Below is an exercise from Iowa State's Qualifying Exams in analysis, and a solution I cooked up for it. I showed it to a friend, and he seems very unhappy with the very end of the proof, though I see nothing wrong with it. Would anyone mind pointing out whether there are any issues? Any help is very appreciated!
Suppose $\{E_n\}$ is a sequence of Lebesgue measurable subsets of $\mathbb{R}$ with $$ \sum_{n=1}^\infty \mu(E_n)<\infty $$ Let $f(x)=\sum_{n=1}^\infty \mu(E_n\cap [0,x])$. Prove that $f$ is absolutely continuous on $[0,\infty)$
We need to show the following: given $\epsilon>0$, there exists some $\delta > 0$ so that $$ \sum_{k=1}^{N} |f(b_k) - f(a_k) | < \epsilon \quad \text{ whenever } \sum_{k=1}^{N} (b_k-a_k) < \delta $$ and the intervals $(a_k,b_k)$ form a partition of $[0,\infty)$ into pairwise disjoint intervals. By assumption, there exists an $n_0 \geq 1$ such that $\sum_{n \geq n_0} \mu(E_n) < \epsilon/2$. Let $x,y \in [0,\infty)$ be such that $x<y$. By monotonicity of the Lebesgue measure, we have that $$ |f(y) - f(x)| = \left| \sum_{n=1}^{\infty} \mu(E_n \cap [0,y]) - \sum_{n=1}^{\infty} \mu(E_n \cap [0,x] )\right| \leq \sum_{n=1}^{\infty} |\mu(E_n \cap [0,y]) - \mu(E_n \cap [0,x])| $$ For each $ n \geq 1$, $E_n \cap [0,x] \subset E_n \cap [0,y]$, each of which have finite measure, and so $$ \mu(E_n \cap [0,y]) - \mu(E_n \cap [0,x]) = \mu ((E_n \cap [0,y]) \setminus (E_n \cap [0,x])) = \mu(E_n \cap [x,y]), $$ for each $ n \geq 1$; consequently, we have that $$ |f(y) - f(x)| \leq \sum_{n=1}^{\infty} \mu(E_n \cap [x,y]). $$ Choose $\delta = \epsilon/2n_{0}$. By monotonicity of the Lebesgue measure, we see that $$ |f(y) - f(x)| \leq \sum_{n=1}^{\infty} \mu(E_n \cap [x,y] \leq \sum_{n=1}^{n_0-1} \mu(E_n \cap [x,y] + \sum_{n \geq n_0} \mu(E_n) < \mu(E_n \cap [x,y] + \epsilon/2. $$ Finally, given any partition $\{x_k,y_k)\}_{k=1^K}$ of $[0,\infty)$ into pairwise disjoint interval with $\sum_{k=1}^{K} (y_k-x_k )< \delta$. Then it follows that $$ | f(y_k) - f(x_k)| < \sum_{n=1}^{n_0-1} \mu(E_n \cap [x_k,y_k]) + \epsilon/2 \leq \sum_{n=1}^{n_0-1}( y_k-x_k )< \epsilon. $$ Hence, $$ \sum_{k=1}^{K} |f(y_k) - f(x_k)| < K \cdot \epsilon, $$ completing the proof.
I know you want someone to look over your proof, but that's a lot of looking. I will suggest a different approach: Briefly, define
$$g(x) = \sum_{n=1}^{\infty} \chi_{E_n}(x).$$
Show $g\in L^1(\mathbb R)$ and that $f(x) = \int_0^x g\,d\mu.$ It is basic then that $f$ is AC on $\mathbb R.$