Absolute continuity of $g(x)=\int_a^x f(t) dt.$ Let $f$ be a nonnegative and Lebesgue measurable function. And difine $g(x):=\displaystyle\int_a^x f(t) dt \ (a\leqq x)$.
Then, prove that $g$ is absolutely continuous.
Let $\epsilon>0.$
I have to determine $\delta=\delta(\epsilon)>0$ and consider $\displaystyle\sum_{k=1}^N |g(x_k)-g(y_k)|$ with $\displaystyle\sum_{k=1}^N |y_k-x_k|<\delta.$
\begin{align}
\sum_{k=1}^N |g(x_k)-g(y_k)|
&=\sum_{k=1}^N \left|\int_{x_k}^{y_k} f(t) dt\right| \\
&\leqq \sum_{k=1}^N \int_{x_k}^{y_k} f(t) dt
\end{align}
If $f$ is bounded, I can do
\begin{align}
\sum_{k=1}^N \int_{x_k}^{y_k} f(t) dt&\leqq M \sum_{k=1}^N (y_k-x_k) \\
&<M\delta \\
&=\epsilon. \ \mathrm{as} \ \delta=\dfrac{\epsilon}{M}.
\end{align}
But I know that Lebesgue measurable function is not necessarily bounded.
How can I prove the absolute continuity of $g$ ?
 A: If $f$ is not bounded, make it bounded except for over a set of small measure!
But how to? We can do it if $f\in L^1$ and the condition $f\in L^1$ is necessary to ensure our $g$ is absolutely continuous. (See the latter part of my answer.)
Let $\varepsilon>0$. Since $\|f\|_1<\infty$, there is $M>0$ such that $\int_A|f|dx<\varepsilon/2$ for $A=\{x: |f(x)|>M\}$. Take $\delta=\varepsilon/2M$ and let $x_0<y_0<\cdots< x_m<y_m$ are points satisfying $\sum_{i=0}^m |y_i-x_i|<\delta$. Then
$$\begin{align}\sum_{i=0}^m |g(y_i)-g(x_i)|&=\sum_{i=0}^m \left|\int_{x_i}^{y_i} f(t)dt\right|\le \sum_{i=0}^m \left|\int_{[x_i,y_i]\cap A} f(t)dt\right|+ \left|\int_{[x_i,y_i]\setminus A} f(t)dt\right|\\ & \le \sum_{i=0}^m \int_{[x_i,y_i]\cap A} |f(t)|dt + M(y_i-x_i)\le \frac{\varepsilon}{2}+M\delta=\varepsilon.\end{align} $$

If $f$ is not $L^1$, then your $g$ is not necessarily absolutely continuous. For example, consider $f(x)=1/x$ over $(0,\infty)$. Then $
g(1/2^n)-g(1/2^{2n})=n\log 2$ while $\frac{1}{2^n}-\frac{1}{2^{2n}}\to 0$ as $n\to\infty$

The OP asked how we can find $M$ such that $\int_{|f|>M}|f|dx<\varepsilon/2$ if $f\in L^1$. Let us try it as follows:
Take $A_n=\{x: n\le |f(x)|<n+1\}$, then $\|f\|_1=\sum_{n=0}^\infty \int_{A_n}|f|dx<\infty$. Especially, the latter sum converges, so for each $\varepsilon>0$ we can find $N\in\mathbb{N}$ such that $\int_N^\infty |f|dx = \sum_{n>N}\int_{A_n}|f|dx<\varepsilon/2$.
