frac{1}{2h} \int_{x-h}^{x+h} \sum_{n=1}^\infty \chi_{G_n}(x) $ (MCT)
Taking $ lim_{h\rightarrow 0} $ of both sides we get $ f'(x) $ on the left and $ \sum_{n=1}^\infty \chi_{G_n}(x) $ a.e. on the right (Lebesgue's Differentiation Theorem).
Now, since $ \sum_{n=1}^\infty m(G_n)<\infty $, we have $ \int_0^1 f' = \int_0^1 \sum_{n=1}^\infty \chi_{G_n}= \sum_{n=1}^\infty \int_0^1 \chi_{G_n}=\sum_{n=1}^\infty m(G_n)<\infty $ $ \Rightarrow f' $ is integrable on $ [0,1] $.
Finally, $ f(x)-f(0)=\int_0^x f' $.
So $ f $ is A.C.
b)
